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lyudmila [28]
3 years ago
12

A strike slip fault is a result of

Chemistry
1 answer:
11111nata11111 [884]3 years ago
8 0
<span>I think its horizontal compression!

Plz give me Brainlyest if i got it right?

</span>
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If you need to reverse the following reaction in order for it to be an intermediate reaction in a Hess's law problem, what would
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When reversing a given reaction, we simply change the sign of the standard enthalpy change value. Therefore, the reaction will become:
H₂O → H₂ + 0.5O₂, ΔH = 286kJ
This is because if a certain amount of energy is released when a reaction occurs, the same amount of energy must be supplied for the reaction to occur in the reverse direction.
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How is it possible for helium to have 8 spectral lines when it only has 2 electrons?
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It has more lines in it compared to hydrogen emission spectrum. It is mainly because the helium atom has more electrons than a hydrogen atom. Therefore, more electrons get excited when we pass a white light beam through a helium sample, and it causes the emission of more spectral lines

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In each pair, choose the larger of the indicated quantities or state that the samples are equal:
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<u>0.4 mol of </u>O_3 molecules has larger mass than 0.4 mol of O atoms

<h3>What are molecules?</h3>

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A neutral pH level isof
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3 years ago
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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.

5 0
3 years ago
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