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3 years ago

Work=? time=5s Power = 40w

Chemistry

1 answer:

antiseptic1488 [7]3 years ago

5 0

Answer:

200 J

Explanation:

Work = Power x time

Work = (40w)(5s)

Work = 200 J

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2 H2S(g) ⇄ 2 H2(g) + S2(g) Kc = 9.3× 10-8 at 400ºC 0.47 moles of H2S are placed in a 3.0 L container and the system is allowed t

Anon25 [30]

Answer: The concentration of hydrogen gas at equilibrium is $1.648\times 10^{-3}M$

Explanation:

We are given:

Initial moles of hydrogen sulfide gas = 0.47 moles

Volume of the container = 3.0 L

The molarity of solution is calculated by using the equation:

$\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of the solution}}$

So, $\text{Initial molarity of hydrogen sulfide gas}=\frac{0.47}{3}=0.1567M$

The given chemical equation follows:

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

Initial: 0.1567

At eqllm: 0.1567-2x 2x x

The expression of $K_c$ for above equation follows:

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

We are given:

$K_c=9.3\times 10^{-8}$

Putting values in above equation, we get:

$9.3\times 10^{-8}=\frac{(2x)^2\times x}{(0.1567-2x)^2}\\\\x=8.24\times 10^{-4}$

So, equilibrium concentration of hydrogen gas = $2x=(2\times 8.24\times 10^{-4})=1.648\times 10^{-3}M$

Hence, the concentration of hydrogen gas at equilibrium is $1.648\times 10^{-3}M$

7 0

3 years ago

What amount of acid is esterified if a mixture consist initially of 1.0mole of ethanoic acid,1.0mole of ethanol and 1.0mole of w

ale4655 [162]

Answer:

0.54 mole

Explanation:

CH3COOH CH3CH2OH CH3COOCH2CH3 H2O

Initial concentration 1.0 mole 1.0 mole 0 mole 1.0mol

Change - x - x + x + x

Equilibrium (1.0 - x) (1.0 - x) x (1.0 + x)

K = [CH3COOCH2CH3]*[H2O]/[CH3COOH]*[CH3CH2OH]

x*(1.0+x)/(1.0-x)(1.0-x) = 4.0

x+x²=4*(1-x)²

x+x² = 4(1² - 2x + x²)

x + x² = 4 - 8x + 4x²

4 - 8x + 4x²- x² - x= 0

3x² - 9x + 4 = 0

x=2.5 , x=0.54

2.5 mole of acid cannot be esterified, because there is only 1.0 mole of acid,

so answer is 0.54 mole.

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4 years ago

0 / 5 pts Question 3 In the reaction of aluminum with iron III oxide, if you start with 54.2 grams of aluminum, how many grams o

adelina 88 [10]

Could you please mark me as brainliest?

Answer: 160.40 g Fe2O3 are needed.

Explanation:

Balanced equation: 2 Al + Fe2O3 —> Al2O3 + 2Fe

54.2 g Al * 1 mol Al / 26.98 g Al * 1 mol Fe2O3 / 2 mol Al * 159.69 g Fe2O3 / 1 mol = 160.40 g Fe2O3 are needed.

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3 years ago

Conversion factors are useful in solving problems in which a given measurement must be expressed in

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Conversion factors are useful in solving problems in which a given measurement must be expressed in some other units of measure.

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2 years ago

A 125.0 mg sample of an unknown, monoprotic acid was dissolved in 100.0 mL of distilled water and titrated with a 0.050 M soluti

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4 years ago