Explanation:
To calculate the amount of water needed to dilute the solution of potassium hydroxide and change its pH from 12.0 to 11.0, we need to use the formula for calculating the pH of a diluted solution.
The formula is:
pH = -log[H+]
where [H+] is the concentration of hydrogen ions in moles per liter.
Since we are diluting the solution by adding water, the concentration of [OH-] (hydroxide ions) will decrease proportionally to the volume of water added. This means that we can use the following equation to calculate the new concentration of [OH-]:
[OH-]1V1 = [OH-]2V2
where V1 is the initial volume of the solution, [OH-]1 is the initial concentration of hydroxide ions, V2 is the final volume of the solution after dilution, and [OH-]2 is the final concentration of hydroxide ions.
We know that the initial pH is 12.0, which means that [OH-]1 = 10^-2.0 M = 0.01 M.
We want to change the pH to 11.0, which means that [OH-]2 = 10^-11.0 M = 1 x 10^-11 M.
We also know that we are adding water to dilute the solution, but we don't know how much water we need to add yet. Let's call this volume of water "Vw".
Using the equation above, we can solve for V2:
[OH-]1V1 = [OH-]2V2
(0.01 M)(100 ml) = (1 x 10^-11 M)(100 ml + Vw)
V2 = (0.01 M)(100 ml)/(1 x 10^-11 M) - Vw
V2 = 10^12 ml - Vw
Now we can use this value for V2 in the pH formula to calculate the new pH:
pH = -log([H+])
[H+] = Kw/[OH-]
Kw is the ion product constant for water, which is equal to 1 x 10^-14 at room temperature.
[H+] = (1 x 10^-14)/(1 x 10^-11)
[H+] = 1 x 10^-3 M
pH = -log(1 x 10^-3)
pH = 3
We want to achieve a pH of 11.0, so we need to add enough water to bring down the pH from 12.0 to 11.0. This means that we need to add enough water so that V2 becomes:
V2 = (0.01 M)(100 ml)/(1 x 10^-11 M) - Vw = 10^11 ml
Therefore, we need to add:
Vw = V2 - initial volume of solution
Vw = (10^11 ml) - (100 ml)
Vw = 99999900 ml or approximately 100 million ml or 100 cubic meters of water.
So, in order to change the pH of a solution of potassium hydroxide with a pH of 12.0 to a pH of 11.0 by adding water only, you would need to add approximately 100 million milliliters or about 100 cubic meters of water.
PLEASE HELP!!!!!!
Reaction A (attached) starts as an orange solution in equilibrium. If SCN- is ADDED to the mixture, what color is the solution most likely to be after adjusting?
- red
- orange
- yellow
- clear
Reaction A (attached) starts as an orange solution in equilibrium. If SCN- is REMOVED from the mixture, what color is the solution most likely to be after adjusting?
- red
- orange
- yellow
- clear
Reaction A (attached) starts as an orange solution in equilibrium. If FeSCN2+ is ADDED to the mixture, what color is the solution most likely to be after adjusting?
- red
- orange
- yellow
- clear
Reaction A is the equilibrium between Fe₃+ and SCN- ions to form FeSCN₂+ ions, which results in an orange color solution.
If SCN- is added to the mixture, according to Le Chatelier's principle, the equilibrium will shift to the right to consume the added SCN- ions. This means that more FeSCN₂+ ions will form, resulting in a darker orange color solution, possibly even turning red if enough SCN- is added.
On the other hand, if SCN- is removed from the mixture, the equilibrium will shift to the left to replace the removed SCN- ions. This means that the FeSCN₂+ ions will dissociate to form Fe₃+ and SCN- ions, resulting in a lighter orange or even yellow color solution.
If FeSCN₂+ is added to the mixture, it would not have any significant effect on the color of the solution, as it is already in equilibrium and adding more product (FeSCN₂+ ions) will not shift the equilibrium in any particular direction. Therefore, the color of the solution would remain orange.
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Helppp Timed Quiz
Calculate the mass of 6. 9 moles of nitrous acid (HNO2). Explain the process or show your work by including all values used to determine the answer
How many milliliters of Sulphur dioxide are formed when 12. 5g of iron sulphide ore (pyrite) reacts with oxygen according to the equation at STP?
