The volume becomes two. You have to use the equation P1 x V1 = P2 x V2
P is pressure and V is volume.
P1 = 50 P2 = 125
V1 = 5 V2 = v (we don't know what it is)
Then set up the equation:
50 times 5 = 125 times v
250 = 125v
the divide both sides by 125 and isolate v
2 = v
Therefore the volume is decreased to 2.
Also, Boyle's Law explains this too: Volume and pressure are inversely related, This means that when one goes up the other goes down (ie when pressure increases volume decreases and vice versa). Becuase the pressure went up from 50 KPa tp 125 KPa the volume had to decrease.
The balanced equation of the reaction is:
O3(g) + NO (g) → O2 (g) + NO2 (g)
Then the ratios of reaction is 1 mol O3 : 1 mol NO : 1 mol O2 : 1 mol NO2
If you have initially 0.05 M of O3 and 0.02 M of NO, the reaction will end when all the NO is consumed.
The by the stoichiometry 0.02 mol of O3 will be consumed in 8 seconds.
And the rate of reaction is change in concetration divided by the time.
The change in concentration in O3 is 0.02 M
Then, the rate respect O3 is 0.02 M / 8 seconds = 0.0025 M/s
the answer to your question is
volume
The grams of glucose are needed to prepare 400g of a 2.00%(m/m) glucose solution g is calculated as below
=% m/m =mass of the solute/mass of the solution x100
let mass of solute be represented by y
mass of solution = 400 g
% (m/m) = 2% = 2/100
grams of glucose is therefore =2/100 = y/400
by cross multiplication
100y = 800
divide both side by 100
y= 8.0 grams
