Propylene glycol, C3H6(OH)2, is sometimes used in automobile antifreeze solutions. If an aqueous solution has
a mole fraction XC₂H(OH)₂ = 0.100, calculate (a) the percent propylene glycol by mass; (b) the molality of the
propylene glycol in the solution.

To determine the gas in g/mL, we need to calculate the volume of the gas first. We can use the formula for the volume of a rectangular room to find the volume of the room:

Volume = length x width x height = 2.00 m x 2.00 m x 5.00 m = 20.00 m³

Next, we need to convert the mass of the gas from grams to kilograms, since density is usually expressed in units of kg/m³. We can do this by dividing the mass by 1000:

Mass = 3175 g ÷ 1000 = 3.175 kg

Finally, we can calculate the density of the gas using the formula:

Density = Mass ÷ Volume

Density = 3.175 kg ÷ 20.00 m³ = 0.1588 kg/m

To convert the density from kg/m³ to g/mL, we need to multiply by 1000 and then divide by 1000:

Density = 0.1588 kg/m³x 1000 g/kg ÷ 1000 mL/L = 0.1588 g/mL

Therefore, the gas has a density of 0.1588 g/mL. Without additional information, we cannot determine the identity of the gas.

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Answer:

see explanations

Explanation:

1. pH = -log [H+]

so [H+] = 10^(-pH) = 10^(-3.57) = 2.7 x 10^(-4) M

2. neutralize implies mol acid = mol base

so [HNO3] = (1 M) * (0.080 L) / (0.250 L) = 0.32 M HNO3

3. mol HNO3 = molarity of HNO3 * volume of HNO3

= 0.32 M HNO3 * 0.250 L HNO3 = 0.080 mol HNO3

Answer:

b) The amount of energy required to eject an electron from an atom.

Answer:

the right answer is first one

The true statement on the data on the graph about the attendance for the play made by Grove Street is B. More people came to the play on Tuesday than on Monday.

The graph shows the number of people who attended a school play by Grove School from Monday to Saturday in a certain week. We see that Monday had the lowest attendance and was followed by Tuesday.

Attendance continued to rise and was higher on Wednesday than Tuesday but then dropped for Thurdsady and Frifay. It was then highest on the Saturday as more parents probably had time to attend.

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Answer: 0.925 L

Explanation:

M=moles /L

L= moles/M

3.70/4 = 0.925 L

Answer: The pincushion moss, the echeveria, and the moon cactus in a terrarium make up a community.

:)

I think this is the answer btw Sorry if i'm wrong I have no clue.

In a terrarium, the pincushion moss, the echeveria, and the moon cactus make up the community.

Moon cactus, also known as Gymnocalycium mihanovichii or Hibotan cactus, is native to South American deserts in places like Brazil and Argentina.

Leucobryum glaucum, commonly known as leucobryum moss or pin cushion moss, is a species of haplolepideous mosses with a wide distribution in eastern North America and Europe.

Echeveria is a large genus of flowering plants in the family Crassulaceae, native to semi-desert areas of Central America, Mexico and northwestern South America.

Therefore, In a terrarium, the pincushion moss, the echeveria, and the moon cactus make up the community.

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Answer:

Correct

pls mrk me brainliest

To identify the limiting reactant, we need to compare the amount of each reactant to the balanced chemical equation. The balanced equation for the reaction between nitrogen and hydrogen to produce ammonia is:

N2 + 3H2 → 2NH3

From the equation, we can see that one mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia.

First, let's convert the mass of nitrogen to moles:

45.25 g N2 x (1 mol N2/28.02 g N2) = 1.612 mol N2

Next, let's convert the volume of hydrogen to moles using the ideal gas law:

PV = nRT

n = PV/RT

n = (52.5 L) (1 atm) / (0.0821 L·atm/mol·K) (273 K) = 2.19 mol H2

Now we can compare the number of moles of each reactant to see which is limiting:

N2 : H2 = 1.612 mol : 2.19 mol

According to the balanced equation, one mole of nitrogen reacts with three moles of hydrogen. So we need 4.836 moles of hydrogen to completely react with 1.612 moles of nitrogen. However, we only have 2.19 moles of hydrogen, which means it is the limiting reactant.

Therefore, the amount of ammonia that can be produced is limited by the amount of hydrogen available.

