Answer:
b, decrease in movement of the molecules
Explanation:
removing the energy will begin making the molecules lock up and stop moving due to the loss of energy.
hope this helped
50 grams or 50,000 mili grams is the mass of solute in 1000 grams of a solution having a concentration of 5 parts per million.
Explanation:
Total mass of solution = 1000 grams or 1000 ml since 1 gram = 1 ml
concentration is 5 parts per million ( 5 mg in 1000 ml solution or 0.005 gram in 1000 ml)
the formula used for parts per million:
parts per million = 
putting the values in the equation:
parts per million = 
0.005 x 1000 = mass of solute
50 grams= mass of solute
converting this into mg
50,000 mg. is the total mass of solute in 5ppm of 1000 ml solution.
Answer:
(C) through the atmosphere
Explanation:
<u>Answer:</u> The rate law of the reaction is ![\text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BHgCl_2%5D%5BC_2O_4%5E%7B2-%7D%5D%5E2)
<u>Explanation:</u>
Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.
For the given chemical equation:

Rate law expression for the reaction:
![\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BHgCl_2%5D%5Ea%5BC_2O_4%5E%7B2-%7D%5D%5Eb)
where,
a = order with respect to 
b = order with respect to 
Expression for rate law for first observation:
....(1)
Expression for rate law for second observation:
....(2)
Expression for rate law for third observation:
....(3)
Expression for rate law for fourth observation:
....(4)
Dividing 2 from 1, we get:

Dividing 2 from 3, we get:

Thus, the rate law becomes:
![\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BHgCl_2%5D%5E1%5BC_2O_4%5E%7B2-%7D%5D%5E2)