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Alina [70]
2 years ago
13

Choose the substance with the higher entropy in each pair. Assume constant temperature, except in part (5)

Chemistry
1 answer:
Nataly [62]2 years ago
3 0

Answer:

I) 1 mol of SO3(g)

2) 1 mol of CO2(g)

3) 3 mol of O2(g)

4) 1 mol of KBr(aq)

5) Seawater at 23°C

6) 1 mol of CCl4(g)

Explanation:

In molecules having greater numbers of atoms, there is an increase the number of ways by which the molecule vibrates thereby leading to a higher number of possible microstates and overall increase in entropy of the system. Hence, 1 mol of SO3(g) has a higher entropy than 1 mol of SO2.

Gases have a higher entropy than liquids and liquids have a higher entropy than gases.

Also, the greater the molecular weight of a molecule, the higher the entropy. Higher number of moles of a gas as well as the increase in temperature of a substance are also factors that lead to higher entropy.

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If 20 grams of Zinc phosphate reacts with excess hydrochloric acid and produces 18 grams of Zinc chloride what is the percent yi
fenix001 [56]

Answer:

Y=85\%

Explanation:

Hello!

In this case, since we know the balanced chemical reaction, we are first able to realize there is a 1:3 mole ratio between zinc phosphate and zinc chloride; it means that we can first compute the moles of the desired product via stoichiometry:

n_{ZnCl_2}=20gZn_3(PO_4)_2*\frac{1molZn_3(PO_4)_2}{386.11gZn_3(PO_4)_2}*\frac{3molZnCl_2}{1molZn_3(PO_4)_2}=0.16gZnCl_2

Next, since those moles are associated with the theoretical yield of zinc chloride, we obtain the corresponding mass:

m_{ZnCl_2}^{theoretical}=0.16molZnCl_2*\frac{136.29gZnCl_2}{1molZnCl_2} =21gZnCl_2

Finally, we compute the percent yield by diving the actual yield (18 g) by the theoretical yield:

Y=\frac{18g}{21g}*100\%\\\\Y=85\%

Best regards!

4 0
3 years ago
An aqueous sodium acetate, NaC2H3O2 , solution is made by dissolving 0.395 mol NaC2H3O2 in 0.505 kg of water. Calculate the mola
liq [111]

<u>Answer:</u> The molality of NaC_2H_3O_2 solution is 0.782 m

<u>Explanation:</u>

Molality is defined as the amount of solute expressed in the number of moles present per kilogram of solvent. The units of molarity are mol/kg. The formula used to calculate molality:

\text{Molality of solution}=\frac{\text{Moles of solute}}{\text{Mass of solvent (in kg)}} .....(1)

Given values:

Moles of NaC_2H_3O_2 = 0.395 mol

Mass of solvent (water) = 0.505 kg

Putting values in equation 1, we get:

\text{Molality of }NaC_2H_3O_2=\frac{0.395mol}{0.505kg}\\\\\text{Molality of }NaC_2H_3O_2=0.782m

Hence, the molality of NaC_2H_3O_2 solution is 0.782 m

8 0
2 years ago
Three factors affecting gas pressure are the amount of gas, temperature, and __________.
Olenka [21]
It's A. volume
Pressure = \frac{moles * const * temperature}{volume}
with const depends on the chosen unit of volume
I think so... 
8 0
3 years ago
Read 2 more answers
The molar mass of glucose is 180.2 g/mol. How many grams of glucose will be produced when 132.0 g of CO2 reacts with an excess o
andriy [413]

Answer:

The mass in grams of glucose produced when 132.0 g of CO2 reacts with an excess of water is 90.1 grams

Explanation:

The chemical equation for the reaction is

6H₂O + 6CO₂  → C₆H₁₂O₆ + 6O₂

From the reaction, it is seen that 6 moles of H₂O reacts ith 6 moles of CO₂ to produce 1 mole of glucose  C₆H₁₂O₆ and 6 moles oxygen gas

The molar mass of CO₂ = 44.01 g/mol

There fpre 132.0 g contains 132.0/44.01 moles or ≅ 3 moles

However since 6 moles of CO₂ produces 1 mole of O₂, then 3 moles of CO₂ will prduce 1/6×3 or 0.5 moles of C₆H₁₂O₆

and since the molar mass (or the mass of one mole) of C₆H₁₂O₆ is 180.2 grams/mole then 0.5 mole of C₆H₁₂O₆ will have a mass of

mass of 1 mole C₆H₁₂O₆ = 180.2 g

mass of 0.5 mole C₆H₁₂O₆ = 180.2 g × 0.5 = 90.1 grams

Mass of glucose produced = 90.1 grams

7 0
3 years ago
The molecular formula of a compound is C7H14O7. what is the empirical formula​
nata0808 [166]
Empirical formula is the simplest way the molecular formula can be wrote so here 7 goes into all of these so it would be CH2O
6 0
2 years ago
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