The freezing point of the solvent in a solution changes as the concentration of the solute in the solution changes (but it does not depend on the identity of either the solvent or the solute(s) particles (kind, size or charge) in the solution).
Generally, pressures lower than 1 atmosphere lower the temperature at which a substance freezes, but for water, a higher pressure gives a lower freezing point. The force from a pressure change figures into the molecular forces already at play in a substance.
Answer:
2.5 L.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and T are constant, and have two different values of V and P:
<em>P₁V₁ = P₂V₂
</em>
P₁ = 5.0 atm, V₁ = 3.5 L.
P₂ = 7.0 atm, V₂ = ??? L.
<em>∴ V₂ = P₁V₁/P₂ </em>= (5.0 atm)(3.5 L)/(7.0 atm) = <em>2.5 L.
</em>
Answer:
C₄H₂N₂
Explanation:
First we<u> calculate the moles of the gas</u>, using PV=nRT:
P = 2670 torr ⇒ 2670/760 = 3.51 atm
V = 300 mL ⇒ 300/1000 = 0.3 L
T = 228 °C ⇒ 228 + 273.16 = 501.16 K
- 3.51 atm * 0.3 L = n * 0.082atm·L·mol⁻¹·K⁻¹ * 501.16 K
Now we<u> calculate the molar mass of the compound</u>:
- 2.00 g / 0.0256 mol = 78 g/mol
Finally we use the percentages given to<em> </em><u>calculate the empirical formula</u>:
- C ⇒ 78 g/mol * 61.5/100 ÷ 12g/mol = 4
- H ⇒ 78 g/mol * 2.56/100 ÷ 1g/mol = 2
- N ⇒ 78 g/mol * 35.9/100 ÷ 14g/mol = 2
So the empirical formula is C₄H₂N₂
0.2346g of C8H18/114.224g x 6.022 x 10^20 mol of C8H18 =
0.01237 or 1.237x10^20 mol of C8H18.