The balanced equation shows that 3 moles of hydrogen produces 2 moles of ammonia. So, with 2.19 moles of hydrogen, we can produce:

2.19 mol H2 x (2 mol NH3/3 mol H2) = 1.46 mol NH3

Now, let's convert the moles of ammonia to liters using the ideal gas law:

PV = nRT

V = nRT/P

V = (1.46 mol) (0.0821 L·atm/mol·K) (273 K) / (1 atm) = 31.2 L

Therefore, 31.2 liters of ammonia gas can be produced from the given amount of nitrogen To identify the limiting reactant, we need to compare the amount of each reactant to the balanced chemical equation. The balanced equation for the reaction between nitrogen and hydrogen to produce ammonia is:

N2 + 3H2 → 2NH3

From the equation, we can see that one mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia.

First, let's convert the mass of nitrogen to moles:

45.25 g N2 x (1 mol N2/28.02 g N2) = 1.612 mol N2

Next, let's convert the volume of hydrogen to moles using the ideal gas law:

PV = nRT

n = PV/RT

n = (52.5 L) (1 atm) / (0.0821 L·atm/mol·K) (273 K) = 2.19 mol H2

Now we can compare the number of moles of each reactant to see which is limiting:

N2 : H2 = 1.612 mol : 2.19 mol

According to the balanced equation, one mole of nitrogen reacts with three moles of hydrogen. So we need 4.836 moles of hydrogen to completely react with 1.612 moles of nitrogen. However, we only have 2.19 moles of hydrogen, which means it is the limiting reactant.

Therefore, the amount of ammonia that can be produced is limited by the amount of hydrogen available.

The balanced equation shows that 3 moles of hydrogen produces 2 moles of ammonia. So, with 2.19 moles of hydrogen, we can produce:

2.19 mol H2 x (2 mol NH3/3 mol H2) = 1.46 mol NH3

Now, let's convert the moles of ammonia to liters using the ideal gas law:

PV = nRT

V = nRT/P

V = (1.46 mol) (0.0821 L·atm/mol·K) (273 K) / (1 atm) = 31.2 L

Therefore, 31.2 liters of ammonia gas can be produced from the given amount of nitrogen and hydrogen.

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The answer is 0 g.

In order to determine how much salt will remain undissolved after adding 123 g of a salt with solubility of 551 g/L to 421 mL of water, we need to first convert the volume of water to liters.421 mL of water = 0.421 L of water

The maximum amount of salt that can dissolve in 1 L of water with a solubility of 551 g/L can be calculated using the formula:

Max amount of salt that can dissolve = Solubility × Volume of solvent Max amount of salt that can dissolve = 551 g/L × 1 L

Max amount of salt that can dissolve = 551 g

Therefore, the maximum amount of salt that can dissolve in 0.421 L of water is:

Max amount of salt that can dissolve = 551 g/L × 0.421 L

Max amount of salt that can dissolve = 231.671 g

Since only 123 g of salt was added to 0.421 L of water, this is less than the maximum amount of salt that can dissolve. Hence, all of the salt will dissolve and there will be no salt left undissolved. Therefore, the answer is 0 g.

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24.9 g of oxygen gas would be required to completely burn 7.33 g of propane.

Oxygen is an element found in air. It is a colorless, odorless gas that makes up about 21% of the Earth's atmosphere. Oxygen is essential for humans and animals to survive, as it is used to produce energy in the cells of living organisms, and to help maintain the balance of other elements in the body.

To answer this question, you need to balance the chemical equation first. The balanced equation is:
[tex]C_3H_8 + 5O_2 - > 3CO_2 + 4H_2O[/tex]
Now that the equation is balanced, you can calculate the mass of oxygen required to burn 7.33 g of propane.
To calculate the mass of oxygen, you need to use the mole ratio of the reactants in the balanced equation. The mole ratio of propane to oxygen is 1:5. Therefore, for every 1 mole of propane, 5 moles of oxygen are required.
Using the mole ratio and the given mass of propane, you can calculate the mass of oxygen required.
7.33 g of propane x (1 mole propane / 44.096 g propane) x (5 moles oxygen / 1 mole propane) x (32.00 g oxygen / 1 mole oxygen) = 24.9 g of oxygen
Therefore, 24.9 g of oxygen gas would be required to completely burn 7.33 g of propane.

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Answer:

Explanation:

mOLARITY IS moles/L

moles = M X L

2.1 X 0.50 = 1.05 moles NaBr or 1.1 moles NaBr to correct sig figs

The only one I know is NaC2HO4. 1 atom in sodium, 2 atoms in carbon, 1 atom in hydrogen and 4 atoms in oxygen completeting the total of 8 atoms in this element.

The first stage in calculating the number of atoms is to determine the number of molecules. To determine the number of moles in an element or compound, reduce the specified mass by the element or compound's molar mass. The number of atoms in 1 mole of a material is or. 023 10 23 atoms.

