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Fittoniya [83]
3 years ago
13

A chemist dissolves 699. mg of pure hydrobromic acid in enough water to make up 180. mL of solution. Calculate the pH of the sol

ution. Be sure your answer has the correct number of significant digits.

Chemistry
1 answer:
Sedaia [141]3 years ago
8 0

Answer: 1.32

Explanation:

First, we must obtain the molar mass of HBr. After that, we try to obtain the concentration of the hydrobromic acid from the formula n=CV since the volume of solution and mass of acid was provided. Recall that n=m/M. If the concentration of acid is thus obtained, we make use of the fact that the concentration of H+ in the acid is equal to the molar concentration of HBr to obtain the pH. The pH is the negative logarithm of the concentration we obtained in the initial step.

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<u>Answer:</u>

<u>For A:</u> The equation is _{92}^{238}\textrm{U}\rightarrow _{90}^{234}\textrm{Th}+_2^4\alpha

<u>For B:</u> The equation is _{94}^{239}\textrm{Pu}\rightarrow _{92}^{235}\textrm{U}+_2^4\alpha

<u>For C:</u> The equation is _{90}^{239}\textrm{Th}\rightarrow _{91}^{235}\textrm{Pa}+_{-1}^0\beta

<u>Explanation:</u>

Alpha decay process is the process in which nucleus of an atom disintegrates into two particles. The first one which is the alpha particle consists of two protons and two neutrons. This is also known as helium nucleus. The second particle is the daughter nuclei which is the original nucleus minus the alpha particle released.

_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha

Beta decay process is defined as the process the neutrons get converted into an electron and a proton. The released electron is known as the beta particle. In this process, the atomic number of the daughter nuclei gets increased by a factor of 1 but the mass number remains the same.

_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta

<u>For A:</u> Uranium-238 emits an alpha particle

The nuclear equation for this process follows:

_{92}^{238}\textrm{U}\rightarrow _{90}^{234}\textrm{Th}+_2^4\alpha

<u>For B:</u> Plutonium-239 emits an alpha particle

The nuclear equation for this process follows:

_{94}^{239}\textrm{Pu}\rightarrow _{92}^{235}\textrm{U}+_2^4\alpha

<u>For C:</u> Thorium-239 emits a beta particle

The nuclear equation for this process follows:

_{90}^{239}\textrm{Th}\rightarrow _{91}^{235}\textrm{Pa}+_{-1}^0\beta

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