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3 years ago

7

An experiment produced 0.10 g CO2 with a volume of 0.056 L at STP. If the accepted density of CO2 at STP is 1.96 g/L, what is th

e approximate percent error?

1 answer:

Mandarinka [93]3 years ago

3 0

The experimental density of CO2 at STP is 0.10/0.056=1.78 g/L. The percent error equals to (1.96-1.78)/1.96*100%=9.18%. So the answer is 9.18%.

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Usually potassium hydrogen phthalate is kept very pure. But Stu Dent thinks the bottle of potassium hydrogen phthalate has been

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Answer:

1.784 g

Explanation:

The equation of the reaction is;

NaOH(aq) + KHC8H4O4(aq) --------> KNaC8H4O4(aq) + H2O(l)

Number of moles of NaOH reacted = 17.47/1000 * 0.5000 M

Number of moles of NaOH reacted =8.735 * 10^-3 moles

From the reaction equation;

1 mole of NaOH reacted with 1 mole of KHC8H4O4

Hence, 8.735 * 10^-3 moles of NaOH reacts with 8.735 * 10^-3 moles of KHP.

So,

Mass of KHP reacted = 8.735 * 10^-3 moles * 204.2 g/mol = 1.784 g

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Answer:

c.- How much of the reactants are needed and how much product will made.

Explanation:

The moles is the matter unit used in chemistry to simplify some calculations, instead of using grams. Also the moles are very useful because the chemical reaction can be balanced.

When a Chemical reaction is balanced, then it can be easily to calculate how many moles are necessary to add in a process to obtain a quantity of grams of a product.

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2NOCl(g) ↔ 2NO(g) + Cl2(g) with K = 1.6 x10–5. In an experiment, 1.00 mole of pure NOCl and 1.00 mole of pure Cl2 are placed in

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Initial concentration of $Cl_2$ = 1 M

The given balanced equilibrium reaction is,

$2NOCl(g)\rightleftharpoons 2NO(g)+Cl_2(g)$

Initial conc. 1 M 0M 1 M

At eqm. conc. (1-2x) M (2x) M (1+x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[NO]^2[Cl_2]}{[NOCl]^2}$

The $K_c$ = $1.6\times 10^{-5}}$

Now put all the given values in this expression, we get :

${1.6\times 10^{-5}}=\frac{(2x)^2\times (1+x)}{(1-2x)^2}$

By solving the term 'x', we get :

$x=0.0019$

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2 years ago