3. Each container holds one litre of water. Write the least amount of salt needed to make each solution saturated. Then draw a picture of the saturated solution in the circle. answer of this.

Answers

Answer 1

To get them to th e point of saturation,

Container A needs 259gm of SaltContain B needs 159gm of saltContainer C needs no additional salt that is 0gms of salt.

Why is this so?

The simple a rationale is that a liter of water will only get saturated after 359 grams of salt has been added to it.

Thus, while A is not saturated and B is almost saturated, container C is super saturated and most likely will contain visible forms of salt crystals.

Note that container B will exhibit the qualities of an unstable solutions.

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A substance is soluble if it can dissolve into another substance Salt is

soluble in water, but does this mean that water can dissolve an unlimited

amount of salt? No, it does not! When water has dissolved all the salt it

can, the resulting solution is described as saturated Before that point, the

solution is unsaturated. A saturated solution of water and salt contains

359 grams of dissolved salt for each litre of water.

Have you ever tasted sea water from the ocean? It is a solution of about

35 g of salt for every litre of water. Now, have you ever tasted the solution

in a jar of pickles? It certainly contains a lot of salt. In fact, the nearly

saturated salt content is exactly what deters harmful micro-organisms and

preserves the pickles.

Each container holds one litre of water. Write the least amount of salt needed to make each solution saturated. Then draw a picture of the saturated solution in the circle. answer of this.

Contain A = A solution of 100g of salt

Contain B = A solution of 200g of salt

Contain c = SAturated Solution (500g)

3. Each Container Holds One Litre Of Water. Write The Least Amount Of Salt Needed To Make Each Solution

Related Questions

How many moles do 235.5 liters of carbon dioxide

Answers

Gas legislation

PV = nRT

where P is atmospheric pressure

Volume (measured in litres)

The number of moles is n.

R = gas constant (0.0821 L atm/(mol K))

Temperature (in Kelvin) equals T.

We need to rearrange the ideal gas law to solve for n:

n = PV / RT

To use this equation, we need to know the pressure, temperature, and volume of the carbon dioxide.

0.002355 moles of carbon dioxide are equal to 235.5 litres of the petrol. As a result, 0.002355 moles of carbon dioxide are equivalent to 0.053 litres.

This is because 1 mole of carbon dioxide is equal to 22.4 litres. This indicates that there are 0.002355 moles of carbon dioxide for every 0.053 litres of carbon dioxide.

Therefore, there are 0.002355 moles of carbon dioxide in 235.5 litres of carbon dioxide.

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Given the reaction:


If there are initially 14.2 moles of iron (III) nitrate nonahydrate, how many moles of steam (water vapor) will be produced?

Answer in mol.

Answers

Total, 127.8 moles of the steam (water vapor) will be produced.

Iron(III) nitrate nonahydrate, or Fe(NO₃)₃·9H₂O, is a compound that contains iron (III) cations (Fe³⁺) and nitrate anions (NO₃⁻) combined with nine water molecules (H₂O) per formula unit. The nine water molecules are referred to as "nonahydrate," which indicates that there are nine water molecules associated with each formula unit of Fe(NO₃)₃. This compound is also commonly known as ferric nitrate nonahydrate.

Balanced equation for the given reaction is;

Fe(NO₃)₃·9H₂O(s) → Fe(NO₃)₃(s) + 9H₂O(g)

According to the equation, one mole of Fe(NO₃)₃·9H₂O produces 9 moles of water vapor (H₂O) when it undergoes the reaction. Therefore, if there are initially 14.2 moles of Fe(NO₃)₃·9H₂O, the number of moles of water vapor (H₂O) produced will be;

14.2 moles of Fe(NO₃)₃·9H₂O × 9 moles of H₂O per 1 mole of Fe(NO₃)₃·9H₂O = 127.8 moles of H₂O

Therefore, 127.8 moles of water vapor will be produced.

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Calculate the percentage composition of elements in the following compounds:
a. Water: H₂O
b. Glucose: C6H12O6
c. Calcium nitrate: Ca(NO3)2
d. Aluminum sulfate: Al2(SO4)3
e. Magnesium phosphate: Mg3(PO4)2​

Answers

The percentage composition of Hydrogen (H) in water (H₂O) is 11.1%

Breakdown of How Percentage Composition is Calculated

Percentage composition is finding the amount of individual elements that made up a compound.

