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3 years ago

11

I have these answered, but I'd like someone to check them for me? Plz and thanks :)

1 answer:

zlopas [31]3 years ago

4 0

1. The third option is the least soluble in water because it is the chain with the most number of hydrocarbons. Next is the second option while the first one is the most soluble.
2. Statements 1 and 2 are true. The third option is not true all the time because it depends on the structure of the compound.

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How many grams of Cl2 are consumed to produce 12.0 g of KCl?

Alla [95]

The reaction is:

Cl2 + 2 KBr --> 2 KCl + Br2

Moles of KCl is

n = m /M = 12 /74 = 0.16 mol

As, twice the moles of KCl is producing from 1 mol of chlorine

mole of Cl2 = 0.16 /2 = 0.08 mol

Mass of Cl2

m /70 = 0.08 = 5.6 g

Hence, 5.6 g mol Cl2 consumed to produce KCl

7 0

3 years ago

Why is it important to keep the two sides of an equation balanced when solving? what other properties do we use to rewrite expre

kompoz [17]

Answer:

<u>Why is it important to keep the two sides of an equation balanced when solving?</u>

If two expressions are equal to each other, and you add the same value to both sides of the equation, the equation will remain equal. When you solve an equation, you find the value of the variable that makes the equation true.

<u>What other properties do we use to rewrite expressions and equations?</u>

State of matter

Please vote for Brainliest and I hope this helps!

4 0

2 years ago

Please help this is my chemistry work

Alja [10]

It’s deceleration hope that helps!

5 0

3 years ago

Read 2 more answers

A runner wants to run 11.7 km. She knows that her running pace is 6.7 mph. How many minutes must she run?

Alona [7]

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3 years ago

A gas mixture with 4 mol of Ar, x moles of Ne, and y moles

maks197457 [2]

Answer:

a) $\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

b) $x=4$

c) $\Delta G_{max}=-2721.9 J/mol$

Explanation:

Gas mixture:

$n_{Ar}= 4 mol$

$n_{Ne}= x mol$

$n_{Xe}= y mol$

$n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}$

$n_{Ne} + n_{Xe}=2*n_{Ar}$

$x + y=8 mol$

$y=8 mol- x$

Mol fractions:

$x_{Ar}=\frac{4 mol}{12 mol}=1/3$

$x_{Ne}=\frac{x mol}{12 mol}=x/12$

$x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12$

Expression of $\Delta G_{mixing}$

$\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)$

$\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]$

$\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

Expression of $\Delta G_{max}$

$\frac{d \Delta G_{mixing}}{dx}=0$

$\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]$

$0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)$

$0=[ln (x/12)-ln ((8-x)/12)$

$ln (x/12)=ln ((8-x)/12)$

$x=(8-x)$

$x=4$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]$

$\Delta G_{max}=-2721.9 J/mol$

4 0

3 years ago