Answer:
THE VOLUME OF 0.200M CALCIUM HYDROXIDE NEEDED TO NEUTRALIZE 35 mL of 0.050 M NITRIC ACID IS 43.75 mL.
Explanation:
Using
Ca VA / Cb Vb = Na / Nb
Ca = 0.0500 M
Va = 35 mL
Cb = 0.0200 M
Vb = unknown
Na = 2
Nb = 1
Equation for the reaction:
Ca(OH)2 + 2HNO3 --------> Ca(NO3)2 + 2H2O
So therefore, we make Vb the subject of the equation and solve for it
Vb = Ca Va Nb / Cb Na
Vb = 0.0500 * 35 * 1 / 0.0200 * 2
Vb = 1.75 / 0.04
Vb = 43.75 mL
The volume of 0.02M calcium hydroxide required to neutralize 35 mL of 0.05 M nitric acid is 43.75 mL
Answer:
95.7 g CO to the nearest tenth.
Explanation:
2C + O2 ---> 2CO
Using relative atomic masses:
24 g C produces 2*12 + 2*16 g CO.
So 41 g produces ( (2*12 + 2*16) * 41 ) / 24
= 95.7 g CO,
We can use the ideal gas equation to determine the temperature with the given conditions of mass of the gas, volume, and pressure. The equation is expressed
PV=nRT where n is the number of moles equal to mass / molar mass of gas. Substituting the given conditions with R = 0.0521 L atm/mol K we can find the temperature
To estimate the molar mass of the gas, we use Graham's law of effusion. This relates the rates of effusion of gases with their molar mass. We calculate as follows:
r1/r2 = √(m2/m1)
where r1 would be the effusion rate of the gas and r2 is for CO2, M1 is the molar mass of the gas and M2 would be the molar mass of CO2 (44.01 g/mol)
r1 = 1.6r2
1.6 = √(44.01 / m1)
m1 = 17.19 g/mol