To calculate, we use the formula:

% composition = [tex]\frac{Atomic Mass}{Molar Mass} x 100[/tex]

(a) Water: H₂O

To calculate the percentage composition of elements in water, we need to find the molar mass of water, which is:

Molar mass of H₂O = (2 × atomic mass of H) + (1 × atomic mass of O)

Molar mass of H₂O = (2 × 1.0 g/mol) + (1 × 16.00 g/mol)

Molar mass of H₂O = 18.0 g/mol

%Hydrogen = 2/18 x100 = 11.11

%Oxygen: 16/18 x 100 = 88.89%

b. Glucose: C₆H₁₂O₆

Molar mass of C₆H₁₂O₆ =  6 (12) + 12(1) + 6(16)

                                       = 72 + 12 + 96

                                       = 180g/mol

Now, we can calculate the percentage composition of elements in glucose:

% Carbon: 72/180 = 40%

% Hydrogen: 12/180 = 7%

Percentage composition of oxygen: (6 × atomic mass of O) / molar mass of C6H12O6 × 100%

Percentage composition of oxygen: (6 × 16.00 g/mol) / 180.18 g/mol × 100%

Percentage composition of oxygen: 53.29%

Therefore, the percentage composition of elements in glucose is 39.99% carbon, 6.72% hydrogen, and 53.29% oxygen.

c. Calcium nitrate: Ca(NO3)2

To calculate the percentage composition of elements in calcium nitrate, we need to find the molar mass of calcium nitrate, which is:

Molar mass of Ca(NO3)2 = atomic mass of Ca + (2 × atomic mass of N) + (6 × atomic mass of O)

Molar mass of Ca(NO3)2 = 40.08 g/mol + (2 × 14.01 g/mol) + (6 × 16.00 g/mol)

Molar mass of Ca(NO3)2 = 164.09 g/mol

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A solution of KNO3 is prepared by dissolving 80 g KNO3 in 100 g water at 60oC. The solution is

Group of answer choices

dilute

saturated

unsaturated

supersaturated

Answers

To determine whether the solution is dilute, saturated, unsaturated, or supersaturated, we need to compare the amount of KNO3 that is dissolved in the solution at 60°C to the maximum amount of KNO3 that can be dissolved in water at 60°C.

The solubility of KNO3 in water at 60°C is 110 g per 100 g of water. Since only 80 g of KNO3 was dissolved in 100 g of water, the solution is unsaturated.

Therefore, the answer is unsaturated.

5.4g of aluminum reacts with sulfuric acid (H₂SO4) to form aluminum sulfate and hydrogen.
a. Write the chemical equation.
b. Find mass of required sulfuric acid.
C. Find volume of the obtained gas.
(AI=23, S = 32, O=16, H =1, 2g of H2 has 22.4L).​

Answers

Answer:

a. The chemical equation for the reaction is:

2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

b. To find the mass of required sulfuric acid, we need to use stoichiometry. We can start by finding the number of moles of aluminum used in the reaction:

Molar mass of Al = 27 g/mol

Number of moles of Al = 5.4 g / 27 g/mol = 0.2 mol

According to the balanced equation, 3 moles of H₂SO₄ are required to react with 2 moles of Al. Therefore, the number of moles of H₂SO₄ required is:

Number of moles of H₂SO₄ = 3/2 x 0.2 mol = 0.3 mol

Molar mass of H₂SO₄ = 2 x 1 g/mol + 32 g/mol + 4 x 16 g/mol = 98 g/mol

Mass of H₂SO₄ required = 0.3 mol x 98 g/mol = 29.4 g

Therefore, 29.4 g of sulfuric acid is required to react with 5.4 g of aluminum.

c. To find the volume of hydrogen gas obtained, we need to use the ideal gas law:

PV = nRT

where P is the pressure of the gas, V is its volume, n is the number of moles of the gas, R is the universal gas constant (0.0821 L atm/mol K), and T is the temperature in Kelvin.

We can start by finding the number of moles of hydrogen gas produced in the reaction. According to the balanced equation, 3 moles of H₂ are produced for every 2 moles of Al. Therefore, the number of moles of H₂ produced is:

Number of moles of H₂ = 3/2 x 0.2 mol = 0.3 mol

Assuming the reaction occurs at standard temperature and pressure (STP), which is 0°C (273 K) and 1 atm, we can use the molar volume of a gas at STP, which is 22.4 L/mol. Therefore:

V = nRT/P = 0.3 mol x 0.0821 L atm/mol K x 273 K / 1 atm = 6.58 L

Therefore, the volume of hydrogen gas produced at STP is 6.58 L.

Explanation:

You have 900,000 atoms of a radioactive substance. After 4 half-lives have past, how many atoms remain?

you cannot have a fraction of an atom, so round the answer to the nearest whole number.

Answers

The number of atoms remaining after 4 half-lives can be calculated using the formula: N = N0 /[tex]2^4[/tex] . Therefore, after 4 half-lives, approximately 56,250 atoms of the radioactive substance remain.