4FeS2+1102=2Fe2O3+8SO2
pls guys
Answer:
so the mass of sulphur dioxide = 0.208334 × 22.4 L = 4.6666816 L = 4666.6816 ml Therefore the volume of sulphur dioxide is 4666.6816 ml .
Explanation:
How many grams of Na₂SO4 were used to prepare 435.2 mL of 0.0156 M Na₂SO4 solution? m(Na₂SO4) =
Answer:
0.965 grams.
Explanation:
To calculate the mass of Na₂SO4, we can use the formula:
mass = moles x molar mass
First, let's calculate the number of moles of Na₂SO4 in 435.2 mL of 0.0156 M solution:
moles = volume (in L) x molarity
moles = 0.4352 L x 0.0156 mol/L
moles = 0.0067872 mol
The molar mass of Na₂SO4 is 142.04 g/mol, so we can calculate the mass of Na₂SO4 as:
mass = moles x molar mass
mass = 0.0067872 mol x 142.04 g/mol
mass = 0.965 g
Therefore, the mass of Na₂SO4 used to prepare 435.2 mL of 0.0156 M Na₂SO4 solution is 0.965 grams.
What 3 elements do you need to calculate the volume of a regular shaped object
We may find the size of a typically Regular shaped object by adding those following phases (area of both the bottom of the item and its height).
The volume of the thing is equivalent to the circumference of the bottom (longest times width times height of the object), or L x W x H. In this illustration, the length units we utilised were centimetres. Constant cross - sectional object volume A solid's volume can be calculated by multiplying its length, breadth, and height all at once.
V is equal to l times w. Regular forms feature interior (inner) slopes which are all equal as well as equal sides. Shapes that are irregular include curves and sides of any size and length. Below are a variety of shapes, both regular and infrequent: Pentagon regularly, regularly.
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How many moles of silver nitrate are needed to produce 6. 75 moles of copper (II) nitrate upon reacting with excess copper?
We would require 13.5 moles of AgNO₃ to make 6.75 moles of Cu(NO₃)₂1
The balanced chemical equation for the reaction between silver nitrate (AgNO₃) and copper (Cu) is:
2AgNO₃ + Cu -> Cu(NO₃)₂ + 2Ag
According to the stoichiometry of the reaction, 2 moles of AgNO₃ react with 1 mole of Cu to produce 1 mole of Cu(NO₃)₂ and 2 moles of Ag. Therefore, if 6.75 moles of Cu(NO₃)₂ are produced, the number of moles of Cu that reacted is also 6.75 moles.
To determine the number of moles of AgNO₃ required, we can use the ratio of moles of Cu to moles of AgNO₃ in the balanced equation. This ratio is 1:2, meaning that for every mole of Cu, 2 moles of AgNO₃ are needed.
Therefore, to produce 6.75 moles of Cu(NO₃)₂, we would need 2 × 6.75 = 13.5 moles of AgNO3.
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Cationic molecular ions are more abundant than Anionic molecular ions. Justify?
The Cationic molecular ions are more abundant than Anionic molecular ions because capacity of cationic to stabilise unpaired electrons make synthesis of cationic molecular ions less energetically demanding.
The capacity of cationic molecular ions to stabilise unpaired electrons makes the synthesis of cationic molecular ions less energetically demanding. The most advantageous reason for the abundance of cationic molecular ions is that the cosmic ray has so much energy that it is largely unaffected by the comparatively little energy necessary to ionise molecules. There is ample of radiation from a broadband source like the Sun to drive the molecules to high energy states from which they can be triggered and spontaneously emit. As a result, cationic molecular ions outnumber anionic ones.
Because of the low temperature and density of interstellar space, the most prevalent cationic molecular ion is H3+. Because its two electrons are sole valence electrons in the system, this H3+ is also thought to be a major source of cationic molecular ions.
Another reason for their prevalence is their capacity to stabilise unpaired electrons, which makes the production of cationic molecule ions less energetically demanding.
