All categories

3 years ago

The rate of disappearance of HBr in the gas phase reaction 2HBr(

1 answer:

3 0

The correct answer would be 0.1505 M/s. Given the rate of disapperance of HBr, we can easily calculate the rate of apearance of the products by looking at the coefficients of the substances in the reaction. In this case, for every two moles of HBr, 1 mole of Br2 is being produced so the reaction rate would be:

- 1/2 (r(HBr)) = rBr2
- 1/2 (-0.301) = rBr2
Rate of appearance of Br2 = 0.1505 M/s

You might be interested in

7. Write the equation for the positron emission of barium-127.

ziro4ka [17]

The reaction is given by

$\\ \rm\Rrightarrow {}^{127}_{56}Ba\longrightarrow {}^{0}_{+1}\beta+{}^{127}_{55}Cs$

Barium goes underneath beta decay to form Ceaseum

Cs is very mellable element
It can melt on your hand

8 0

2 years ago

Which statement describes the blood type of a person with the alleles Ai?

lyudmila [28]

Answer:

d i dont now but im sure my answer is d im not good to write english word

5 0

3 years ago

If a certain gas occupies a volume of 10. L when the applied pressure is 5.0 atm , find the pressure when the gas occupies a vol

lorasvet [3.4K]

Answer:

20L is the new volume

Explanation:

In this case, moles and T° from the gas remain constant. This is the formula we must apply, to solve this:

P₁ . V₁ = P₂ . V₂

5 atm . 10 L = P₂ . 2.5L

P₂ = (5 atm . 10 L) / 2.5L →20L

5 0

3 years ago

The activation energy of a certain uncatalyzed reaction is 64 kJ/mol. In the presence of a catalyst, the Ea is 55 kJ/mol. How ma

Ksivusya [100]

Answer:

About 5 times faster.

Explanation:

Hello,

In this case, since the Arrhenius equation is considered for both the catalyzed reaction (1) and the uncatalized reaction (2), one determines the relationship between them as follows:

$\frac{k_1}{k_2}=\frac{Aexp(-\frac{Ea_1}{RT} )}{Aexp(-\frac{Ea_2}{RT})} \\\frac{k_1}{k_2}=\frac{exp(-\frac{Ea_1}{RT} )}{exp(-\frac{Ea_2}{RT})}$

By replacing the corresponding values we obtain:

$\frac{k_1}{k_2}=\frac{exp(-\frac{55000J/mol}{8.314J/molK*673.15K} )}{exp(-\frac{64000J/mol}{8.314J/molK*673.15K} )} =4.8$

Such result means that the catalyzed reaction is about five times faster than the uncatalyzed reaction.

Best regards.

4 0

3 years ago

Scattered.<br> The particles. <br> are not

exis [7]

the change in the direction of motion of a particle because of a collision with another particle. As defined in physics, a collision can occur between particles that repel one another, such as two positive (or negative) ions, and need not involve direct physical contact of the particles

5 0

3 years ago