Radioactive decay is the process by which a nucleus of an atom loses energy by emitting ionizing radiation. The rate of decay of a radioactive substance is measured by its half-life, which is the time it takes for half of the radioactive atoms in a sample to decay.

N = N0 /[tex]2^n[/tex]

where: N0 = initial number of atoms N = final number of atoms

Substituting the given values,

N = 900,000 / [tex]2^4[/tex]

N = 56,250

Rounding to the nearest whole number,

N ≈ 56,250 atoms

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Calculate the oxidation number of oxygen in magnesium pyrophosphate Mg2p2o7

Answers

The oxidation state of oxygen in magnesium pyrophosphate (Mg2P2O7) is +5.

o calculate the oxidation state of oxygen in magnesium pyrophosphate (Mg2P2O7), we first need to know the oxidation states of the other atoms in the molecule.

The magnesium ion (Mg2+) has a fixed oxidation state of +2 in this compound, and the phosphate ion (PO43-) has an overall oxidation state of -3.

We can set up an equation to solve for the oxidation state of oxygen (O) in the pyrophosphate ion:

2(+2) + 2(O) + 7(-2) = 0

Simplifying the equation gives:

4 + 2O - 14 = 0

2O = 10

O = +5

Therefore, the oxidation state of oxygen in magnesium pyrophosphate (Mg2P2O7) is +5.

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The oxidation number of oxygen in magnesium pyrophosphate is -2.

How to calculate oxidation number?

Oxidation number refers to the hypothetical charge of an atom within a molecule.

For monoatomic ions, the oxidation number always has the same value as the net charge corresponding to the ion.

Oxidation no. of Mg = +2Oxidation no. of P = +5Oxidation no of O = x

0 = +2(2) + 5(2) + X(7)

0 = 14 + 7x

-14 = 7x

x = -2

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CO, (9) +2NH_(9) - CO(NH,) (s) +H, O(1)
a. What is the maximum mass of urea, CO(NH), that can be manufactured from the reaction of 2.20 moles of CO2 with sufficient amount of ammonia.

Answers

The mass of the ammonia that is required is  258 g.

What is the stoichiometry of the reaction?

The quantitative correlations between the reactants and products in a chemical reaction are the focus of the chemistry subfield known as stoichiometry.

We have to know that;

1 mole of CO2 produces 1 mole of urea

2.2 moles of CO2 produces 2.2 urea

Given that the number of moles of urea = 455 g/60 g/mol

= 7.58 moles

Now;

2 moles of NH3 produces 1 mole of urea

x moles of NH3 produces 7.58 moles of urea

x = 7.58 * 2/1

= 15.16 moles

Mass of the ammonia =  15.16 moles * 17 g/mol

= 258 g

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Based on the information provided, which solution is a base and weak electrolyte

Answers

An example of a composition which fulfills the qualifications of being both a base and a weak electrolyte is ammonia (NH3).

How to explain the electrolyte

A base is any constituent which voluntarily receives protons (H+) in an associated chemical reaction while an electrolyte denotes any material that can conduct electricity through liquids or in melted state.

Upon dissolution in water, it is apt to accept a proton from such and thus create the acidic ion known as ammonium (NH4+). Nonetheless, due to its scarce dissociation into hydroxide (OH-) and ammonium ions, it is deemed a weak electrolyte.

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(25 pts) Explain in depth the relationship between LIMITING reactant and THEORETICAL yield

Answers

Answer: I got you fam

Explanation:

A limiting reactant is a reactant in a chemical reaction that limits the amount of product created.

So for example if there are elements X and Y reacting to create product XY, once say element X runs out, the reaction stops, even though there is still more of the reactant Y. So there is 0 g of element X remaining, and maybe 2 g left of element Y. X is the limiting reactant since it limits the amount of product made.

Theoretical yield is the maximum amount of product that could be made in an experiment. This occurs if a reaction is 100% effective (and experimentally, this doesn't usually happen, which is why it is called theoretical).

A student used 0.17 grams of Alka Seltzer for their experiment. What's the mass of sodium bicarbonate (NaHCO3) in their sample?

Answers

Answer:

To find the mass of sodium bicarbonate (NaHCO3) in a sample of Alka Seltzer that weighs 0.17 grams, we need to know the percentage of NaHCO3 in Alka Seltzer. This information is usually provided on the label of the product. Once we know the percentage of NaHCO3, we can use the following formula to calculate the mass of NaHCO3 in the sample:

mass of NaHCO3 = sample mass (in grams) x % NaHCO3 / 100

For example, if the percentage of NaHCO3 in Alka Seltzer is 50%, then the mass of NaHCO3 in a 0.17 gram sample would be:

mass of NaHCO3 = 0.17 x 50 / 100 = 0.085 grams

Therefore, the mass of sodium bicarbonate in the student's 0.17 gram sample of Alka Seltzer depends on the percentage of NaHCO3 in the Alka Seltzer, which should be provided on the product label.

