I turned in this exact assignment today haha
the blood vessels dilate to draw body heat away from the body and towards the surface of the skin
sweat glands release sweat, when sweat evaporates it releases heat
Answer : The rate law for the overall reaction is, ![Rate=k[NO]^2[H_2]](https://tex.z-dn.net/?f=Rate%3Dk%5BNO%5D%5E2%5BH_2%5D)
Explanation :
Rate law : It 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.
As we are given the mechanism for the reaction :
Step 1 : 
(slow)
Step 2 : 
(fast)
Overall reaction : 
The rate law expression for overall reaction should be in terms of 
.
As we know that the slow step is the rate determining step. So,
The slow step reaction is,
The expression of rate law for this reaction will be,
Hence, the rate law for the overall reaction is,
Answer:
P₂ = 2.7 atm
Explanation:
Given data:
Initial temperature = 30°C
Initial pressure = 3.00 atm
Final temperature = -5°C
Final pressure = ?
Solution:
Initial temperature = 30°C = 30 + 273 = 303 K
Final temperature = -5°C = -5 + 273 = 268 K
According to Gay-Lussac Law,
The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.
Mathematical relationship:
P₁/T₁ = P₂/T₂
Now we will put the values in formula:
3.0 atm / 303 K = P₂/268 K
P₂ = 3.0 atm × 268 K / 303 K
P₂ = 804 atm. K /293 K
P₂ = 2.7 atm
The approximate alcohol content is 210 ml.
Explanation:
It can be deduced from the question that each bottle is of 1000ml or 1 litre.
The first bottle is one half full means it has 500 ml of solution and it has 20% alcohol in it. So volume of alcohol in the solution is
20/100*500
=100 ml
The first bottle is one fifth full, so the volume of mixture is 1/5th of 1000ml
so it is 200ml having 30% alcohol
30/100*200
= 60 ml
The third bottle is one tenth full so its volume is 1/10*1000
100 ml. having 50% of alcohol
50/100*100
50 ml.
The alcohol content obtained from all these 3 litres is:
100+60+50
= 210 ml of alchohol is obtained from 800 ml of mixture.
.2135 mol of Ba(OH)2 are needed to neutralize .427 mol of 2HC2H3O2