An element is a particle. Since the two terms are identical, the answer is always one, and only one, if you're searching for the number of atoms in an element.

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Full Question: What’s the elements found in these formulas?

NaC2HO4

H2F5BLi

2He2PSO4

3He2O4PH

the mass of carbon in 100.0 g of vinegar is 40.01 g, the mass of hydrogen is 6.70 g, and the mass of oxygen is 53.29 g.

To calculate the mass of each element in 100.0 g of vinegar, we need to use the percent composition of the compound. The percent composition tells us the percentage of each element in the compound by mass.

The percent composition of vinegar is:

Carbon (C): 40.01%

Hydrogen (H): 6.70%

Oxygen (O): 53.29%

To calculate the mass of each element, we need to multiply the percent composition by the total mass of the compound (100.0 g).

Mass of carbon (C) = 40.01% x 100.0 g = 40.01 g

Mass of hydrogen (H) = 6.70% x 100.0 g = 6.70 g

Mass of oxygen (O) = 53.29% x 100.0 g = 53.29 g

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a.  Total number of ions in 47. 8 g of SrF2 = 1.635 ions.

b.  Mass (kg) of 4. 90 mol of CuCl2 · 2 H2O = 0.833 kg

c.  Mass (mg) of 2. 67 1022 units of Bi(NO3)3 · 5 H2O = 21,700 mg

a. Total number of ions in 47. 8 g of SrF2

The molar mass of SrF2 per gram = 87.62 g/mol

The number of moles of SrF2 in 47.8 g is calculated by using the formula,

The number of moles = n = (m/M) = 47.8 g / 87.62 g/mol

The number of moles  = 0.545 mol

Total number of ions present in the molecule SrF2 = (1 Sr2+ ion/mol) x (0.545 mol) + (2 F- ions/mol) x (0.545 mol)

The Total number of ions = 0.545 + 2(0.545)

The total number of ions = 1.635 ions.

b. Mass (kg) of 4. 90 mol of CuCl2 · 2 H2O

Total number of ions in CuCl2 · 2 H2O = (63.55 g/mol Cu + 2 x 35.45 g/mol Cl + 2 x 18.02 g/mol H + 16.00 g/mol O) + 2 x 18.02 g/mol H + 2 x 16.00 g/mol O

Total number of ions = 170.48 g/mol

The mass of 4.90 mol of CuCl2 · 2 H2O is:

Molar mass (m) = nM = 4.90 mol x 170.48 g/mol

Molar mass (m)  = 833.8 g = 0.833 kg

c.  Mass (mg) of 2.67 10^22 units of Bi(NO3)3 · 5 H2O

Total number of ions in Bi(NO3)3 · 5 H2O = (208.98 g/mol Bi + 3 x 62.01 g/mol N + 9 x 16.00 g/mol O) + 5 x 18.02 g/mol H + 5 x 16.00 g/mol O

Total number of ions = 485.09 g/mo

Total number of moles (n) = N/NA = (2.67 × 10^22)/6.022 × 10^23

Total number of moles = 0.0444 mol

The mass of 0.0444 mol of Bi(NO3)3 · 5 H2O is:

Molar mass (m) = nM = 0.0444 mol x 485.09 g/mol

Molar mass = 21.7 g =  21,700 mg

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The specific components and design of a battery can vary depending on the type of battery and its intended use.

The basic components of a battery include:

Anode: This is the negative electrode of the battery where the oxidation reaction takes place.

Cathode: This is the positive electrode of the battery where the reduction reaction takes place.

Electrolyte: This is the medium that allows ions to flow between the anode and cathode, completing the circuit and generating the electrical energy.

Separator: This is a porous material that separates the anode and cathode, preventing direct contact between the electrodes while still allowing ions to flow between them.

Terminal: These are the points on the battery where the electrical energy is transferred out of the battery to power a device or charge another battery.

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a) A gas was produced

b) Reaction 1 does not takes place in a beaker

c) The reactions are balanced

d) The law of conservation of mass can be used to show that a reaction is balanced.

A combustion reaction, for instance, might not be able to continue if there isn't enough fuel or oxygen in the beaker to support it.

Moreover, a beaker is unlikely to contain an ignition source, like as a spark or flame, which is typically required to start the reaction.

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Answer:

There are 8 oxygen atoms in the formula 4H3O2.

Explanation:

Answer:

There are 6 oxygen atoms in 4H3O2

Explanation:

There are 3 oxygens but since oxygen is a diatomic atom you multiply it by 2 and you get 6.