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explain spd and s- block element
Explanation:
SPD (s-block elements) are the elements located in the leftmost two columns of the periodic table. These elements include the alkali metals (Group 1A) and the alkaline earth metals (Group 2A). These elements are the most reactive of all elements and are characterized by their low ionization energies, high reactivities and relatively large atomic radii. The s-block elements are the only elements to form covalent bonds with other atoms. They also tend to form ionic bonds with other elements and have a tendency to form multiple bonds with each other. Because of their low ionization energies, they are very reactive, which is why they are often used in industrial applications, such as in the production of fertilizers, explosives, and drugs. The s-block elements are also important in the biochemical processes of living organisms. For example, sodium, potassium and calcium ions are essential in the regulation of the nervous system and muscular contraction. Magnesium is also important for energy production and muscle relaxation. In summary, the s-block elements are the most reactive of all the elements and have many important uses in industry and biochemistry. They are characterized by their low ionization energies, high reactivities, and relatively large atomic radii.
How many liters of a 0. 352M solution of CaSO4 would contain 62. 1g CaSO4 (molar mass of Calcium Sulfate is 135. 14g/mol)
A. 0. 16
B. 0. 77
C. 1. 35
D. 176. 4
1.3 Liters of a 0. 352M solution of CaSO4 would contain 62. 1g CaSO₄. So, option C is correct answer.
The solution has a concentration of 0.35M, it means that there ar 0.35M moles of CaSO₄ are there is the solution.
So, we can write is as,
Concentration = Moles/volume in Liters, this can be written as,
Concentration = Mass/Molar mass/volume in Liters
The molar mass of CaSO₄ is given to be 135.14 g/mol and the given mass of the compound is 62.1 grams. Now, putting all values,
0.35 = 62.1/(135.14V)
Volume = 1.35 Liters.
So, the volume of the solution required is found to be 1.35L hence option C is correct.
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While investigating the properties of a new substance created in class, Sally and Roberto
record the following observations:
The new substance is solid.
The new substance forms into thin flat sheets.
The new substance is smooth.
The new substance will burn.
The new substance looks like it will tear easily.
The new substance looks like it will dissolve in acid easily.
Which of their statements would be an inference based on the observations they have
made?
Every substance has some physical properties and chemical properties. The new substance is solid and the new substance forms into thin flat sheets can be considered as the inference based on the observations.
A physical property is defined as an any property that is measurable, whose value describes a state of a physical system. The changes in the physical properties of a system can be used to describe the changes between momentary states of the substance. These properties are referred to as observables. Physical properties are not modal properties.
A chemical property is defined as any of a material's properties that becomes evident during or after a chemical reaction that is, any quality that can be established only by changing a substance's chemical identity.
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In another experiment to determine ksp of the salt ab2, 500. 0 ml of 0. 100 m a2 was added to 500. 0 ml of 0. 205 m b-. After the precipitate settled, a portion of the supernatant liquid was filtered and analyzed and found to contain 2. 3 x 10-7 m a2. Calculate the ksp
A portion of the supernatant liquid was filtered and analysed after the precipitate settled. the ksp is 5.0 ×
[tex] {10}^{ - 4} [/tex]
The balanced chemical equation for the dissolution of AB2 in water is:
AB2 (s) ⇌ A2+ (aq) + 2B- (aq)
The solubility product expression for AB2 is:
Ksp = [A2+][B-]
[A2+] = 0.100 M × 0.5000 L / (0.5000 L + 0.5000 L) = 0.0500 M
[B-] = 0.205 M × 0.5000 L / (0.5000 L + 0.5000 L) = 0.1025 M
[A2+]_ppt = [A2+]_initial - [A2+]_filtered
[A2+]_ppt = 0.0500 M - 2.3 × 10M
[A2+]_ppt = 0.0500 M
[B-]_ppt = 2 × [A2+]_ppt
[B-]_ppt = 2 × 0.0500 M
[B-]_ppt = 0.100 M
Now, we can use these values to calculate the Ksp:
Ksp = [A2+][B-]^2
Ksp = (0.0500 M)(0.100 M)
Ksp = 5.0 ×
[tex] {10}^{ - 4} [/tex]
Therefore, the Ksp of AB2 is 5.0 ×
[tex] {10}^{ - 4} [/tex]
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An 11 M solution containing 449. 4 g of ammonium sulfate would contain how many liters?
Answer:
309.1 mL
What is molarity?
The most common way to express solution concentration is molarity (M), which is defined as the amount of solute in moles divided by the volume of solution in litres: M = moles of solute/litres of solution.
