How many moles of liquid water must freeze to remove 100 kJ of heat? (ΔHf = –334 J/ g.

1 answer:

5 0

First figure out how many grams must freeze and then convert the grams to moles.
Hf = -334 J/g. Convert this to KJ/g by dividing by 1000. (There are 1000 Joules in a kJ). 
Hf = -334 J/g ÷ 1000 J/kj = -0.334 kJ/g 
Now, divide 100 kJ by -0.334 kJ/g (see how the units are lining up?) 
100 kJ ÷ -0.334 kJ/g = 299 g 
Now convert this to moles by dividing by the molecular weight of water (18.0g/mole). 
299 ÷ 18.0 = 16.6 moles

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The reaction of NO2 with ozone produces NO3 in a second-order reaction overall.

Brilliant_brown [7]

Answer : The rate of reaction is,

$Rate=4.77\times 10^{-19}M/s$

The appearance of $NO_3$ is, $4.77\times 10^{-19}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$NO_2(g)+O_3(g)\rightarrow NO_3(g)+O_2(g)$