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3 years ago

3. The speed of a reaction can be increased by increasing reactant concentration or decreasing

Chemistry

2 answers:

Sloan [31]3 years ago

5 0

Answer:

true

because with the both states we increase the surface of reaction

Ksivusya [100]3 years ago

4 0

It would probably be true

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State Hess' law of constant heat summation.

AveGali [126]

Answer:

-74.6 kj/mol

Explanation:

you can see the answer at the pic

8 0

2 years ago

Lime is added to a lake to neutralize the effects of acid rain. The pH value of the

OverLord2011 [107]

Answer:

The change in concentration of H+ ions is -9.99×10⁻⁵

Explanation:

When pH rises, we talk about neutralization and, there is a decline in proton's concentration.

pH lower than 7 is acid

pH higher than 7 is basic

Protons are neutralized by hydroxides to make water as this:

H₃O⁺ + OH⁻ ⇄ H₂O

pH = - log [H₃O⁺]

[H₃O⁺] = 10^-pH

In conclussion: [H₃O⁺]₁ = 10⁻⁴ → 1×10⁻⁴

[H₃O⁺]₂ = 10⁻⁷ → 1×10⁻⁷

1×10⁻⁷- 1×10⁻⁴

The change in concentration of H+ ions is -9.99×10⁻⁵

4 0

3 years ago

If the rate of diffusion of H2S is about one Fourth times faster than that of SO2 by what factors is the rate of diffusion of am

Rudiy27

Explanation:

rate of diffusion of H2S is about one Fourth times faster than that of SO2 by what factors is the rate of diffusion of ammonia faster than that

8 0

2 years ago

Ammonia is formed according to the reaction below. A chemist mixes 21 grams of nitrogen gas and 18 grams of hydrogen gas in a 2.

Rama09 [41]

Answer : The mass of hydrogen gas consumed will be, 4.5 grams

Explanation : Given,

Mass of $N_2$ = 21 g

Mass of $H_2$ = 18 g

Molar mass of $N_2$ = 28 g/mole

Molar mass of $H_2$ = 2 g/mole

First we have to calculate the moles of $N_2$ and $H_2$ .

$\text{Moles of }N_2=\frac{\text{Mass of }N_2}{\text{Molar mass of }N_2}=\frac{21g}{28g/mole}=0.75moles$

$\text{Moles of }H_2=\frac{\text{Mass of }H_2}{\text{Molar mass of }H_2}=\frac{18g}{2g/mole}=9moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

From the given balanced reaction, we conclude that

As, 1 mole of $N_2$ react with 3 moles of $H_2$

So, 0.75 moles of $N_2$ react with $3\times 0.75=2.25$ moles of $H_2$

From this we conclude that, $H_2$ is an excess reagent because the given moles are greater than the required moles and $N_2$ is a limiting reagent because it limits the formation of product.

The moles of hydrogen gas consumed = 2.25 mole

Now we have to calculate the mass of hydrogen gas consumed.

$\text{Mass of }H_2=\text{Moles of }H_2\times \text{Molar mass of }H_2$

$\text{Mass of }H_2=(2.25mole)\times (2g/mole)=4.5g$

Therefore, the mass of hydrogen gas consumed will be, 4.5 grams

5 0

3 years ago

For a titration Maria added 38.4 mL of 0.250 M H2SO4 to 23.5

Novosadov [1.4K]

Answer:

0.409 m

Explanation:

6 0

3 years ago