To calculate volume required for the solution, we first must find number of moles (n) required. To do this, we can divide mass in grams, by molar mass. Molar mass can be found on a standard IUPAC Periodic Table (International Union of Pure and Applied Science).
n[(NH₄)₂SO₄] = m/MM = 449.4/[(14.01+1.008×4)×2+32.07+16.00×4]
= 3.4006 mol
Now we have moles of solute, as well as molarity, (from the question - 11M) Thus, we can calculate litres of solution.
Volume = moles ÷ molarity = 3.4006 / 11 = 0.309 L = 309.1 mL
Suppose a vessel contains ClCH2CH2Cl at a concentration of 0. 810M. Calculate the concentration of ClCH2CH2Cl in the vessel 6. 20 seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits
the concentration of ClCH2CH2Cl in the vessel 5.80 seconds later would be 1.49 M.
The second-order rate law for the given reaction is:
rate = k[ClCH2CH2Cl]
where k is the rate constant.
We can use the integrated rate law for a second-order reaction to solve for the concentration of ClCH2CH2Cl at a later time:
1/[ClCH2CH2Cl]t = 1/[ClCH2CH2Cl]0 + kt
where [ClCH2CH2Cl]t is the concentration of ClCH2CH2Cl at time t, [ClCH2CH2Cl]0 is the initial concentration, and k is the rate constant.
Plugging in the values given in the problem, we get:
1/[ClCH2CH2Cl]t = 1/1.22 + (0.743 M-1 s-1)(5.80 s)
1/[ClCH2CH2Cl]t = 0.8205
[ClCH2CH2Cl]t = 1.220 M / 0.8205
[ClCH2CH2Cl]t = 1.487 M
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limiting reactant and percent yield
A limiting reagent is a chemical reactant that controls the amount of product produced. The limiting reagent provides the lowest product yield estimated from the available reagents (reactants).
What is the relationship between limiting reactant and percent yield?No additional product will develop once the limiting reactant has been completely consumed. The theoretical yield of the reaction is the maximum quantity of product that might be created depending on the limiting reactant. The 'percent yield' is calculated by comparing the actual yield to the theoretical yield..
To calculate the "expected yield" of the product, multiply the reaction equivalents for the limiting reagent by the product's stoichiometric factor. This is the estimated number of millimoles. To get the predicted mass of the product, multiply this value by the MW of the product.
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how many grams of solid barium sulfate form when 38.0 ml of 0.160 m barium chloride reacts with 54.0 ml of 0.065 m sodium sulfate? aqueous sodium chloride forms also
The amount of solid barium sulfate formed is 1.42g and 7.02 moles of sodium chloride are formed.
When 38.0 mL of 0.160 M barium chloride reacts with 54.0 mL of 0.065 M sodium sulfate, aqueous sodium chloride, and solid barium sulfate form. To determine the number of grams of solid barium sulfate formed, we must first calculate the amount of barium chloride and sodium sulfate used.
The amount of barium chloride used is equal to the product of the molarity and the volume, or 38.0 mL x 0.160 M = 6.08 mmol.
Similarly, the amount of sodium sulfate used is 54.0 mL x 0.065 M = 3.51 mmol.
The reaction between barium chloride and sodium sulfate is an exchange reaction:
BaCl₂(aq) + Na₂SO₄(aq) -> 2NaCl(aq) + BaSO₄(s)
We know that for every mole of barium chloride that is used, one mole of barium sulfate is formed, and for every mole of sodium sulfate used, two moles of sodium chloride are formed.
Therefore, 6.08 mmol of barium sulfate is formed and 7.02 mmol of sodium chloride is formed.
The number of grams of solid barium sulfate formed is equal to the product of the molar mass of barium sulfate and the number of moles of barium sulfate formed.
233.43 g/mol x 6.08 mmol = 1419.25 mg = 1.42g
In conclusion, when 38.0 mL of 0.160 M barium chloride reacts with 54.0 mL of 0.065 M sodium sulfate, 1.42g of solid barium sulfate, and 7.02 mmol of aqueous sodium chloride form.
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Chemistry 1001 solutions: a special type of mixture
Answer key
Solutions are mixtures that consist of two or more substances that are evenly distributed throughout the mixture.
Solutions are special because all of the substances in the solution are completely dissolved and cannot be seen with the unaided eye. Chemistry 1001 solutions are typically mixtures of solvents and solutes. A solvent is the substance that does the dissolving, and the solute is the substance that is dissolved.
