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3 years ago

I need help help me please

Chemistry

1 answer:

Rzqust [24]3 years ago

3 0

Answer:

C is the correct answer.

Explanation:

We can eliminate some answers immediately:

A is incorrect because nitrogen and hydrogen do not change their identity, they just combine into a new molecule that contains them. Notice how there are 2 nitrogen atoms and 6 hydrogen atoms on both sides of the equation.

B is incorrect because the first law of thermodynamics tells us that matter and energy will always be conserved, so the mass of the products must be equal to the mass of the reactants, not less than the mass of the reactants.

C is correct because balanced chemical equations help chemists predict how much product will form from certain amounts of reactants.

D is incorrect because this equation is showing the exact opposite. Molecular hydrogen and nitrogen can combine stoichiometrically to form ammonia.

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Classify each of the following properties as physical or chemical: A. Ice melts in the sun. B. Copper is a shiny metal. C. Paper

vampirchik [111]

Answer:

The answer to your question is below

Explanation:

Physical properties are the properties in which there is no change in the nature of the substance but there is a change in the physical state.

Chemical properties are the properties in which the substance changes its nature. A new substance is formed.

A. Ice melts in the sun. Physical p.

B. Copper is a shiny metal. Physical p.

C. Paper can burn. Chemical p.

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3 years ago

The acid-dissociation constant for benzoic acid (C6H5COOH) is 6.3×10−5. Calculate the equilibrium concentration of H3O+ in the s

raketka [301]

Answer : The equilibrium concentration of $H_3O^+$ in the solution is, $2.1\times 10^{-3}M$

Explanation :

The dissociation of acid reaction is:

$C_6H_5COOH+H_2O\rightarrow H_3O^++C_6H_5COO^-$

Initial conc. c 0 0

At eqm. c-x x x

Given:

c = $7.0\times 10^{-2}M$

$K_a=6.3\times 10^{-5}$

The expression of dissociation constant of acid is:

$K_a=\frac{[H_3O^+][C_6H_5COO^-]}{[C_6H_5COOH]}$

$K_a=\frac{(x)\times (x)}{(c-x)}$

Now put all the given values in this expression, we get:

$6.3\times 10^{-5}=\frac{(x)\times (x)}{[(7.0\times 10^{-2})-x]}$

$x=2.1\times 10^{-3}M$

Thus, the equilibrium concentration of $H_3O^+$ in the solution is, $2.1\times 10^{-3}M$

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2 years ago

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Dennis_Churaev [7]

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2 years ago

You have studied the gas-phase oxidation of HBr by O2: 4 HBr(g) + O2(g) → 2 H2O(g) + 2 Br2(g) You find the reaction to be first

Vlada [557]

Answer:

Explanation:

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