Explanation:

The heat of fusion of water is 79.9 cal/g. If a 7.2 g piece of ice melts in 105 g of water at 34.3 deg C in an insulated bottle, what is the final temperature of the water?

The heat of fusion of water is 79.9 cal/g. If a 7.2 g piece of ice melts in 105 g of water at 34.3 deg C in an insulated bottle, what is the final temperature of the water?

Type your answer...

Answers

The heat of fusion of water is 79.9 cal/g. If a 7.2 g piece of ice melts in 105 g of water at 34.3 deg C in an insulated bottle, 35071.6 °C is the  final temperature of the water.

The physical concept of temperature indicates in numerical form how hot or cold something is. A thermometer is used to determine temperature. Thermometers are calibrated using a variety of temperature scales, which historically defined distinct reference points or thermometric substances.

The most popular scales were the Celsius scale, sometimes known as centigrade, with the unit symbol °C, the scale of Fahrenheit (°F), or the Kelvin scale (K), with the latter being mostly used for scientific purposes.

Δ T = T(initial) - T(final)

T(final)= m × c × q - T(initial)

T(final)= 79.9 x 4.184 x 105 - 30.0

T(final)= 35071.6 °C

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Reaction Ai Sodium Bicarbonate and Hydrochloric Acid
Experimental Data
(a) Mass of evaporating dish
watch glass
(b) Mass of evaporating dish watch glass sodium bicarbonate
(c) Mass of sodium bicarbonate used
(d) Mass of evaporating dish watch glass sodium chloride
(e) Mass of sodium chloride collected (experimental yield)
.
Mole Ration and Reaction Story
.
.
Data Analysis
1) Use your data to determine the experimental mole-to-mole ratio between sodium bicarbonate and sodium chloride
Show your work for each
NaHCOS
Convert the mass of sodium bicarbonate used to moles
100.69
1
mole
9
Convert the mass of sodium chloride collected in moder
Nac
2g
104.2
3.bg
mole
g
Divide both of your results from the preceding two steps by the lower mole value to determine the simplest mole-to-
mole ratio between sodium bicarbonate and sodium chloride.

Answers

The reaction between sodium carbonate and hydrochloric acid

Na2CO3 + 2HCl = 2NaCl + CO2 + H2O

How to solve

For reaction A

Mass of sodium bicarbonate used = (Mas of evaporating dish + watch glas + sodium bicarbonate) - (Mas of evaporating dish + watch glas)

= 46.582 - 46.263

= 0.319 g

Mass of sodium chloride = (mas of evaporating dish + watch glas + sodium chloride) - (Mas of evaporating dish + watch glas)

= 46.473 - 46.263

= 0.210 g

Moles of sodium bicarbonate (NaHCO3) used = mas/molecular weight

= (0.319 g) / (84 g/mol)

= 0.00380 mol

Moles of sodium chloride (NaCl) used = mas/molecular weight

= (0.210 g) / (58.44 g/mol)

= 0.00359 mol

Mol ratio of NaHCO3 : NaCl = 0.00380 : 0.00359

Divide by 0.00359

Simple mol ratio

NaHCO3 : NaCl = 1.06 : 1

After rounding

Mol ratio of NaHCO3 : NaCl = 1 : 1

Moles of NaHCO3 = moles of NaCl = 0.00359 mol

Theoretical yield of NaCl = moles x molecular weight

= 0.00359 mol x 58.44 g/mol

= 0.210 g

the percent yield of sodium chloride

= actual yield x 100 / theoretical yield

= 0.210*100/0.210

= 100%

the reaction between sodium bicarbonate and hydrochloric acid

NaHCO3 + HCl = NaCl + CO2 + H2O

For reaction B

Mass of sodium carbonate used = (Mas of evaporating dish + watch glas + sodium carbonate) - (Mas of evaporating dish + watch glas)

= 51.677 - 51.368

= 0.309 g

Mass of sodium chloride = (mas of evaporating dish + watch glas + sodium chloride) - (Mas of evaporating dish + watch glas)

= 51.671 - 51.368

= 0.303 g

Moles of sodium carbonate (Na2CO3) used = mas/molecular weight

= (0.309 g) / (106 g/mol)

= 0.00292 mol

Moles of sodium chloride (NaCl) used = mas/molecular weight

= (0.303 g) / (58.44 g/mol)

