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fiasKO [112]
3 years ago
5

What happens to the total amount of matter and energy during a chemical reaction ?

Chemistry
1 answer:
larisa86 [58]3 years ago
7 0
It would be conserved
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Carbon has 4 valence electrons, oxygen has 6
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Calculate the ph of a solution at 25. 0 °C that contains 2. 95 × 10^-12 m hydronium ions.
Makovka662 [10]

The pH of a solution at 25. 0 °C that contains 2. 95 × 10^-12 m hydronium ions is 13.5.

<h3>What is pH? </h3>

pH is defined as the concentration of the hydrogen bond which is released or gained by the species in the solution which depicts the acidity and basicity of the solution.

<h3>What is pOH? </h3>

pOH is defined as the concentration of the hydronium ion present in solution.

pOH value is inversely proportional to the value of pH.

pH value increases, pOH value decreases and vice versa.

Given,

Total H+ ions = 2.95 ×10^(-12)M

<h3>Calculation of pH</h3>

pH = -log[H+]

By substituting the value of H+ ion in given equation

= log(2.95× 10^(-12) )

= 13.5

Thus we find that the pH of a solution at 25. 0 °C that contains 2. 95 × 10^-12 m hydronium ions is 13.5.

learn more about pH:

brainly.com/question/12942138

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8 0
1 year ago
First-order reaction that results in the destruction of a pollutant has a rate constant of 0.l/day. (a) how many days will it ta
Klio2033 [76]

Rate equation for first order reaction is as follows:

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}

Here, k is rate constant of the reaction, t is time of the reaction, A_{0} is initial concentration and A_{t} is concentration at time t.

The rate constant of the reaction is 0.1 day^{-1}.

(a) Let the initial concentration be 100, If 90% of the chemical is destroyed, the chemical present at time t will be 100-90=10, on putting the values,

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{10}=23.03 days

Thus, time required to destroy 90% of the chemical is 23.03 days.

(b) Let the initial concentration be 100, If 99% of the chemical is destroyed, the chemical present at time t will be 100-99=1, on putting the values,

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{1}=46.06 days

Thus, time required to destroy 99% of the chemical is 46.06 days.

(c)  Let the initial concentration be 100, If 99.9% of the chemical is destroyed, the chemical present at time t will be 100-99.9=0.1, on putting the values,

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{0.1}=69.09 days

Thus, time required to destroy 99.9% of the chemical is 69.09 days.

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2 years ago
When are kids ready to assume adult responsibilities?
Serga [27]

Answer:

18 years old or when they become mature.

Explanation:

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cual es la moralidad del etanol CH3CH2OH en un vino que contine 11% alcohol si su densidad es de 0.7899 g/cm3
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Explanation:

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