Answer:
a. The original temperature of the gas is 2743K.
b. 20atm.
Explanation:
a. As a result of the gas laws, you can know that the temperature is inversely proportional to moles of a gas when pressure and volume remains constant. The equation could be:
T₁n₁ = T₂n₂
<em>Where T is absolute temperature and n amount of gas at 1, initial state and 2, final states.</em>
<em />
<em>Replacing with values of the problem:</em>
T₁n₁ = T₂n₂
X*7.1g = (X+300)*6.4g
7.1X = 6.4X + 1920
0.7X = 1920
X = 2743K
<h3>The original temperature of the gas is 2743K</h3><h3 />
b. Using general gas law:
PV = nRT
<em>Where P is pressure (Our unknown)</em>
<em>V is volume = 2.24L</em>
<em>n are moles of gas (7.1g / 35.45g/mol = 0.20 moles)</em>
R is gas constant = 0.082atmL/molK
And T is absolute temperature (2743K)
P*2.24L = 0.20mol*0.082atmL/molK*2743K
<h3>P = 20atm</h3>
<em />
Answer:
Mass = 6.183 g
Solution:
Step 1: Calculate number of moles of Boric acid using following formula,
Molarity = Moles ÷ Volume
Solving for Moles,
Moles = Molarity × Volume
Putting Values,
Moles = 0.05 mol.L⁻¹ × 2.0 L
Moles = 0.1 mol
Step 2: Calculate Mass of Boric Acid using following formula,
Moles = Mass ÷ M.mass
Solving for Mass,
Mass = Moles × M.mass
Putting values,
Mass = 0.1 mol × 61.83 g.mol⁻¹
Mass = 6.183 g
Flask used to prepare this solution is called as Volumetric flask. Take 2 L volumetric flask, add 6.183 g of Boric acid and fill it to the mark with distilled water.
Answer:
Partial pressure of nitrogen gas is 0.98 bar.
Explanation:
According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gases.




where,
= total pressure = 3.9 bar
= partial pressure of nitrogen gas
= partial pressure of oxygen gas
= partial pressure of argon gases
= Mole fraction of nitrogen gas = 0.25
= Mole fraction of oxygen gas = 0.65
= Mole fraction of argon gases = 0.10
Partial pressure of nitrogen gas :

Partial pressure of oxygen gas :

Partial pressure of argon gas :

Answer:
the same chemistry student has a weight of 155lbs what is the student weight in grams? (16 oz= 1 lb, 1 oz= 28.34g)
Explanation:1 lb = 16oz, so multiply your pounds by 16 to get you ounces of the student, then multiply by 28.34 to get grams
155 X 16 X 28.34 = 70283.2