= 0.00518 mol

Mol ratio of

Na2CO3 : NaCl = 0.00292 : 0.00518

Divide by 0.00292

Simple mol ratio

Na2CO3 : NaCl = 1 : 1.78

After rounding

Mol ratio of Na2CO3 : NaCl = 1 : 2

Moles of NaCl = 2 x moles of Na2CO3

= 2 x 0.00292 = 0.00584 mol

Theoretical yield of NaCl = moles x molecular weight

= 0.00584 mol x 58.44 g/mol

= 0.341 g

the percent yield of sodium chloride

= actual yield x 100 / theoretical yield

= 0.303*100/0.341

= 88.86%

the reaction between sodium carbonate and hydrochloric acid

Na2CO3 + 2HCl = 2NaCl + CO2 + H2O

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We would like to find RDS using Tapel slope. Provide Tafel slope when we assume each step is RDS, alpha a=0.5 * Target Reaction : Cu oxidation [mV]

Mechanism1 Cu-> Cu2+ +2e-
Mechanism2 Cu-> Cu+ +e-
Cu+->Cu2+ +e-​

Answers

The Tafel slope for Mechanism 2 is the sum of the slopes of both steps, presuming that each step in Mechanism 2 is RCS::

b2 = b2_1 + b2_2 = 0.1184 + 0.1184 = 0.2368 V

How to solve

To identify the rate-controlling step (RCS) utilizing the Tafel slope, we initially need to calculate the Tafel slope for each suggested mechanism when the electron transfer coefficient (alpha, α) equals 0.5.

The target reaction involves Cu oxidation.

Mechanism 1:

Cu -> Cu²⁺ + 2e⁻

Mechanism 2:

Cu -> Cu⁺ + e⁻

Cu⁺ -> Cu²⁺ + e⁻

The Tafel slope (b) can be computed with the following formula:

b = (2.303 * R * T) / (α * n * F)

Where:

R signifies the gas constant (8.314 J/mol K)

T represents the temperature in Kelvin (let's assume 298 K, standard room temperature)

α denotes the electron transfer coefficient (0.5)

n is the number of electrons exchanged in the RCS

F is the Faraday constant (96,485 C/mol)

For Mechanism 1, n = 2 (since 2 electrons are exchanged in the rate-controlling step):

b1 = (2.303 * 8.314 * 298) / (0.5 * 2 * 96,485) = 0.0592 V

For Mechanism 2, we must examine both steps. Let's initially evaluate the Tafel slope for each step.

Step 1 (n = 1):

b2_1 = (2.303 * 8.314 * 298) / (0.5 * 1 * 96,485) = 0.1184 V

Step 2 (n = 1):

b2_2 = (2.303 * 8.314 * 298) / (0.5 * 1 * 96,485) = 0.1184 V

The Tafel slope for Mechanism 2 is the sum of the slopes of both steps, presuming that each step in Mechanism 2 is RCS::

b2 = b2_1 + b2_2 = 0.1184 + 0.1184 = 0.2368 V

Having obtained the Tafel slopes for both mechanisms, we can now compare them to the experimental Tafel slope to ascertain which mechanism is more likely the RCS.

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NEED HELP FIGURING HOW MANY MOL!! PLEASE QUICK!!THANK YOU SO MUCH

Answers

The number of moles of the gas by the ideal gas law is 0.18 moles.

What is the ideal gas law?

The behavior of an ideal gas, a hypothetical gas made up of randomly moving particles with little volume and no intermolecular interactions, is described by the ideal gas law.

Although intermolecular interactions and non-zero particle volume prevent gases from always behaving in an ideal manner, the ideal gas law is nevertheless a good approximation for many gases under some circumstances.

We know that;

PV = nRT

We have ;

P = 1.2 atm

V = 3.4 L

T = 10 + 273 = 283 K

n = ?

n = PV/RT

n = 1.2 * 3.4/0.082 * 283

n =4.08 /23.2

n = 0.18 moles

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According to the law of conservation of mass, how much zinc was produced if Calcium = 25 g, Zinc Carbonate = 125 g, and Calcium Carbonate = 95 g.


HELPP

Answers

All the balanced chemical equations obey the law of conservation of mass. The numbers which are used to balance the chemical equation are called the coefficients. So here the mass of zinc is 55 g.

According to the law of conservation of mass, the mass can neither be created nor be destroyed but it can be converted from one form to another. The reactants appear on the left hand side and the products appear on the right hand side.

The amount of products is equal to the amount of reactants according to the law of conservation of mass.

