Answer:
1. n = 0.174mol
2. T= 26.8K
3. P = 1.02atm
4. V = 126.88L
Explanation:
1. P= 2.61atm
V = 1.69L
T = 36.1 °C = 36.1 + 273= 309.1K
R = 0.082atm.L/mol /K
n =?
n = PV / RT = (2.61x1.69)/(0.082x309.1)
n = 0.174mol
2. P = 302 kPa = 302000Pa
101325Pa = 1atm
302000Pa = 302000/101325 = 2.98atm
V = 2382 mL = 2.382L
T =?
n = 3.23 mol
R = 0.082atm.L/mol /K
T= PV /nR = (2.98x2.382)/(3.23x0.082) = 26.8K
3. P =?
V = 0.0250 m³ = 25L
T = 288K
n = 1.08mol
R = 0.082atm.L/mol /K
P = nRT/V = (1.08x0.082x288)/25 = 1.02atm
4. P = 782 torr
760Torr = 1 atm
782 torr = 782/760 = 1.03atm
V =?
T = 303K
n = 5.26 mol
R = 0.082atm.L/mol /K
V = nRT/P
V = (5.26x0.082x303)/1.03 = 126.88L
Answer:
Explanation:
Hello there!
In this case, since the combustion of B2H6 is:
Thus, since there is 1:2 mole ratio between the reactant and product, the produced grams of the latter is:
Best regards!
Answer:
3.1atm
Explanation:
Given parameters:
Volume of gas = 2L
Number of moles = 0.25mol
Temperature = 25°C = 25 + 273 = 298K
Unknown:
Pressure of the gas = ?
Solution:
To solve this problem, we use the ideal gas equation.
This is given as;
PV = nRT
P is the pressure
V is the volume
n is the number of moles
R is the gas constant = 0.082atmdm³mol⁻¹K⁻¹
T is the temperature
P =
Now insert the parameters and solve;
P = = 3.1atm
Answer:
Explanation:
Dado que:
masa de oxígeno gaseoso = 100 g
presión = 1 atm
temperatura = 273 K
(a)
número de moles de oxígeno contenidos en el matraz = masa de oxígeno / masa molar de oxígeno
= 100 g / 16 gmol⁻¹
= 6.25 moles
(b) El número de moléculas de oxígeno es el siguiente:
Dado que 1 mol de oxígeno gaseoso contiene 6.023 * 10²³ moléculas de oxígeno.
Entonces, 6.25 moles contendrán:
= (6.25 × 6.023 * 10²³) moléculas de oxígeno.
≅ 3.764 × 10²³ moléculas de oxígeno.
(c) El número de átomos de oxígeno es:
= 2 × 3.764 × 10²³
= 7.528 × 10²³ átomos de oxígeno
(d) Usando la ecuación de gas ideal
PV = nRT
El volumen ocupado por el oxígeno =
Volumen ocupado por oxígeno =
Volumen ocupado por oxígeno= 14185.76 m³