Examples of solvents include water and alcohol, and examples of solutes include salts and sugar.
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Voge and Morgan (I. E. C. Proc. Design Dev. 11 454 1972) studied the dehydrogenation of n-butenes to butadiene over an iron oxide catalyst in the presence of steam. At 640°C the kinetics can be represented by a two consecutive first-order reaction network. Butene 1k→Butadiene 2k→ CO2 + cracked products with k2 = 0. 8 k1 a. In a preliminary series of experiments the following level of conversion was measured in an isothermal bed at 640°C. Equivalent spherical diameter Conversion 3. 4 mm 36. 9 5. 1 mm 31. 7 9. 5 mm 25. 6 Use these data to compute the Thiele modulus for the three pellets. Note that this is the conversion at the end of a reactor, not the local one. B. Determine the yield of butadiene using catalyst particles with a diameter of 1. 0 and 7. 0 mm in a reactor of the same length. Detrmine first the impact o n the conversion and then on the yield. C. . What will be the effect of increasing both rate constants by a factor of 4? The feed is pure butene and steam
A. Thiele modulus for the three pellets is 1.78, 2.74, and 6.04 for conversion 3. 4 mm 36.9, 5.1 mm 31.7, and 9.5 mm 25.6 respectively
B. The yield of butadiene using catalyst particles with a diameter of 1. 0 and 7. 0 mm in a reactor of the same length is 0.118 and 0.073 respectively.
C. The Thiele modulus (φ) will rise for all the pellets if both rate constants are increased by a factor of 4, as the reactions speed up.
a) To calculate the Thiele modulus (φ) for each pellet, we can use the following equation:
φ² / (6 x Th) = -(1/2)ln(1-X)
where,
φ is the particle diameter
Th is the Thiele modulus
X is the conversion.
The conversion for the first pellet, which has a 3.4 mm diameter, is 36.9%. Thus,
Th = φ² / [tex](6\times(-1/2)ln(1-X))[/tex]
Th = [tex]3.4^2 / (6\times(-1/2)ln(1-0.369))[/tex]
Th = 1.78
The conversion for the first pellet, which has a 5.1 mm diameter, is 31.7%. Thus,
Th = φ² / [tex](6\times(-1/2)ln(1-X))[/tex]
Th = [tex]5.1^2 / (6\times(-1/2)ln(1-0.317))[/tex]
Th = 2.74
The conversion for the first pellet, which has a 9.5 mm diameter, is 25.6%. Thus,
Th =φ²/ [tex](6\times(-1/2)ln(1-X))[/tex]
Th = [tex]9.5^2 / (6\times(-1/2)ln(1-0.256))[/tex]
Th = 6.04
B) In order to compute the butadiene yield utilizing catalyst particles with diameters of 1.0 and 7.0 mm, we must first know the conversion at the reactor's end.
We may infer from the information provided that the conversion is a diminishing function of pellet size.
Hence, we may interpolate the conversion values for the two particle sizes:
Conversion at 1.0 mm
= 36.9 x (1 - ((1.0 - 3.4) / (9.5 - 3.4))
= 26.6%
Conversion at 7.0 mm
= 25.6 x (1 - ((7.0 - 3.4) / (9.5 - 3.4))
= 16.5%
Thus, we must account for both the conversion and the selectivity when calculating the butadiene yield.
The second reaction's selectivity ([tex]S_2[/tex]) may be represented as
[tex]k_2 / (k_1 + k_2)[/tex] = 0.8 / 1.8
= 0.444 since the reactions occur one after the other.
For the 1.0 mm particle, the yield of butadiene can be calculated as:
Yield
= Conversion x Selectivity
= 0.266 x 0.444
≈ 0.118
For the 7.0 mm particle, the yield of butadiene can be calculated as:
Yield
= Conversion x Selectivity
= 0.165 x 0.444
≈ 0.073
c) The Thiele modulus (φ) will rise for all the pellets if both rate constants are increased by a factor of 4, as the reactions speed up. For the same reactor duration and circumstances, this will lead to a drop in conversion and yield.
Using the Thiele modulus equation from section (a), we can recalculate the conversion and yield using the new rate constants to determine the magnitude of this effect.
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When 16. 5 g of calcium carbonate react with an excess of hydrochloric acid, calcium chloride, carbon dioxide, and water vapor are produced. Write the equation. What is the theoretical yield of carbon dioxide?