25 + 125 = 95 + Zn

Zn = 55 g

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7. What mass of solid NH4Cl and what volume of 1.00 mol-L¹ NaOH solution should be used to
prepare 1 L of a buffer solution of pH 9.00? Suppose the overall concentration of the buffer is 0.125
mol-L¹. (Answer V = 45 mL)

Answers

STEP BY STEP SOLUTION :

To prepare a buffer solution of pH 9.00, we need to use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

Where [A-]/[HA] is the ratio of the concentrations of the conjugate base and acid of the buffer, respectively. Since we are given the pH and the overall concentration of the buffer, we can solve for the ratio [A-]/[HA]:

9.00 = pKa + log([A-]/[HA])pKa = 9.25 (the pKa of NH4Cl)9.00 = 9.25 + log([A-]/[HA])log([A-]/[HA]) = -0.25[A-]/[HA] = 0.56

Next, we can use the definition of the concentration of a solution to find the concentration of NH4Cl needed to make a 0.125 mol-L^-1 buffer solution:

0.125 mol-L^-1 = [NH4Cl] + [NaOH]

Since the NaOH solution is 1.00 mol-L^-1, we can assume that the contribution of NaOH to the total concentration of the buffer is negligible, and so:

0.125 mol-L^-1 = [NH4Cl]

Finally, we can use the molar mass of NH4Cl to find the mass of NH4Cl needed to prepare 1 L of the buffer solution:

mass NH4Cl = molar mass * molesmass NH4Cl = (14.01 g-mol^-1 + 1.01 g-mol^-1 + 35.45 g-mol^-1) * 0.125 molmass NH4Cl = 6.63 g

So we need to use 6.63 g of NH4Cl and enough volume of 1.00 mol-L^-1 NaOH solution to make a total volume of 1 L. To find the volume of NaOH solution needed, we can use the definition of molarity:

molality = moles of solute / volume of solution (in liters)

Rearranging this equation, we get:

volume of solution = moles of solute / molality

Since we are adding NaOH solution to NH4Cl to make a total volume of 1 L, the molality of NaOH solution is also 0.125 mol-L^-1. Therefore:

volume of NaOH solution = moles of NaOH / molality of NaOHvolume of NaOH solution = (1 L - volume of NH4Cl solution) * 0.125 mol-L^-1

Substituting the values we know:

volume of NaOH solution = (1 L - 0.45 L) * 0.125 mol-L^-1volume of NaOH solution = 0.056 L = 56 mL

So we need to use 6.63 g of NH4Cl and 56 mL of 1.00 mol-L^-1 NaOH solution to prepare 1 L of a buffer solution of pH 9.00.

The pKa of NH4Cl is 9.25. Therefore, the pH of the buffer can be calculated using the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

where [A-]/[HA] is the ratio of the concentration of the conjugate base to the concentration of the weak acid. Since NH4Cl is a salt of a weak acid (NH4+) and a strong base (Cl-), the weak acid in this case is NH4+.

Rearranging the Henderson-Hasselbalch equation gives:

[A-]/[HA] = 10^(pH - pKa)

Substituting the given values into the equation:

[A-]/[HA] = 10^(9.00 - 9.25) = 0.562

Since the overall concentration of the buffer is 0.125 mol-L¹, we can set up the following two equations:

[A-] + [HA] = 0.125 mol-L¹

[V1] [C1] = [V2] [C2]

where V1 is the volume of NaOH solution, C1 is the concentration of NaOH solution, V2 is the total volume of the buffer solution (1 L), and C2 is the concentration of NH4Cl.

Since NH4Cl is a 1:1 electrolyte, [A-] = [NH3] and [HA] = [NH4+]. Therefore, we can rewrite the first equation as:

[NH3] + [NH4+] = 0.125 mol-L¹

Substituting [A-]/[HA] = 0.562 into the equation gives:

[NH3] = 0.562 [NH4+]

Substituting this into the equation [NH3] + [NH4+] = 0.125 mol-L¹ gives:

[NH4+] = 0.125 / (1 + 0.562) = 0.0517 mol-L¹

[C2] = 0.0517 mol-L¹

The molar mass of NH4Cl is 53.49 g-mol¹. Therefore, the mass of NH4Cl needed to prepare 1 L of a 0.0517 mol-L¹ solution is:

m = C × V × M

where m is the mass of NH4Cl, C is the concentration of NH4Cl, V is the volume of

Units 1 and 2 1 Choose the correct option. a) An example of a chemical change is: A melting chocolate B leaving an iron nail to rust C dissolving salt in water D allowing solid air freshener to sublimate b) Which is not a property of physical change? A Atoms and molecules are conserved. B Atoms and molecules are rearranged. C The changes are permanent. D The changes are reversible. c) Which is not a property of chemical change? A A chemical reaction takes place. B There are large changes in energy. C The products have different properties from the reactants. D They are examples of phase changes. d) Which is a wrong statement? A To prove the Law of Constant Composition you add all the atomic masses of atoms in a compound. B The Law of Constant Composition states that all samples of a chemical compound have the same elemental composition. During a chemical reaction the atoms are conserved. C D During a chemical reaction, the sum of the formula masses of the products must equal the sum of the formula masses of the reactants​

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Chemistry and Physics are both great topics

How should the Key change for a weak base?

