Calculate the molarity of a solution obtained dissolving 10.0 g of cobalt(Ⅱ) bromide hexahydrate in enough water to make 450 mL
1 answer:
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A. The concentration ratio between trials 2 and 1 is: 0.3 / 0.1 = 3.
B. The ratio between reaction rates is: 0.054 / 0.006 = 9.
C. We solve the equation: (concentration ratio)^x = (reaction rate ratio):
, which gives x = 2, and the exponent in the rate law is 2.
D. The rate law will have the equation: r = k[NO2]^2
E. To solve for k, we can substitute the values for [NO2] = 0.1, rate = 0.006. This gives:
, which yields k = 0.6 L/mol-s.
F. Since the exponent is 2, this is a second-order reaction.
B. The ratio between reaction rates is: 0.054 / 0.006 = 9.
C. We solve the equation: (concentration ratio)^x = (reaction rate ratio):
D. The rate law will have the equation: r = k[NO2]^2
E. To solve for k, we can substitute the values for [NO2] = 0.1, rate = 0.006. This gives:
F. Since the exponent is 2, this is a second-order reaction.
Answer : The correct option is, (a) 345 K
Explanation :
The conversion used for the temperature from degree Celsius to Kelvin is:
where,
= temperature in Kelvin
= temperature in centigrade
As we are given the temperature in degree Celsius is, 72
Now we have to determine the temperature in Kelvin.
Therefore, the temperature in Kelvin is, 345 K
Answer: a) is the limiting reagent
b) 0.27 g of will be produced.
Explanation:
To calculate the moles :
According to stoichiometry :
3 moles of require = 2 moles of
Thus 0.003 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and
is the excess reagent.
b) As 3 moles of give = 2 moles of
Thus 0.003 moles of give =
of
Mass of
Thus 0.27 g of will be produced.