Answer:
One mole of calcium carbonate reacts with 2 moles of HCl forms one mole of calcium chloride, water and carbon dioxide each. The theoretical yield of calcium chloride from 16.5 g of calcium carbonate is 18.3 g
Explanation:
If you have 0.857 moles of gas with a volume of 1.97 L at a pressure of 0.942 atm, what is the temperature of the gas in K?
Considering the ideal gas law, if you have 0.857 moles of gas with a volume of 1.97 L at a pressure of 0.942 atm, the temperature is 26.407 K.
Definition of ideal gas lawIdeal gas refers to a hypothetical gas composed of molecules that do not attract or repel each other and are approximated by point particles that themselves have no volume.
An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T), related by the ideal gas law. This law is an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar gas constant:
P×V = n×R×T
Temperature of he gasIn this case, you know:
P= 0.942 atmV= 1.97 Ln= 0.857 molesR= 0.082 (atmL)÷(molK)T= ?Replacing in the ideal gas law:
0.942 atm× 1.97 L = 0.857 moles× 0.082 (atmL)÷(molK)× T
Solving:
(0.942 atm× 1.97 L)÷ (0.857 moles× 0.082 (atmL)÷(molK))= T
26.407 K= T
Finally, the temperature is 26.407 K.
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If 48.0 g of NaCl react with 19.0 g of H₂SO4, what mass of Na₂SO4 will be produced?
equation is
2NaCI + H2SO4--Na2SO4 + 2HCI
Answer: 355
Explanation: Given the molecular weights:
M
r
N
a
O
H
=
40
g
m
o
l
M
r
N
a
2
S
O
4
=
142
g
m
o
l
The analogy of the moles will be held constant:
n
N
a
O
H
n
N
a
2
S
O
4
=
2
1
n
N
a
O
H
n
N
a
2
S
O
4
=
2
For each one, substitute:
n
=
m
M
r
Therefore:
n
N
a
O
H
n
N
a
2
S
O
4
=
2
m
N
a
O
H
M
r
N
a
O
H
m
N
a
2
S
O
4
M
r
N
a
2
S
O
4
=
2
200
40
x
142
=
2
200
⋅
142
40
x
=
2
200
⋅
142
=
2
⋅
40
x
x
=
200
⋅
142
2
⋅
40
=
100
⋅
142
40
=
10
⋅
142
4
=
1420
4
=
=
710
2
Given the molecular weights:
M
r
N
a
O
H
=
40
g
m
o
l
M
r
N
a
2
S
O
4
=
142
g
m
o
l
The analogy of the moles will be held constant:
n
N
a
O
H
n
N
a
2
S
O
4
=
2
1
n
N
a
O
H
n
N
a
2
S
O
4
=
2
For each one, substitute:
n
=
m
M
r
Therefore:
n
N
a
O
H
n
N
a
2
S
O
4
=
2
m
N
a
O
H
M
r
N
a
O
H
m
N
a
2
S
O
4
M
r
N
a
2
S
O
4
=
2
200
40
x
142
=
2
200
⋅
142
40
x
=
2
200
⋅
142
=
2
⋅
40
x
x
=
200
⋅
142
2
⋅
40
=
100
⋅
142
40
=
10
⋅
142
4
=
1420
4
=
=
710
2
= 355 GMS
Steps of Scientific learning
The steps in scientific learning are observations, hypotheses, predictions, testing, analysis, and conclusion.
Steps in scientific learningThe steps of scientific learning are as follows:
Observations - Gathering information about a particular phenomenon.Hypothesis - Developing a tentative explanation based on the observations.Predictions - Deriving predictions from the hypothesis that can be tested through experimentation.Testing - Conducting experiments to test the predictions.Analysis - Analyzing the results of the experiments to determine if they support or refute the hypothesis.Conclusion - Drawing a conclusion based on the results and determining whether further research is necessary.More on scientific learning can be found here: https://brainly.com/question/7508826
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How many moles of CS2 would be products if 12.4 moles of CO was also produced
Answer:
Explanation:
need the equation
3Cl2 + 2N2 → 2N2Cl3
Molar masses:
chlorine gas = 70. 9 g/mol
nitrogen gas = 28. 02 g/mol
dinitrogen trichloride = 134. 27 g/mol
a. 1. 98 mol of chlorine reacts with 1. 98 moles of nitrogen. What is the limiting reactant? (chlorine)
b. What mass of dinitrogen trichloride is produced from the reaction? (177. 2364)
c. How many moles of the excess reactant is left over? (0. 66)
d. If the percent yield of dinitrogen trichloride is 95. 5%, what is the actual yield of the product? (169. 2607)
0.0279 mole of the excess reactant is left over if 1.98 mole of chlorine react with 1.98 moles of nitrogen.
