I got <span>0.00125 hm . Hope I helped, and good luck(: </span>
Answer:
- <em>The volume of 14.0 g of nitrogen gas at STP is </em><u><em>11.2 liter.</em></u>
Explanation:
STP stands for standard pressure and temperature.
The International Institute of of Pure and Applied Chemistry, IUPAC changed the definition of standard temperature and pressure (STP) in 1982:
- Before the change, STP was defined as a temperature of 273.15 K and an absolute pressure of exactly 1 atm (101.325 kPa).
- After the change, STP is defined as a temperature of 273.15 K and an absolute pressure of exactly 105 Pa (100 kPa, 1 bar).
Using the ideal gas equation of state, PV = nRT you can calculate the volume of one mole (n = 1) of gas. With the former definition, the volume of a mol of gas at STP, rounded to 3 significant figures, was 22.4 liter. This is classical well known result.
With the later definition, the volume of a mol of gas at STP is 22.7 liter.
I will use the traditional measure of 22.4 liter per mole of gas.
<u>1) Convert 14.0 g of nitrogen gas to number of moles:</u>
- n = mass in grams / molar mass
- Atomic mass of nitrogen: 14.0 g/mol
- Nitrogen gas is a diatomic molecule, so the molar mass of nitrogen gas = molar mass of N₂ = 14.0 × 2 g/mol = 28.0 g/mol
- n = 14.0 g / 28.0 g/mol = 0.500 mol
<u>2) Set a proportion to calculate the volume of nitrogen gas:</u>
- 22.4 liter / mol = x / 0.500 mol
- Solve for x: x = 0.500 mol × 22.4 liter / mol = 11.2 liter.
<u>Conclusion:</u> the volume of 14.0 g of nitrogen gas at STP is 11.2 liter.
Follow Avogadro’s Number
1 mole = 6.02 x 10^23
So we can do it
4.77x10^25/6.02x10^23 = 79.2 mole
Depends on what compound we’re talking about here
This is a freezing point depression problem, so it will use the equation:
ΔT = i Kf<span> m
</span>
i = 1 (naphthalene does not dissociate further when dissolved)
Kf = 7.10 C/m (a constant for paradichlorobenzene, which you'd be given)
m = moles of solute / kg of solvent; moles of solute = 4.78 / 128.2 = 0.0373; kg of solvent = 0.032; m = 0.0373 / 0.032 = <span>1.166m
</span>ΔT = 1 x 7.10 x 1.166 = 8.279 C
This means the normal freezing point of pure paradichlorobenzene is decreased by 8.279 C; the normal freezing point (again, something you'd be given) is 53.5 C, so the new freezing point would be 53.5 - 8.279 = 45.221 C.
