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hichkok12 [17]
3 years ago
12

A 10.50 gram sample of a compound is decomposed to yield 3.40 g Na, 2.37 g S, and 4.73 g O. What is the mass percentage of each

element in this compound?
Chemistry
1 answer:
9966 [12]3 years ago
6 0

Answer:

Na = 32.4% , % S = 22.6% and %O = 45.0%

Explanation:

% Na = 3.4/10.5. × 100%

= 32.4%

%S = 2.37/10.5 × 100%

= 22.6%

% O= 4.73/10.5 × 100%

= 45.0%xplanation:

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Answer:

Aristotle believes that all material substances are matter and form.

Explanation:

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Solution (A) contains 0.01 M of NH4Cl; solution
ArbitrLikvidat [17]

Answer:

NH4Cl, NaCl, Ba(OH)2, NaOH

Explanation:

NH4Cl is an acidic salt formed by the neutralization of a strong acid (HCl) with a weak base (NH3). Hence, it will habe a PH<7 (the lowest PH).

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Hence, we have

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3rd choice

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2 years ago
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For the reaction 2 H2S(g) D 2 H2 (g) + S2 (g), Kp = 1.5 × 10−5 at 800.0°C. If the initial partial pressures of H2 and S2 in a cl
Usimov [2.4K]

Answer: The approximate equilibrium partial pressure of H_2S is 3.92  atm

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

The given balanced equilibrium reaction is,

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

K_p=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}

1.5\times 10^{-5}=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}

On reversing the reaction:

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

initial pressure  4.00atm    2.00 atm       0

eqm          (4.00-2x)atm      (2.00-x) atm      2x atm

K_p=\frac{[H_2S]^2}{[H_2]^2\times [S_2]}

K_p'=\frac{1}{K_p}=0.67\times 10^5

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

0.67\times 10^5=\frac{2x]^2}{[4.00-2x]^2\times [2.00-x]}

x=1.96

[H_2S]=2x=2\times 1.96=3.92 atm

Thus approximate equilibrium partial pressure of H_2S is 3.92 atm

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Answer:

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