The particles in a suspension can be separated from heterogeneous mixtures by passing the mixture through a filter.

Suspensions are heterogeneous mixtures that contain particles of a larger size that are suspended in a liquid medium. These particles are large enough to be visible to the ordinary eye and can be separated from the mixture through the use of a filter.

When a suspension is passed through a filter, the particles in the mixture are trapped in the filter medium, and the liquid passes through, leaving the solid particles behind.

This is due to the physical properties of the particles in suspension, which are large enough to be filtered out. In contrast, the particles in solutions and colloids are much smaller, and cannot be separated through filtration.

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Silver sulfide (Ag2S) is the common tarnish on silver objects. The mass of silver sulfide that can be made from 1.53 x 10^-3g of hydrogen sulfide (H2S) obtained from a rotten egg is 3.67 g.

Let's understand this in detail:

The balanced chemical equation for the reaction is:

2 AgNO3 + H2S → Ag2S + 2 HNO3

The molar mass of H2S is 34.08 g/mol.

1.53 x 10^-3g of H2S = (1 mol / 34.08 g) * (1.53 x 10^-3g) = 4.49 x 10^-5 mol H2S.

From the balanced equation, it is clear that one mole of Ag2S is formed from one mole of H2S.

Therefore, the number of moles of Ag2S formed from 1.53 x 10^-3g of H2S is 4.49 x 10^-5 mol.

The molar mass of Ag2S is 247.8 g/mol. Therefore, the mass of Ag2S formed from 1.53 x 10^-3g of H2S is

(247.8 g/mol) * (4.49 x 10^-5 mol) = 0.0111 g or 11.1 mg.

To convert to grams, divide by 1000:11.1 mg ÷ 1000 = 0.0111 g or 11.1 mg = 1.11 x 10^-2

Therefore, the mass of silver sulfide that can be made from 1.53 x 10^-3g of hydrogen sulfide obtained from a rotten egg is 3.67 g (approx).

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Answer:

Relative atomic mass is the average mass of an atom of an element, compared to 1/12th the mass of carbon 12 atom. Relative molecular mass is the ratio of the average mass of one molecular of an element or compound to1/2 of the mass of an atom of carbon 12.

This is calculated by dividing the mass of aluminum (10.7 g) by its molar mass (26.98 g/mol), resulting in 0.395 mol.

To find the amount of pure substance (in moles) in 10.7 g of aluminum, we need to divide the given mass by the molar mass of aluminum.

The molar mass of aluminum is given as 26.98 g/mol.

Therefore, the amount of pure substance (in moles) in 10.7 g of aluminum can be calculated as:

moles = mass/molar mass

moles = 10.7 g/26.98 g/mol

moles = 0.396 moles (rounded to three significant figures)

So, there are approximately 0.395 moles of pure substance (aluminum) in 10.7 g of aluminum.

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Answer:

To calculate the number of moles of NaOH needed to create a 2 M solution in 8 liters of solution, we need to use the formula:

moles = molarity x volume (in liters)

Here, the molarity (M) is given as 2 M, and the volume (V) is given as 8 liters.

So,

moles = 2 M x 8 L

= 16 moles

Therefore, 16 moles of NaOH are needed to create a 2 M solution in 8 liters of solution.

Explanation:

C₆H₁₂O₆ has a molar mass of 180.18 g/mol.

The molar mass of a compound is the mass of one mole of the compound and is expressed in grams per mole (g/mol). To calculate the molar mass of a compound, we sum the atomic masses of all the atoms in the compound.

For NO₂, we have:

Molar mass of N = 14.01 g/mol

Molar mass of O = 16.00 g/mol

Therefore, the molar mass of NO₂ = 14.01 g/mol + 2(16.00 g/mol) = 46.01 g/mol.

For C₆H₁₂O₆ (glucose), we have:

Molar mass of C = 12.01 g/mol

Molar mass of H = 1.01 g/mol

Molar mass of O = 16.00 g/mol

Therefore, the molar mass of C₆H₁₂O₆  = 6(12.01 g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol) = 180.18 g/mol.

Knowing the molar mass is useful in calculating various other properties of the compound, such as the number of moles present in a given mass or the mass of a given number of moles.

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The molar mass of NO2 is 46 g/mol.

The molar mass of C6H12O6 is 180 g/mol.

The atomic masses of all the atoms in a compound's formula are added to determine its molar mass. In order to compute the molar mass of NO2, the atomic masses of one nitrogen atom (14.01 g/mol) and two oxygen atoms (2 x 16.00 g/mol) are added together, yielding a result of 46.01 g/mol.

