Answer:
140 K
Explanation:
Step 1: Given data
- Initial pressure of the gas (P₁): 3 atm
- Initial temperature of the gas (T₁): 280 K
- Final pressure of the gas (P₂): 1.5 atm
- Final temperature of the gas (T₂): ?
Step 2: Calculate the final temperature of the gas
We have a gas whose pressure is reduced. If we assume an ideal behavior, we can calculate the final temperature of the gas using Gay-Lussac's law.
T₁/P₁ = T₂/P₂
T₂ = T₁ × P₂/P₁
T₂ = 280 K × 1.5 atm/3 atm = 140 K
Answer:
The number of formula units in 3.81 g of potassium chloride (KCl) is approximately 3.08 × 10²²
Explanation:
The given parameters is as follows;
The mass of potassium chloride produced in the chemical reaction (KCl) = 3.81 g
The required information = The number of formula units of potassium chloride (KCl)
The Molar Mass of KCl = 74.5513 g/mol
Therefore, we have;
1 mole of a substance, contains Avogadro's number (6.022 × 10²³) of formula units
Therefore;
0.051106 moles of KCl contains 0.051106 × 6.022 × 10²³ ≈ 3.077588 × 10²² formula units
From which we have, the number of formula units in 3.81 g of potassium chloride (KCl) ≈ 3.08 × 10²² formula units.
Answer:
24.309 g/mol
Explanation:
To get the atomic mass, all we have to do is calculate with the masses of the three isotope, the real quantity present, taking account of the percent and then, do a sum of these three values. Like a pondered media.
For the first isotope:
23.99 * (78.99/100) = 18.95 g/mol
For the second isotope:
24.99 * (10/100) = 2.499 g/mol
For the last isotope:
25.98 * (11.01/100) = 2.86 g/mol
Now, let's sum all three together
AW = 18.95 + 2.499 + 2.86
AW = 24.309 g/mol
Answer:
V = 85.619 L
Explanation:
To solve, we can use the ideal gas law equation, PV = nRT.
P = pressure (645 mmHg)
V = volume (?)
n = amount of substance (3.00 mol)
R = ideal gas constant (62.4 L mmHg/mole K)
T = temperature (295K)
Now we would plug in the appropriate numbers into the equation using the information given and solve for V.
(645)(V) = (3.00)(62.4)(295)
(V) = (3.00)(62.4)(295)/645
V = 85.619 L