Answers

Weak bases partially ionize in water to produce hydroxide ions. Because the ionization is not complete, the concentration of OH⁻ in a weak base solution is typically much less than the initial base concentration.

Using six carbon atoms as an example, write the condensed structural formula and the names of the following functional groups: alcohol, ether, aldehyde, ketone, carboxylic acid, ester, amine and amide.

Answers

For alcohol, ether, aldehyde, ketone, carboxylic acid, ester, amine and amide, the condensed structural formulas of six carbon members of the series are shown below.

What is the condensed structural formula?

The condensed structural formula for an alcohol with six carbon atoms is C6H13OH

The condensed structural formula for an ether with six carbon atoms is C6H14O

The condensed structural formula for an aldehyde with six carbon atoms is C6H10O

The condensed structural formula for a ketone with six carbon atoms is C6H10O

The condensed structural formula for a carboxylic acid with six carbon atoms is C6H10O2

The condensed structural formula for an ester with six carbon atoms is C6H12O2

The condensed structural formula for an amine with six carbon atoms is C6H15N

The condensed structural formula for an amide with six carbon atoms is C6H13NO

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What is the molarity 10.0g of Cr(NO3)3 in 325 mL of solution

Answers

Answer:

Explanation:

molar mass Cr(NO3)3 = 238 g/mol

Convert 325 ml to liters:  325 mls x 1 L / 1000 mls = 0.325 L

Convert 10.0 g to moles:  10.0 g x 1 mol / 238 g = 0.0420 moles

Molarity = moles/liters = 0.0420 moles / 0.325 L = 0.129 M (3 sig. figs.)

We've figured out what part of the salt causes the flame to change color, so now let's measure the wavelengths created with four metals.

Use the ruler under the "tools" icon in the upper right of the video player to measure the wavelengths of light released by each compound.

Answers

The wavelength of one of the spectral lines for strontium chloride SrCl₂ is approximately 460.7 nanometers (nm).

When strontium chloride SrCl₂ is heated, it emits a characteristic red color, which indicates that it produces spectral lines in the red part of the visible spectrum. The most intense spectral line for SrCl₂ is at approximately 460.7 nm, which corresponds to the transition from the 5² electronic configuration to the 4d state.

This transition is responsible for the red color observed when strontium chloride is introduced to a flame. The wavelength of a spectral line is related to the energy of the transition and is given by

λ = hc ÷ E

where λ is the wavelength, h is Planck's constant, c is the speed of light, and E is the energy of the transition. In the case of SrCl₂, the energy of the transition from 5s² to 4d is approximately 2.69 eV, which corresponds to a wavelength of 460.7 nm.

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The complete question is:

What is the wavelength of one of the spectral lines for strontium chloride SrCl₂?

Question 10 (1 point)
What mass of iron (III) nitrate, Fe (NO3)3, is needed to prepare a 125 mL solution of
0.250 M?
7.56 grams
60.5 grams
2.68 grams
30.2 grams

Answers

The mass of iron (III) nitrate needed to prepare a 125 mL solution of 0.250 M is approximately 7.56 grams.

To calculate the mass of iron (III) nitrate needed to prepare a 125 mL solution of 0.250 M, we can use the formula:

mass (in grams) = volume (in liters) x concentration (in moles/liter) x molar mass (in grams/mole)

First, let's convert the volume from milliliters (mL) to liters (L):

125 mL = 0.125 L

Next, we can use the given concentration of 0.250 M to calculate the number of moles of Fe(NO3)3 needed:

moles = concentration x volume

moles = 0.250 mol/L x 0.125 L

moles = 0.03125 mol

Finally, we can use the molar mass of Fe(NO3)3 to convert from moles to grams:

mass = moles x molar mass

mass = 0.03125 mol x 241.86 g/mol

mass ≈ 7.56 g

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An OBJECT absorbs like between the light wavelengths of 430 - 400 nm. What is the color of the OBJECT?

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The color of the object appears as yellow/orange.

What is the color?

The visible light spectrum spans a wavelength range of 400 to 700 nanometers (nm), with shorter wavelengths appearing as blue or violet and longer wavelengths as red.

This object is absorbing light with wavelengths between 430 and 400 nm, indicating that it is absorbing light that is visible in the blue and violet spectrum.

We know that this color that we see is actually a complementary color to blue/violet from the color wheel.

