Answer:
Becomes 1/4 as much
Explanation:
If a and b are directly proportional, this means we may write an equality between them if some proportionality constant, k, is included, so that:
The equation shows that:
And since k is a constant, this would imply that increasing a by some fraction would also require b to increase by the same fraction.
In the context of this problem, we're given that a becomes 1/4 as much, meaning b should also become 1/4 as much to keep the same proportionality constant, k.
Answer:
Explanation:
Hello,
The law of mass action, allows us to know the required amounts, thus, for this chemical reaction it is:
![\frac{1}{-3} \frac{d[D]}{dt} =\frac{1}{-1} \frac{d[E]}{dt} =\frac{1}{-2} \frac{d[F]}{dt} =\frac{1}{5} \frac{d[G]}{dt} =\frac{1}{4} \frac{d[H]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B-3%7D%20%5Cfrac%7Bd%5BD%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B-1%7D%20%5Cfrac%7Bd%5BE%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B-2%7D%20%5Cfrac%7Bd%5BF%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B5%7D%20%5Cfrac%7Bd%5BG%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B4%7D%20%5Cfrac%7Bd%5BH%5D%7D%7Bdt%7D)
Now, we answer:
(a)
![\frac{d[H]}{dt}=4*\frac{1}{-3} *(-0.12M/s)=0.16M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BH%5D%7D%7Bdt%7D%3D4%2A%5Cfrac%7B1%7D%7B-3%7D%20%2A%28-0.12M%2Fs%29%3D0.16M%2Fs)
(b)
![\frac{d[E]}{dt}=-1*\frac{1}{5} *(0.2M/s)=-0.04M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BE%5D%7D%7Bdt%7D%3D-1%2A%5Cfrac%7B1%7D%7B5%7D%20%2A%280.2M%2Fs%29%3D-0.04M%2Fs)
(c) Since no initial data is specified, we could establish the rate of the reaction as based of the law of mass action:
![r=\frac{1}{-3} \frac{d[D]}{dt} =\frac{1}{-1} \frac{d[E]}{dt} =\frac{1}{-2} \frac{d[F]}{dt} =\frac{1}{5} \frac{d[G]}{dt} =\frac{1}{4} \frac{d[H]}{dt}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B1%7D%7B-3%7D%20%5Cfrac%7Bd%5BD%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B-1%7D%20%5Cfrac%7Bd%5BE%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B-2%7D%20%5Cfrac%7Bd%5BF%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B5%7D%20%5Cfrac%7Bd%5BG%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B4%7D%20%5Cfrac%7Bd%5BH%5D%7D%7Bdt%7D)
Thus, any of the available expressions are suitable to quantify the rate of the reaction.
Best regards.
Answer:
(b) Calculate the molarity of a solution of 4.8 mole of HCl in 600 mL of solution. ... (g) Calculate the mass of Na2CO3 that must be used to make 700 mL of a 0.136 M Na2CO3 ... (h) What mass of NaOH is needed to make 200 mL of a 0.300 M NaOH solution? ... However, when we are reacting solutions we have to convert.
Explanation:
No because it changes its composition to ozone
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Hope this helps :))