The mole is defined as the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon 12. It is expressed as "mol". It contains 6.022 X1023 entities like particles, atoms, ions, molecules, etc. of the given substance. It measures the number of atoms, ions, or molecules. It is a standard scientific unit for measuring large quantities of very small entities such as atoms, molecules, or other specified particles. It explains the number of atoms or other particles in a mole is the same for all substances.
If 1. 98 mole of chlorine reacts with 1. 98 moles of nitrogen. then, the we can calculate the amount of reactant left over with the help of the concept for mole.
Moles = 1.98 / 70.9
= 0.0279
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The correct question is,
3Cl2 + 2N2 → 2N2Cl3
Molar masses:
chlorine gas = 70. 9 g/mole
nitrogen gas = 28. 02 g/mole
dinitrogen trichloride = 134. 27 g/mole
1. 98 mole of chlorine reacts with 1. 98 moles of nitrogen. How many moles of the excess reactant is left over?
If you have gaseous carbon dioxide at 45°C and 6 atm of pressure, and you begin to cool it,
it will eventually turn into what physical state?
As a result, when the gas is cooled to a temperature below 45 °C, it will condense into a liquid condition. The pressure affects the precise temperature at which the gas will condense.
What happens to carbon dioxide when the pressure is raised above 1 atm?The slope of the solid-liquid equilibrium line for carbon dioxide is positive; when pressure is increased, the melting point rises.
What happens when sulphur is heated to 200 C at 1 atm from 80 C?Three triple points exist (a). (b) In an atmosphere, monoclinic is more stable than rhombic. (c) Sulfur changes state from liquid to vapour and back to liquid when heated from 80°C to 200°PC.
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What's the difference between a reflected in defecated sound wave
A. A reflected sound wave changes direction as it passes through an opening a diffracted sound wave bounces back to the place of origin
B. A reflected sound wave bounces back to the place if origin, a diffracted sound wave is absorbed as it passes through an opening
C. a reflected sound wave is absorbed by the medium, a diffracted sound wave changes direction as it passes through an opening
D. A reflected sound wave bounces back to the place of origin, a diffracted sound wave changes direction as it passes through an opening
Answer:
D
Explanation:
The difference between a reflected and diffracted sound wave is:
A. A reflected sound wave changes direction as it passes through an opening, whereas a diffracted sound wave bounces back to the place of origin.
Reflection occurs when a sound wave hits a surface and bounces back in the opposite direction. When a sound wave is reflected, it changes direction but does not necessarily change its wavelength or frequency. This is why we can hear echoes in a room with reflective surfaces.
Diffraction, on the other hand, occurs when a sound wave passes through an opening or around an obstacle and changes direction. When a sound wave is diffracted, it spreads out and changes its wavelength and frequency. This is why we can hear sound around corners or through a partially open door.
Therefore, option D is the correct answer: A reflected sound wave bounces back to the place of origin, a diffracted sound wave changes direction as it passes through an opening.
To earn full credit for your answers, you must show the appropriate formula, the correct substitutions , and your answer including the correct units.
If a country were doubling its population every 13 years, what would its growth rate be?
BoldItalicUnderline
The growth rate of the country's population if it was doubling it's population every 13 years is 5. 38%.
How to find the growth rate ?To calculate the growth rate of a population that is doubling every 13 years, we can use the rule of 70, which states that the approximate number of years it takes for a population to double can be calculated by dividing 70 by the annual growth rate.
So if a population is doubling every 13 years, we can find the annual growth rate as follows:
70 / 13 = 5.38
Therefore, the annual growth rate of this population would be approximately 5.38%.
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A student performs a neutralization reaction involving an acid and a base in an open polystyrene coffee-cup calorimeter. How would the calculated value of h differ from the actual value if there was significant heat loss to the surroundings
The calculated value of ∆H will be lower than the actual value due to heat loss. This is because the heat lost to the surroundings is not accounted for in the calculation, resulting in a lower measured temperature change than what would have occurred if no heat was lost.
