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3 years ago

Ten pints of 15 % salt solution are mixed with 15 pints of 10 % salt solution. What is the

1 answer:

7 0

$Concentration\ rate=\frac{mass\ of\ product}{mass\ of\ substance}*100\%\\\\
15\%=\frac{m_1}{ms_1}*100\%\\ms_1=10pints\\
15\%=\frac{m_1}{10}*100\%\\
\frac{15}{100}=\frac{m_1}{10}\ \ |*10\\
m_1=\frac{150}{100}=1,5pints\\\\
10\%=\frac{m_2}{15}*100\%\\
0,1=\frac{m_2}{15}\ \ |*15\\
1,5=m_2\\
Resulting\ solution:\\
ms=ms_1+ms_2=15+10=25pints\\m=m_1+m_2=1,5+1,5=3pints
\\\frac{m}{ms}*100\%=\frac{3}{25}*100\%=12\%$

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What is the value for ∆Soreaction for the following reaction, given the standard entropy values?

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To get the ∆S of the reaction, we simply have to add the ∆S of the reactants and the ∆S of the products. Then, we get the difference between the ∆S of the products and the ∆S of the products. If the ∆S is negative, then the reaction spontaneous. If the otherwise, the reaction is not spontaneous.

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3 years ago

Which of the following reactions have a positive ΔSrxn? Check all that apply.

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Answer:

The reactions that have a positive ΔS rxn are the first and the fourth choices:

2A(g) + B(s) → 3C(g)

2A(g) + 2B(g) → 5C(g)

Explanation:

ΔS rxn is the change of entropy of the chemical reaction.

ΔS rxn = S after reaction - S before reaction.

Therefore, a positive ΔS rxn means that the entropy after the reaction is greater than the entropy before the reaction.

You may use some assumptions to predict whether a reaction will lead an increase or decrease of the entropy.

First, assume that all the non-shown conditions, such as temperature and pressure, are constant.

Under that assumption, and from the meaning of entropy as a measure of the disorder or randomness of a system you can predict the sign of the change of entropy.

2A(g) + B(s) → 3C(g)

1) The solid compounds, B(s) in this case, are very ordered and so they have low entropy.

2) Gas molecules are highly disordered (scattered), and the greater the number of molecules of the gas the larger the entropy, S).

Hence, since the product side shows 3 gas molecules and the reactant side shows 2 gas molecules and 1 solid molecule, you predict that the products have a larger entropy than the reactants, meaning an increase in entropy: ΔS rxn is positive.

2A(g) + B(g) → C(g)

Using the same reasoning, 3 gas molecules in the reactant side have more entropy than 1 molecule in the product side, and so the reaction leads to a decrease in the entropy: ΔS rxn is negative

A(g) + B(g) → C(g)

Again, 2 gas molecules in the reactant side have more entropy than 1 molecule in the product side, and so the reaction leads to a decrease in the entropy: ΔS rxn is negative

2A(g) + 2B(g) → 5C(g)

With the same reasoing, 5 molecules in the product side, lets you predict that will have more entropy than 4 molecules in the reactant side, and, the entropy will increase: ΔS rxn is positive.

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4 years ago

What is the volume of 33.25g of butane gas at 293 C and 10.934 kPa?

PSYCHO15rus [73]

Answer:

V = 240.79 L

Explanation:

Given data:

Volume of butane = ?

Temperature = 293°C

Pressure = 10.934 Kpa

Mass of butane = 33.25 g

Solution:

Number of moles of butane:

Number of moles = mass/ molar mass

Number of moles = 33.25 g/ 58.12 g/mol

Number of mole s= 0.57 mol

Now we will convert the temperature and pressure units.

293 +273 = 566 K

Pressure = 10.934/101 = 0.11 atm

Volume of butane:

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

V = nRT/P

V = 0.57 mol × 0.0821 atm.L/ mol.K ×566 K / 0.11 atm

V = 26.49 L/0.11

V = 240.79 L

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3 years ago

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So ACTUALLY an "electromotive force", but of your answer choices.

D. Electrical Field Energy

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3 years ago

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