Chemistry, 14.11.2019 19:31 joshuac15

Asample of 5.10 ml of diethylether (c2h5oc2h5; density = 0.7134 g/ml) is introduced into a 6.00 -l vessel that already contains a mixture of n2 and o2, whose partial pressures are pn2 = 0.752 atm and po2 = 0.206 atm. the temperature is held at 35.0 °c, and the diethylether totally evaporates.
a) calculate the partial pressure of the diethylether.
b) calculate the total pressure in the container.

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Answer from: 20copstevens

A) partial pressure of diethylether = 0.206 atm

B) Total pressure = 1.164 atm

Explanation:

Given: Density of diethylether = 0.7134 g/mL, Volume of diethylether= 5.10 mL, Total volume = 6 L

Molar mass of diethylether = 74.12 g/mol

A) As we know, Density = given mass ÷ volume

⇒ given mass of diethylether = density × volume = 0.7134 g/mL ×5.10 mL = 3.64 g

As the number of moles of diethylether = given mass ÷ molar mass = 3.64 g ÷ 74.12 g/mol = 0.049 mol

To find out the pressure of diethylether, we use the ideal gas equation:

PV = nRT

Here, P is the partial pressure of the gas, n is the number of moles of gas = 0.049 mol, V is the total volume = 6 L, R is the gas constant = 0.08206 L·atm/(mol·K) and T is the temperature = 35°C = 35 + 273 = 308 K (0 °C = 273 K)

⇒ P = nRT ÷ V = (0.049 mol × 0.08206 L·atm/(mol·K) × 308 K) ÷ 6 L

⇒ P = 0.206 atm = partial pressure of diethylether : pC₂H₅OC₂H₅

B) Given: partial pressure of: pN₂= 0.752 atm, pO₂ = 0.206 atm and pC₂H₅OC₂H₅ = 0.206 atm

As the Total pressure of all the gases = sum of partial pressure of the gases

⇒ Total pressure = pN2 + pO₂ + pC₂H₅OC₂H₅ = 0.752 atm + 0.206 atm 0.206 atm = 1.164 atm

Answer from: kailahgranger

Explanation:

(A) The given data is as follows.

Volume of diethyl ether = 5.10 ml

Density of diethyl ether = 0.7134 g/ml

Partial pressure of nitrogen = 0.752 atm

Partial pressure of oxygen = 0.206 atm

Temperature = $35^{o}C$ = (35 + 273) K = 308 K

As, Density = $\frac{mass}{volume}$

0.7134 g/ml = $\frac{mass}{5.10 ml}$

mass = 3.63 g

Also, No. of moles = $\frac{mass}{\text{molar mass}}$

Hence, moles of diethyl ether will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{3.63 g}{74.12 g/mol}$

= 0.0489 mol

Hence, using the ideal gas equation we will find the partial pressure of diethyl ether as follows.

PV = nRT

$P \times 6 L = 0.0489 \times 0.0821 L atm/mol K \times 308 K$

P = 0.206 atm

Hence, the partial pressure of diethyl ether is 0.206 atm.

(B) As it is known that total pressure of a mixture of gases is equal to the sum of partial pressures of all the gases present in the mixture.

Therefore, we will calculate the total pressure as follows.

$P_{total} = P_{N_{2}} + P_{O_{2}} + P_{C_{2}H_{5}OC_{2}H_{5}}$

= 0.752 atm + 0.206 atm + 0.206 atm

= 1.164 atm

Hence, the total pressure in the container is 1.164 atm.

Answer from: Quest

може ли да зареждате повече информация, моля

Answer from: Quest

the dissociation of nickel hydroxide is follows:

$ni(oh)_2 > ni^2^+ + 2 oh^-$

$ksp for no(oh)2 = 5.5 * 10^-^1^6$

$5.5 * 10^-^1^6 = s * (2 s)^2 = 4 s^3$

$s^3 = 137.5 * 10^-^1^8$

$s = 5.14 * 10^-^6 = [oh^-]$

the concentration of hydroxide ion is

$5.14 * 10^-^6 = [oh^-]$

the equilibrium constant value of nickel hydroxide is = 6.1 * 10^-^1^6 and thus equilibrium equation can be written as:

$k = [ ni^2^+] * [oh^-]^2$

$6.1 * 10^-^1^6 = [ ni^2] * [5.14 * 10^-^6]^2$

$[ ni^2^+] = 0.23 * 10^-^4 mol /l$

thus, concentration of ni is 0.23 * 10^-^4 mol /l

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Asample of 5.10 ml of diethylether (c2h5oc2h5; density = 0.7134 g/ml) is introduced into a 6.00 -l...

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