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3 years ago

1. How many moles of oxygen gas are required to react with 15.56 moles of hydrogen gas? (2 pts.)

Chemistry

1 answer:

loris [4]3 years ago

7 0

Answer:

7.78moles of O2

Explanation:

2H2 (g) + O2 (g) → 2H2O (g)

From the balanced equation,

2 moles of H2 required 1mole of O2.

Therefore, 15.56 moles of H2 will require = 15.56/2 = 7.78moles of O2

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Describe the process by which Ag+ ions are precipitated out of solution. 4. In your testing, several precipitates are formed, and then dissolved as complexes.

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Given the following information, what is the concentration of H2O(g) at equilibrium? [H2S](eq) = 0.671 M [O2](eq) = 0.587 M Kc =

MAVERICK [17]

Answer: The equilibrium concentration of water is 0.597 M

Explanation:

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{c}$

For a general chemical reaction:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_{eq}$ is written as:

$K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

The concentration of pure solids and pure liquids are taken as 1 in the expression.

For the given chemical reaction:

$2H_2S(g)+O_2(g)\rightleftharpoons 2S(s)+2H_2O(g)$

The expression of $K_c$ for above equation is:

$K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}$

We are given:

$[H_2S]_{eq}=0.671M$

$[O_2]_{eq}=0.587M$

$K_c=1.35$

Putting values in above expression, we get:

$1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}$

$[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M$

Hence, the equilibrium concentration of water is 0.597 M

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3 years ago

Calculate the ph of a solution formed by mixing 200.0 ml of 0.30 m hclo with 300.0 ml of 0.20 m kclo. the ka for hclo is 2.9 × 1

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Answer:

The pH of the solution will be 7.53.

Explanation:

Dissociation constant of KClO= $K_a=2.8\times 10^{-8}$

Concentration of acid in 1 l= 0.30 M

Then in 200 ml = $0.30 M\times 0.200 L=0.06 M$

The concentration of acid, HClO=[acid]= 0.006 M

Concentration of salt in 1 L = 0.20 M

Then in 300 ml = $0.20 M\times 0.300 L=0.06 M$

The concentration of acid, KClO=[salt]= 0.006 M

The pH of the solution will be given by formula :

$pH=pK_{a}^o+\log\frac{[salt]}{[acid]}$

$pH=-\log[2.8\times 10^{-8}]+\frac{[0.06 M]}{[0.06 M]}$

The pH of the solution will be 7.53.

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3 years ago

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Strontium-90 is a radioisotope that will decrease in mass by one-half every 29 years. How many years will it take for a 10.0-gra

Lelechka [254]

Answer: It will take 29 years for a 10.0-gram sample of strontium-90 to decay to 5.00 grams

Explanation:

Radioactive decay process is a type of process in which a less stable nuclei decomposes to a stable nuclei by releasing some radiations or particles like alpha, beta particles or gamma-radiations. The radioactive decay follows first order kinetics.

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

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$t_{\frac{1}{2}=\frac{0.693}{\lambda}$

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3 years ago

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professor190 [17]

Answer:

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