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Kruka [31]
2 years ago
14

For the following equilibrium system, which of the following changes will form more CaCO3? CO2(g) + Ca(OH)2(s) ⇌ CaCO3(s) + H2O(

l) ΔH o rxn = −113 kJ a. Decrease temperature at constant pressure (no phase change). b. Increase volume at constant temperature. c. Increase partial pressure of CO2. d. Remove one-half of the initial CaCO3.
Chemistry
1 answer:
Talja [164]2 years ago
8 0

Answer: d. Remove one-half of the initial CaCO3.

Explanation: Le Chatelier's principle states that changes on the temperature, pressure, concentration and volume of a system will affect the reaction in an observable way. So in the reaction above:

A decrease in temperature will shift the equilibrium to the left because the reaction is exothermic, which means heat is released during the reaction. In other words, when you decrease temperature of a system, the equilibrium is towards the exothermic reaction;

A change in volume or pressure, will result in a production of more or less moles of gas. A increase in volume or in the partial pressure of CO2, the side which produces more moles of gas will be favored. In the equilibrium above, the shift will be to the left.

A change in concentration will tip the equilibrium towards the change: in this system, removing the product will shift the equilibrium towards the production of more CaCO3 to return to the equilibrium.

So, the correct answer is D. Remove one-half of the initial CaCO3.

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Movement of the ___<br> creates the London dispersion forces.
Tanzania [10]

Answer: electrons

Explanation: moving electrons cause momentarily charge

Distribution on molecule. This distribution induces similar distribution to

Adjacent molecule.

7 0
2 years ago
Suppose you had a balloon containing 1 mole of helium at STP and a balloon containing 1 mole of oxygen at STP. Which statement(s
AleksAgata [21]

Answer:

The true statement  is option A.

Explanation:

Using ideal gas equation:

PV = nRT

where,

P = Pressure of gas =  1 atm

V = Volume of gas = ?

n = number of moles of gas = 1 mol

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 273.15 K

V=\frac{nRT}{P}=\frac{1 mol\times 0.0821 atm L/mol K\times 273.15 K}{1 atm}

V = 22.42 L

This means that 1 mole of an ideal gas at STP occupies 22.42 liters of volume.

So, 1 mole of helium gas and 1 mole of oxygen gas will have same value of volume in their respective balloons at STP.

7 0
2 years ago
The concentration of sodium chloride in an aqueous solution that is 2.23 M and that has a density of 1.01 g/mL is __________% by
amm1812

Answer:

The concentration of sodium chloride in an aqueous solution that is 2.23 M and that has a density of 1.01 g/mL is 12.90% by mass

Explanation:

2.23 M aqueous solution of NaCl means there are 2.23 moles of NaCl in 1000 mL of solution.

We know that density is equal to ratio of mass to volume.

Here density of solution is 1.01 g/mL.

So mass of 1000 mL solution = (1.01\times 1000) g = 1010 g

molar mass of NaCl = 58.44 g/mol

So mass of 2.23 moles of NaCl = (2.23\times 58.44) g = 130.3 g

% by mass  is ratio of mass of solute to mass of solution and then  multiplied by 100.

Here solute is NaCl.

So % by mass of 2.23 M aqueous solution of NaCl = \frac{130.0}{1010}\times 100% = 12.90%

3 0
2 years ago
Hydrogen was collected over water using the approach in the manual. The water temperature was 220C and the measured pressure ins
olasank [31]

Answer:

Pressure of hydrogen gas = 695.2 mmHg

Explanation:

Given:

Water temperature = 22°C

Pressure inside the tube = 715 mmHg

Find:

Pressure of hydrogen gas

Computation:

Using vapor pressure of water table

Water pressure at 22°C = 19.8 mmHg

Pressure inside the tube = Pressure of hydrogen gas + Water pressure at 22°C

715 = Pressure of hydrogen gas + 19.8

Pressure of hydrogen gas = 715 - 19.8

Pressure of hydrogen gas = 695.2 mmHg

3 0
2 years ago
A goldsmith melts 12.4 grams of gold to make a ring. The temperature of the gold rises from 26°C to 1064°C, and then the gold me
DiKsa [7]

Problem One

You will use both m * c * deltaT and H = m * heat of fusion.

Givens

m = 12.4 grams

c = 0.1291

t1 = 26oC

t2 = 1204

heat of fusion (H_f) = 63.5 J/grams.

Equation

H = m * c * deltaT + m * H_f

Solution

H = 12.4 * 0.1291 * (1063 - 26) + 12.4 * 63.5

H = 1660.1 + 787.4

H = 2447.5 or 2447.47 is the exact answer. I have to leave the rounding to you. I have no idea where to round it although I suspect 2450 would be right for 3 sig digs.

Problem Two

Formula and Givens

t1 = 14.5

t2 = 50.0

E = 5680

c = 4.186

m = ??

E = m c * deltaT

Solution

5680 = m * 4.186 * (50 - 14.5)

5680 = m * 4.186 * (35.5)

5680 = m * 148.603 * m

m = 5680 / 148.603

m = 38.22 grams That isn't very much. Be very sure you are working in joules. You'd leave that many grams in the kettle after drying it thoroughly.

m = 38.2 to 3 sig digs.

8 0
2 years ago
Read 2 more answers
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