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jekas [21]
11 months ago
15

when you combust 100.0 g 100.0 g of propane at 500. k 500. k and 1.00 atm 1.00 atm in a closed container, you expected to collec

t 279 l 279 l of carbon dioxide. instead, when you collect the gas, it measures 651 l 651 l in total. which of the following accounts for this difference?
Chemistry
1 answer:
bixtya [17]11 months ago
6 0

for the following accounts Theoretical yield was calculated incorrectly

The random error that results from measuring the masses of the reactants and the product (+ or - 0.001 g?) could cause experimental error in a yield measurement. Or it can be a deliberate mistake brought on by impurities that are still in the final product (and artificially inflating its mass). Because the real yield is frequently lower than the theoretical value, percent yield is typically lower than 100%. This may be due to incomplete or conflicting reactions or sample loss during recovery.

Learn more about Theoretical yield here:

brainly.com/question/22702379

#SPJ4

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TITRATION is the process of reaching equilibrium between acids and bases.
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<u>Answer:</u> The unknown salt is NaF

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles of salt = 0.050 moles

Volume of solution = 0.500 L

Putting values in above equation, we get:

\text{Molarity of salt}=\frac{0.050mol}{0.500L}\\\\\text{Molarity of salt}=0.1M

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pH + pOH = 14

We are given:

pH = 8.08

pOH=14-8.08=5.92

  • To calculate pOH of the solution, we use the equation:

pOH=-\log[OH^-]

Putting values in above equation, we get:

5.92=-\log[OH^-]

[OH^-]=10^{-5.92}=1.202\times 10^{-6}M

The unknown salt given are formed by the combination of weak acid and strong acid which is NaOH

The chemical equation for the hydrolysis of X^- ions follows:

                    X^-(aq.)+H_2O(l)\rightleftharpoons HX(aq.)+OH^-(aq.);K_b

<u>Initial:</u>              0.1

<u>At eqllm:</u>        0.1-x                           x              x

Concentration of OH^-=x=1.202\times 10^{-6}M

The expression of K_b for above equation follows:

K_b=\frac{[OH^-][HX]}{[X^-]}

Putting values in above expression, we get:

K_b=\frac{(1.202\times 10^{-6})\times (1.202\times 10^{-6})}{(1-(1.202\times 10^{-6}))}\\\\K_b=1.445\times 10^{-11}M

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K_w=K_b\times K_a

where,

K_w = Ionic product of water = 10^{-14}

K_a = Acid dissociation constant

K_b = Base dissociation constant = 1.445\times 10^{-11}

Putting values in above equation, we get:

10^{-14}=1.445\times 10^{-11}\times K_a\\\\K_a=\frac{10^{-14}}{1.445\times 10^{-11}}=6.92\times 10^{-4}

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K_a\text{ for HCl}=1.3\times 10^{6}

K_a\text{ for HClO}=3.0\times 10^{-8}

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