Glucose is a typical monosaccharide and has the chemical formula C6H12O6. The atomic masses of every atom in the molecule must be added in order to determine the molar mass. Hydrogen, oxygen, and carbon all have atomic masses of 1.01 g/mol, 12.01 g/mol, and 16.00 g/mol, respectively. dividing the atomic mass of each element by the number of atoms in that element.

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Answer:

1. 79.2 g of CO2

2. the reaction is spontaneous and favors the formation of products

3. CH2O

Explanation:

To solve the problem, we first need to calculate the limiting reagent by comparing the amount of glucose and oxygen available for the reaction. We will assume that the reaction goes to completion.

The balanced equation shows that for every 1 mole of glucose, 6 moles of oxygen are required. Therefore, the moles of oxygen required for 9.0 grams of glucose is:

moles of glucose = mass/molar mass = 9.0/180.16 = 0.0499 mol

moles of oxygen = 6 x moles of glucose = 6 x 0.0499 = 0.2994 mol

Since we have 0.2994 moles of oxygen available, and only 0.2000 moles of oxygen are required to react with 0.0499 moles of glucose to produce 0.0270 moles of water (according to the balanced equation), oxygen is the limiting reagent.

Using the balanced equation, we can now calculate the moles of CO2 produced:

moles of water produced = mass/molar mass = 5.4/18.02 = 0.2997 mol

moles of CO2 produced = 6 x moles of water produced = 6 x 0.2997 = 1.7982 mol

Finally, we can calculate the mass of CO2 produced:

mass of CO2 produced = moles x molar mass = 1.7982 x 44.01 = 79.2 g

Therefore, the mass of CO2 produced when 9.0 grams of glucose completely reacts with 9.6 grams of oxygen to produce 5.4 grams of water is 79.2 g.

To compare the entropy of the reactants to the entropy of the products, we can use the equation:

ΔS = ΣS(products) - ΣS(reactants)

The entropy of a substance depends on its state and temperature, and can be looked up in tables. At standard conditions (298 K and 101.3 kPa), the molar entropy of glucose, oxygen, CO2, and liquid water are:

S(C6H12O6) = 212.8 J/(mol K)

S(O2) = 205.0 J/(mol K)

S(CO2) = 214.8 J/(mol K)

S(H2O) = 69.9 J/(mol K)

Using the above values and the balanced equation, we can calculate the entropy change:

ΔS = (6 x S(CO2) + 6 x S(H2O)) - (S(C6H12O6) + 6 x S(O2))

ΔS = (6 x 214.8 + 6 x 69.9) - (212.8 + 6 x 205.0)

ΔS = 287.4 J/(mol K)

Since ΔS is positive, the entropy of the products is greater than the entropy of the reactants. This means that the reaction is spontaneous and favors the formation of products.

The empirical formula for glucose can be determined by dividing the subscripts by their greatest common factor. In this case, the empirical formula is:

C6H12O6 ÷ 6 = CH2O

Therefore, the empirical formula for glucose is CH2O.

N2 (g) + 3H2 (g) 2NH3 is the balanced chemical equation of N2 + H2 NH3 (g). Based on the rule of conservation of mass, it is possible to make the atoms on both the reactant and product sides equal.

In a balanced chemical equation, the total number of atoms of an element present in a species is equal to the product of the stoichiometric coefficient and the number of atoms of the element in one molecule of the species.

The total number of oxygen atoms in the reactive species '2O2', for example, is four. Use these easy principles to balance equations on your own: Verify that all of the equation's formulas are valid. Just deal with one ingredient at a time. Balancing adds up to a lot of money.

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Answer: An object's speed and motion can be changed by a force acting on it. This is described by Newton's Second Law of Motion, which states that the acceleration of an object is directly proportional to the force applied to it, and inversely proportional to its mass. Therefore, the greater the force applied to an object, the greater its acceleration will be, and the greater the change in speed or motion will be. Additionally, the direction of the force also plays a role in changing an object's motion, as it can cause the object to change direction or rotate.

Your welcome (:

Explanation:

Glass Because wood and other materials can be electrified through induction, but glass can't.

Is specific heat capacity C or Q?

The equation q = mcT may be used to compute the amount of heat acquired or lost by a sample (q), where m is the sample's mass, c is the specific heat, and T is the temperature change.

The quantity of heat required to raise the temperature of a given amount of stuff by one degree Celsius is referred to as heat capacity. The heat capacity of one gram of a material is known as its specific heat capacity (or specific heat), whereas the heat capacity of one mole is known as its molar heat capacity.

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