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How many different genus groups are there? List them

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This is the answer naming them would be hard-

A bag of frozen broccoli weighs 306.0 grams. You microwave it and notice a lot is steam so you weigh after microwaving and it is 275.0 grams. What happened to the percent mass of water? Show your work

Answers

There are different methods to calculate the concentration of a solution. Mass percentage is one among them. Mass percentage is mainly used to calculate the concentration of a binary solution. Here mass percent of water is 10.13.

Mass percentage of a particular component in a solution is equal to mass in grams of that component present per 100 g of the solution. For example, a 5% aqueous solution of urea means 5g of urea in 100 g of its aqueous solution.

Mass percentage = Mass of the component / Total mass of solution × 100

Mass of water = 306.0 - 275.0 = 31

% Mass = 31 / 306.0 × 100 = 10.13%

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A soft drink contains 33g of sugar in 349g of H2O. What is the concentration of sugar in the soft drink in mass precent?

Answers

The concentration of the sugar in the soft drink in mass percent is 8.64%

How do i determine the concentration in mass percent?

First, we shall determine the mass of the solution. Details below:

Mass of sugar = 33 gramsMass of water = 349 gramsMass of solution =?

Mass of solution = Mass of sugar + mass of water

Mass of solution = 33 + 349

Mass of solution = 382 grams

Finally, we shall determine the mass percent of the sugar in the solution. Details below:

Mass of sugar = 33 gramsMass of solution = 382 gramsPercentage of sugar =?

Percentage = (mass of sugar / mass of solution) × 100

Percentage of sugar = (33 / 382) × 100

Percentage of sugar = 8.64%

Thus, we can conclude that the mass percent of the sugar is 8.64%

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How many mL of hydrogen chloride gas will be produced from 25.0 g of BaCl2 at STP? BaCl2(s) + H2SO4(aq) → BaSO4(aq) + 2HCl(g)

Answers

5380 mL of hydrogen chloride gas will produce 25.0 g of  BaCl₂ at STP.

To solve this problem, we need to use stoichiometry to determine the amount of hydrogen chloride gas produced from the given amount of BaCl₂.

First, we need to balance the chemical equation:

BaCl₂(s) + H₂SO₄(aq) → BaSO₄(aq) + 2HCl(g)

According to the balanced equation, 1 mole of BaCl₂ produces 2 moles of HCl.

The molar mass of BaCl₂ is 208.23 g/mol.

So, the number of moles of BaCl₂ in 25.0 g is:

n(BaCl₂) = mass ÷ molar mass = 25.0 g ÷ 208.23 g/mol = 0.120 mol

From the balanced equation, 1 mole of BaCl₂ produces 2 moles of HCl. Therefore, the number of moles of HCl produced is:

n(HCl) = 2 × n(BaCl₂) = 2 × 0.120 mol = 0.240 mol

At STP (standard temperature and pressure), 1 mole of any gas occupies 22.4 L.

Therefore, the volume of HCl gas produced at STP is:

V(HCl) = n(HCl) × 22.4 L/mol = 0.240 mol × 22.4 L/mol = 5.38 L

However, the question asks for the volume of hydrogen chloride gas in mL, so we need to convert the answer to mL:

1 L = 1000 mL

Therefore, V(HCl) = 5.38 L × 1000 mL/L = 5380 mL

So, 25.0 g of BaCl₂ at STP will produce 5380 mL of hydrogen chloride gas.

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Make a drawing representing the reaction that occurs between calcium nitrate and sodium oxalate.

Answers

The reaction between calcium nitrate (Ca(NO₃)₂) and sodium oxalate (Na₂C₂O₄) can be represented by the following chemical equation:

Ca(NO₃)₂ + Na₂C₂O₄ → CaC₂O₄ + 2NaNO₃

This is a double displacement reaction, where the calcium ion (Ca²⁺) from calcium nitrate and the oxalate ion (C₂O₄²⁻) from sodium oxalate switch places to form calcium oxalate (CaC₂O₄) and sodium nitrate (NaNO₃). The balanced chemical equation shows that one mole of calcium nitrate reacts with one mole of sodium oxalate to form one mole of calcium oxalate and two moles of sodium nitrate.

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Given the following data
C2H4 (g) + 3O2 (g) -> 2CO2 (g) + 2H2O (l) H = –1411.0 kJ
2C2H6 (g) + 7O2 (g) -> 4CO2 (g) + 6H2O (l) H = –3119.8 kJ
2H2 (g) + O2 (g) -> 2H2O (l) H = –571.7 kJ

calculate H for the reaction
C2H4 (g) + H2 -> C2H6 (g)

Answers

Answer:H = 5

Explanation: You would get this answer if you divide by 2 then multiple by 7

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