If there is significant heat loss to the surroundings during a neutralization reaction in an open polystyrene coffee-cup calorimeter, the calculated value of ∆H (enthalpy change) would be lower than the actual value.
The enthalpy change for the reaction is calculated using the following equation: ∆H = q / n
where q is the heat released or absorbed by the reaction, n is the number of moles of the limiting reactant, and ∆H is the enthalpy change per mole of the limiting reactant.
In an open calorimeter, heat can be lost to the surroundings through conduction, convection, and radiation. The extent of heat loss depends on various factors such as the ambient temperature, the specific heat capacity of the surroundings, and the rate of heat transfer.
To minimize heat loss, the experiment should be conducted as quickly as possible, with good stirring to ensure uniform mixing of the reactants, and the calorimeter should be well-insulated to reduce heat loss to the surroundings. If these precautions are not taken, the calculated value of ∆H will be lower than the actual value due to heat loss.
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Given the following formula, calculate how many grams of C*O_{2} be produced from 11.89g C3H8. C_{3}*H_{8} + 5O_{2} -> 3C*O_{2} + 4H_{2}*O 11.89g ?g
To solve this issue, we must use stoichiometry to calculate the relationship between the production of CO2 and the amount of C3H8.
Determine how many grammes of C*O 2 will result from 11.89 grammes of C3H8 using the following formula: C 3*H 8 + 5O 2 -> 3C. *O {2} + 4H {2}*O 11.89g ?Using its molar mass, we must first determine the moles of C3H8: C3H8 has a molar mass of 44.11 g/mol, which is calculated by multiplying 3 (12.01 g/mol) by 8 (1.01 g/mol). C3H8 moles are equal to 11.89 g / 44.11 g/mol, or 0.27 mol. Next, we apply the balanced chemical equation's C3H8 to CO2 mole ratio: Three molecules of CO2 are produced from one mol of C3H8. the following will be the moles of CO2 produced: The formula for moles of CO2 is: 0.27 mol C3H8 x (3 mol CO2/1 mol C3H8) = 0.81 mol CO2. Finally, we use its molar mass to translate the moles of CO2 into grammes: Molar mass of CO2 equals 12.01 g/mol plus 2 (16.00 g/mol), or 44.01 g/mol. mass of CO2 = 44.01 g/mol x 0.81 mol CO2 to get 35.64 g CO2. As a result, 11.89 g of C3H8 will result in 35.64 g of CO2.
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consider the structures given below. which of these structures would be the most stable contributor to the resonance hybrid formed when anisole undergoes nitration? question blank 1 of 3 e how does anisole direct
The most stable contributor to the resonance hybrid formed when anisole undergoes nitration is structure C.
This is because the negative charge is delocalised to both the oxygen atom and the ring structure of the anisole, which makes the molecule more stable. Anisole directs nitration through resonance, where the delocalised electrons present in the aromatic ring contribute to the electron density around the electrophile at centre and facilitate the electrophilic attack.
Thus, structure C, which has the most delocalised electrons and the most resonance stabilisation, will be the most stable contributor to the resonance hybrid formed when anisole undergoes nitration.
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The most stable contributor to the resonance hybrid formed when anisole undergoes
nitration
is Structure A, which is
ortho
-substituted.
This is because anisole is a meta-directing compound, meaning that it prefers the intermediate formed when a nitronium ion attacks the
meta
-position of an aromatic ring.
Structure A is
ortho-substituted
, meaning that it is the most likely intermediate formed when anisole undergoes nitration.
Structure A is the most stable intermediate of the three structures, since it is the only one with two equivalent aromatic carbocations in the
resonance
hybrid.
The reason why anisole is meta-directing is because of the electron-withdrawing oxygen atom, which increases the electron density of the ring opposite to it (the
meta
-position).
This increased electron density of the meta-position makes it more likely for a nitronium ion to attack the meta-position, as opposed to the ortho- and para-positions.
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The diagram shows a cross section of a thermos the vacuum between the outer and inner bottles is a space containing no air of other matter
Answer:
which diagram are you talking about?
The answer is
It prevents particles from transferring thermal energy between the outer and inner bottles