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Natasha2012 [34]
3 years ago
7

What are the units of k in the following rate law?

Chemistry
1 answer:
Ludmilka [50]3 years ago
8 0

Answer:

B. \frac{1}{M^{2} s }

Explanation:

The unit for rate is M/s while the unit for each molecule should be M. You can find the unit for k by putting the units for rate and the molecules into the equation

rate= k{X][Y]

M/s= k * M^{2} * M^{1}

k= (M/s) / (M^{3})

k= \frac{1}{M^{2} s }

You can also use this predetermined formula to solve this problem faster: k= \frac{M^{1-n} }{s }

Where n is the number of molecule. There are 3 molecule(2X and 1Y) so n=3, so

k= \frac{M^{1-n} }{s }

k= \frac{M^{1-3} }{s }= \frac{m^{-2}}{s}= \frac{1}{M^{2} s }

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Given the Henry’s law constant for O2(4.34*109Pa) at 25° C, calculate the molar concentration of oxygen in air-saturated and O2s
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This is an incomplete question, here is a complete question.

The Henry's law constant for oxygen dissolved in water is 4.34 × 10⁹ g/L.Pa at 25⁰C.If the partial pressure of oxygen in air is 0.2 atm, under atmospheric conditions, calculate the molar concentration of oxygen in air-saturated and oxygen saturated water.

Answer : The molar concentration of oxygen is, 2.67\times 10^2mol/L

Explanation :

As we know that,

C_{O_2}=k_H\times p_{O_2}

where,

C_{O_2} = molar solubility of O_2 = ?

p_{O_2} = partial pressure of O_2 = 0.2 atm  = 1.97×10⁻⁶ Pa

k_H = Henry's law constant  = 4.34 × 10⁹ g/L.Pa

Now put all the given values in the above formula, we get:

C_{O_2}=(4.34\times 10^9g/L.Pa)\times (1.97\times 10^{-6}Pa)

C_{O_2}=8.55\times 10^3g/L

Now we have to molar concentration of oxygen.

Molar concentration of oxygen = \frac{8.55\times 10^3g/L}{32g/mol}=2.67\times 10^2mol/L

Therefore, the molar concentration of oxygen is, 2.67\times 10^2mol/L

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Answer:

True

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Neutral solutions maintain a pH of 7. Water and human blood are great examples of neutral solutions. Acids mixed with bases can be neutralized and given a pH of 7.

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