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3 years ago

How many moles of Chlorine gas are required to produce 5.35 moles of NaCl in this reaction?

Chemistry

1 answer:

meriva3 years ago

5 0

Answer:

2.675 moles of Chlorine gas are required

Explanation:

Chlorine gas, Cl2, reacts with NaBr to produce NaCl and Br2 as follows:

Cl2 + 2NaBr → 2NaCl + Br2

<em>Where 1 mole of chlorine gas produce 2 moles of NaCl</em>

<em />

Thus, to produce 5.35 moles of NaCl are required:

5.35mol NaCl * (1mol Cl2 / 2mol NaCl) =

<h3>2.675 moles of Chlorine gas are required</h3>

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Assume that silver and gold form ideal, random mixtures. Calculate the mass of pure Ag needed to cause an entropy increase of 20

KengaRu [80]

Answer:

$m_{Ag}=2,265.9g$

Explanation:

Hello!

In this case, since the definition of entropy in a random mixture is:

$\Delta S=-n_TR\Sigma[x_i*ln(x_i)]$

For this silver-gold mixture we write:

$\Delta S=-(n_{Au}+n_{Ag})R\Sigma[\frac{n_{Au}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Au}}{n_{Au}+n_{Ag}} )+\frac{n_{Ag}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Ag}}{n_{Au}+n_{Ag}} )]$

By knowing the moles of gold:

$n_{Au}=100g*\frac{1mol}{197g} =0.508mol$

It is possible to write the aforementioned formula in terms of the variable $x$ representing the moles of silver:

$20\frac{J}{mol}=-(0.508+x)8.314\frac{J}{mol*K} \Sigma[\frac{0.508}{0.508+x} *ln(\frac{0.508}{0.508+x} )+\frac{x}{0.508+x} *ln(\frac{x}{0.508+x} )]$

Which can be solved via Newton-Raphson or a solver software, in this case, I will provide you the answer:

$x=n_{Ag}=21.0molAg$

So the mass is:

$m_{Ag}=21.0mol*\frac{107.9g}{1mol}\\ \\m_{Ag}=2,265.9g$

Best regards!

3 0

3 years ago

A 32.8 g iron rod, initially at 22.4 C, is submerged into an unknown mass of water at 63.1 C, in an insulated container. The fin

laila [671]

Answer:

mass water = 32.4 g

Explanation:

specific heat iron = 0.450 J/g°C

specific heat water = 4.18 J/g°C

32.8 x 0.450 ( 59.1 - 22.4) + mass water x 4.18 ( 59.1- 63.1)=0

541.7 - mass water x 16.7 = 0

mass water = 32.4 g

3 0

3 years ago

If 50.0 g of formic acid (hcho2, ka = 1.8 x 10^-4) and 30.0 g of sodium formate (NaCHO2) are dissolved to make 500 ml. of soluti

gizmo_the_mogwai [7]

If 50.0 g of formic acid (HCHO2, ka = 1.8 x 10^-4) and 30.0 g of sodium formate (NaCHO2) are dissolved to make 500 ml. of solution, the ph of this solution is 3.35.

Therefore, option C is the correct option.

Given,

Given mass of sodium formate = 30 g

Given mass of formic acid = 50 g

Volume of sodium formate = 500 ml

Volume of formic acid = 500ml

Molar mass of sodium formate = 68 g

Molar mass of formic acid = 46 g

<h3>To calculate concentration of sodium formate and formic acid</h3>

The ratio of number of moles and the volume of solution is Molar concentration of substance.

Concentration of sodium formate

Cb = 30/(68×500)

= 0.00088m

Concentration of formic acid

Ca = 50/(46×500)

= 0.00217m

Now,

by using Henderson hesselbalch equation,

pH = pKa + log(Cb/Ca)

pKa = -log(1.8 × 10^(-4))

= 3.75

pH = 3.75 + log(0.00088/0.00217)

pH = 3.75 - 0.392

pH = 3.35

Thus, we calculated that the value of pH of solution of formic acid and sodium formate is 3.35.

learn more about pH:

brainly.com/question/13423434

#SPJ4

3 0

2 years ago

Which element is malleable and can conduct electricity in the solid phase?

enot [183]

None of them
1st property of nonmetals "They don't conduct electricity. Except Carbon

6 0

3 years ago

How do u say department store, chase, and market in Spanish?

Afina-wow [57]

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6 0

3 years ago

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