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goldenfox [79]
3 years ago
14

Which of the following explains the increase in reaction rate caused by raising the temperature or raising the pressure of a sys

tem?
a) surface area theory
b) activation energy theory
c) potential energy theory
d) collision theory
Chemistry
1 answer:
ddd [48]3 years ago
4 0

I think its B

~sorry if its wrong

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When a mixture of oxygen and hydrogen is ignited in a test tube, the
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The extend of sewage treatment required mostly depends on the (A) Control of suspended solid (B) BOD of the sewage (C) Both a an
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The extent of sewage treatment required mostly depends on the control of suspended solids and BOD of the sewage. Hence, option C is correct.

<h3>What is Inorganic matter?</h3>

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Hence, option C is correct.

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8 0
2 years ago
What volume of 0.152 M KMnO4 solution would completely react with 20.0 mL of 0.381 M FeSO4 solution according to the following n
ANEK [815]

<u>Answer:</u> The volume of permanganate ion (potassium permanganate) is 10.0 mL

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     .....(1)

Molarity of ferrous sulfate solution = 0.381 M

Volume of solution = 20.0 mL = 0.020 L   (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

0.381M=\frac{\text{Moles of ferrous sulfate}}{0.020L}\\\\\text{Moles of ferrous sulfate}=(0.381mol/L\times 0.020L)=0.00762mol

For the given chemical equation:

5Fe^{2+}+8H^++MnO_4^-\rightarrow 5Fe^{3+}+Mn^{2+}+4H_2O

By Stoichiometry of the reaction:

5 moles of iron (II) ions (ferrous sulfate) reacts with 1 mole of permanganate ion (potassium permanganate)

So, 0.00762 moles of iron (II) ions (ferrous sulfate) will react with = \frac{1}{5}\times 0.00762=0.00152mol of permanganate ion (potassium permanganate)

Now, calculating the volume of permanganate ion (potassium permanganate) by using equation 1, we get:

Molarity of permanganate ion (potassium permanganate) = 0.152 M

Moles of permanganate ion (potassium permanganate) = 0.00152 mol

Putting values in equation 1, we get:

0.152mol/L=\frac{0.00152mol}{\text{Volume of permanganate ion (potassium permanganate)}}\\\\\text{Volume of permanganate ion (potassium permanganate)}=\frac{0.00152mol}{0.152mol/L}=0.01L=10.0mL

Hence, the volume of permanganate ion (potassium permanganate) is 10.0 mL

3 0
3 years ago
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