Answer:
0.047 %
Explanation:
Step 1: Given data
- Partial pressure of ozone (pO₃): 0.33 torr
- Total pressure of air (P): 695 torr
Step 2: Calculate the %v/v of ozone in the air
Air is a mixture of gases. We can find the %v/v of ozone (a component) in the air (mixture) using the following expression.
<em>%v/v = pO₃/P × 100%</em>
%v/v = 0.33 torr/695 torr × 100%
%v/v = 0.047 %
Answer:
0.0011 mol/L.s
Explanation:
The average rate of disappearing of the reagent is the variation of the concentration of it divided by the time that this variation is being measured. The reaction rate, is proportional to the coefficient of the substance, so, for a generic reaction:
aA + bB --> cC + dD
rate = -(1/a)Δ[A]/Δt = -(1/b)Δ[B]/Δt = (1/c)Δ[C]/Δdt = (1/d)Δ[D]/dt
The minus sign is because of the reagent is desapering, so:
rate = -(1/2)*(0.0209 - 0.0300)/(10 - 6)
rate = 0.0011 mol/L.s
Answer:
Explanation:
Burning fossil fuels releases the carbon dioxide stored millions of years ago. ... The concentration of carbon dioxide in the atmosphere has increased more in the northern hemisphere where more fossil fuel burning occurs. Since the Industrial Revolution the concentration globally has increased by about 40 % .
Answer:
n = 0.26 mol.
Explanation:
Given,
Pressure, P = 99.7 kPa = 1 atm
where 101.325 kPa = 1 atm
P = 0.984 atm
Temperature, T = 297 K
Volume = 6.452 L
Now, using ideal gas equation
PV = n RT
0.984 x 6.452 = n x 0.08206 x 297
n = 0.26 mol.
Answer:
0.0222 mole of NaOH is needed to react with NH4F
Explanation:
NH4F + NaOH --> NaF + NH3 + H2O
Data given
Mass of NH4F =0.821g, Concentration of NaOH= 1M, volume of NaoH =25ml
But mole = (CV)/1000
given mole of NaoH = (1 * 25)/1000 = 0.025moles of NaOH used
Molar mass of NH4F = 37g/mol
mole of NH4F used 0.821 / 37 = 0.0222 mole NH4F
Determine the excess and limiting reactant,
NaOH is in excess
0.025 - 0.0222 = 0.0028 mole NaOH excess
0.0222 mole of NaOH is required to react with NH4F