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3 years ago

14

A helium balloon contains 0.40 g of helium. given the molecular weight of helium is 4.0 g/mol, how many moles of helium does it

contain?

1 answer:

Rasek [7]3 years ago

8 0

Answer is 0.10 mol.

<em>Explanation;</em>

n = m/M

Where n is number of moles (mol), m is the mass of substance (g) and M is the molar mass of the substance (g/mol).

n = ?

m = 0.40 g

M = 4.0 g/mol

From substitution,

n = 0.40 g /4.0 g/mol

n = 0.10 mol

Hence, the balloon has 0.10 mol of He.

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In a first-order decomposition reaction, 50.0% of a compound decomposes in 10.5 min.

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In this reaction 50% of the compound decompose in 10.5 min thus, it is half life of the reaction and denoted by symbol $t_{1/2}$ .

(a) For first order reaction, rate constant and half life time are related to each other as follows:

$k=\frac{0.6932}{t_{1/2}}=\frac{0.6932}{10.5 min}=0.066 min^{-1}$

Thus, rate constant of the reaction is $0.066 min^{-1}$ .

(b) Rate equation for first order reaction is as follows:

$k=\frac{2.303}{t_{1/2}}log\frac{[A_{0}]}{[A_{t}]}$

now, 75% of the compound is decomposed, if initial concentration $[A_{0} ]$ is 100 then concentration at time t $[A_{t} ]$ will be 100-75=25.

Putting the values,

$0.066 min^{-1}=\frac{2.303}{t}log\frac{100}{25}=\frac{2.303}{t}(0.6020)$

On rearranging,

$t=\frac{2.303\times 0.6020}{0.066 min^{-1}}=21 min$

Thus, time required for 75% decomposition is 21 min.

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For elements in the third row of the periodic table and beyond, the octet rule is often not obeyed. a friend of yours says this

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For the equilibrium

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Answer:

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$S_2=0.002 M$

Explanation:

We are given that for the equilibrium

$2H_2S\rightleftharpoons 2H_2(g)+S_2(g)$

$k_c=9.0\times 10^{-8}$

Temperature, $T=700^{\circ}C$

Initial concentration of

$H_2S=0.30M$

$H_2=0.30 M$

$S_2=0.150 M$

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles of reactant reduced and form product

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$H_2S=0.30+2x$

$H_2=0.30-2x$

$S_2=0.150-x$

At equilibrium

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$K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}$

Substitute the values

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$9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}$

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By solving we get

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$H_2=0.30-2(0.148)=0.004 M$

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