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Ipatiy [6.2K]
2 years ago
11

Which statement best describes the law of conservation of mass? (1 point)

Chemistry
1 answer:
lisov135 [29]2 years ago
6 0

Matter cannot be created or destroyed in a chemical reaction best describe law of conservation of mass.

<h3>What is law of conservation of mass?</h3>

The law of conservation of mass states that for any system closed Mass can neither be created not destroy. the mass of the system must remain constant over time,.

Therefore, Matter cannot be created or destroyed in a chemical reaction best describe law of conservation of mass.

Learn more about law of conservation of mass here.

brainly.com/question/15289631

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How many moles of oxygen are in 1 mole of fe(no3)3?
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One mole of Fe(NO3)3, or iron(III) nitrate, has three moles of nitrate molecules, which have three moles of oxygen atoms each. We can show this mathematically:
1 mole Fe(NO3)3 * (3 moles NO3)/(1 mole Fe(NO3)3) = 3 moles NO3
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9 moles of Oxygen in one mole Fe(NO3)3
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1. D)
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How does the volume of water change the solubility of sodium chloride in simple words?
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The reaction C 4 H 8 ( g ) ⟶ 2 C 2 H 4 ( g ) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ / mol. 262 kJ/mol. At 600.0 K,
ludmilkaskok [199]

Answer: 4.3\times 10^{-13}s^{-1}

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 600.0K = 6.1\times 10^{-8}s^{-1}

K_2 = rate constant at 775.0 = ?

Ea = activation energy for the reaction = 262 kJ/mol = 262000J/mol

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 600.0K

T_2 = final temperature = 775.0K

Now put all the given values in this formula, we get

\log (\frac{6.1\times 10^{-8}}{K_2})=\frac{262000}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{775.0K}]

\log (\frac{6.1\times 10^{-8}s^}{K_2})=5.150

(\frac{6.1\times 10^{-8}}{K_2})=141253.8

Therefore, the value of the rate constant at 775.0 K is 4.3\times 10^{-13}s^{-1}

5 0
2 years ago
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