Chemistry, 09.04.2020 22:12 emmmmmmaaaa

Suppose 4.61 g of zinc was allowed to react with hydrochloric acid to produce zinc chloride and
hydrogen gas. How much zinc chloride should you get? (Expected)
Suppose that you actually recovered 8.56 g of zinc chloride. What is your percent yield?

Answers: 3

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Answer from: Quest

when [oh-] is 10-5 m, and [h3o+] is 10-9 m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.

Answer from: Quest

answer : (a) volume of the gas = 8.67 × $10^{-3}$ l

(b) second virial coefficient b = -1.21226 l/mole

solution : given, t = 300k

p = 20atm

z = 0.86

n = 8.2 mmoles = 8.2 × $10^{-3}$ moles

r = 0.082 l atm/k/mol

find, (a) volume of gas and (b) second virial coefficient b at 300k.

formula used : $z=\frac{pv}{nrt}$

and $z=1+\frac{b}{v}$

(a) $z=\frac{pv}{nrt}$

$v=\frac{znrt}{p}$

= $\frac{0.86\times 8.2\times[tex]10^{-3}$ \times 0.082\times 300}{20}[/tex]

= 8.67 × $10^{-3}$ l

(b) $z=1+\frac{b}{v}$

$b=(z-1)v$

= (0.00867-1) × 8.659

= - 8.5839 l/mole

Answer from: Quest

answer///carbon;

Answer from: Quest

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 : $4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

equation 2 : $2x (s) + 3cl_{2} (g) \rightarrow 2xcl_{3} (s) h rxn = -800 kj/mol$

equation 3 : $4xcl_{3} (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) + 6cl_(2} h rxn = -200 kj/mol$

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 : $4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+800) \times 2 kj/mol$

let us add equation 1 & 4

$4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

$4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+1600) kj/mol$

$4xcl_{3} )s) + 3o_{2} \rightarrow 2x_{2}o_{3} (s) + 6cl_{2} (g) h rxn = (+1000) kj/mol$

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

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Suppose 4.61 g of zinc was allowed to react with hydrochloric acid to produce zinc chloride and