Well there are 1,000 milligrams in 1 gram so 9.815 grams of sugar is equivalent to 9,815 milligrams of sugar. So a bar of milk chocolate contains 9,815 milligrams of sugar.
Total atoms is 9 ( 2 carbon atoms, 5 hydrogen atoms, 1 oxygen atom and 1 hydrogen atom = 9 atoms)
Element is 3 ( Carbon, Hydrogen and Oxygen)
The gas has already created enough pressure to become a gas, which is the most expandable it could turn in to. Thats what i think, hope it helps.
A chemist (from Greek chēm (ía) alchemy; replacing chemist from Medieval Latin alchimista[1]) is a scientist trained in the study of chemistry. Chemists study the composition of matter and its properties. Chemists carefully describe the properties they study in terms of quantities, with detail on the level of molecules and their component atoms. Chemists carefully measure substance proportions, reaction rates, and other chemical properties. The word 'chemist' is also used to address Pharmacists in Commonwealth English.
<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)
