So in one hour half the amount remains (that's what half life means). In two hours 1/4 (or half of half) would remain and in three hours 1/8 would remain.
So the answer is 1/8
Molality is one way of expressing concentration of a solute in a solution. It is expressed as the mole of solute per kilogram of the solvent. To calculate for the molality of the given solution, we need to convert the mass of solute into moles and divide it to the mass of the solvent.
Molality = 29.5 g glucose (1 mol / 180.16 g ) / .950 kg water
Molality = 0.1724 mol / kg
<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)
Answer:
The solute is the substance being dissolved.
The solvent is the substance dissolving the solute.
Therefore, the salt is the solute and the water is the solvent.
Explanation:
The salt is the solute.