Chemical compounds have chemically bonded molecules so that they exhibit different properties (i.e. chemical) compared to the individual molecules comprising the compound. Mixtures are simply the combinations of different molecules and compounds that are not chemically bonded together, and can therefore be separated by physical means. Mixtures usually retain the properties of its components. The hydrogen and oxygen molecules in a mixture do not form strong bonds between each other. The molecules of both gases are only contained in the same space or volume and the individual molecules retain their chemical properties.
A compound containing hydrogen and oxygen molecules exhibit different chemical (and even physical) properties compared to the individual molecules themselves.
Water for example, is a compound with 2 hydrogen atoms and 1 oxygen atom, chemically-bonded together. Hydrogen gas is highly flammable, water is not. Oxygen gas is an essential reagent for combustion (or burning) reactions, water is not.
Thus, throwing a lighted match to a gaseous mixture of hydrogen

and oxygen

would create fire, or even an explosion (since hydrogen is flammable and oxygen feeds the reaction). Throwing a match to water vapor

would not create fire.
If no subscript appears, one atom of that element is present.
<u>Answer:</u> The rate law expression is
and value of 'k' is 
<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[NO]^a[O_2]^b](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BNO%5D%5Ea%5BO_2%5D%5Eb)
where,
a = order with respect to nitrogen monoxide
b = order with respect to oxygen
- <u>Expression for rate law for first observation:</u>
....(1)
- <u>Expression for rate law for second observation:</u>
....(2)
- <u>Expression for rate law for third observation:</u>
....(3)
Dividing 1 from 2, we get:

Dividing 1 from 3, we get:

Thus, the rate law becomes:
![\text{Rate}=k[NO]^2[O_2]^1](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BNO%5D%5E2%5BO_2%5D%5E1)
Now, calculating the value of 'k' by using any expression.
Putting values in equation 1, we get:
![8.55\times 10^{-3}=k[0.030]^2[0.0055]^1\\\\k=1.727\times 10^3M^{-2}s^{-1}](https://tex.z-dn.net/?f=8.55%5Ctimes%2010%5E%7B-3%7D%3Dk%5B0.030%5D%5E2%5B0.0055%5D%5E1%5C%5C%5C%5Ck%3D1.727%5Ctimes%2010%5E3M%5E%7B-2%7Ds%5E%7B-1%7D)
Hence, the rate law expression is
and value of 'k' is 
To find heat, you need the following formula
heat= specific heat x mass x ΔT
specific heat (for water)= 4.184--> according to the table
mass= 100 grams
ΔT= 75 -10= 65 °C
let's plug the values.
specific heat= 4.184 x 100 x 65= 27200 joules or 27.2 Kilojoules