The density of Argon gas at a pressure of 753 mmHg and a temperature of 35 °C is equal to 1.59 g/L.
The state of a quantity of gas is calculated by its pressure, volume, and temperature. The ideal gas law can be explained as the product of the volume and pressure of gas is equal to the multiplication of the universal gas constant and absolute temperature.
The mathematical equation for an ideal gas can be written as follows:
PV = nRT
PV =(m/M) RT
PM/RT = m/V
d = PM/RT
Where n is the moles of gas, T is the temperature of the gas, V is the volume of the gas, and R is the gas constant.
Given, the temperature of argon gas, T = 35 °C = 35 +273 = 308 K
The pressure of the argon gas, P = 753 mmHg = 1.01 atm
The molar mass of the Argon gas, M = 40 g/mol
Substitute V, R, P, and T in the ideal gas equation, we get:
The density of Argon gas, d = PM/RT
d= 1.01 ×40/(0.082 × 308)
d = 1.59 g/L
Therefore, the density of Ar gas is 1.59 g/L.
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Answer: The reaction must proceed in the reverse direction.
Explanation:
is the constant of a certain reaction at equilibrium while is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.
For the given chemical reaction:
The expression of for above equation follows:
We are given:
There are 3 conditions:
When ; the reaction is product favored.
When ; the reaction is reactant favored.
When ; the reaction is in equilibrium.
As, , the reaction will be favoring reactant side or the reaction must proceed in the reverse direction.
Hence, the reaction must proceed in the reverse direction.
red
black
white
platelets
round
Answer:
The answer is C!
Explanation:
I took the test ;)
Given that the molar mass of CO2 is 44.01 g/mol, how many liters of oxygen is required at STP to produce 88.0 g of CO2 from this reaction?
44.8 L
45.00 L
89.55 L
89.6 L
89.6 L of O₂
The balanced chemical equation is as,
CH₄ + 2 O₂ → CO₂ + 2 H₂O
As at STP, one mole of any gas (Ideal gas) occupies exactly 22.4 L of Volume. Therefore, According to equation,
44 g ( 1 mol) CO₂ is produced by = 44.8 L (2 mol) of O₂
So,
88 g CO₂ will be produced by = X L of O₂
Solving for X,
X = (88 g × 44.8 L) ÷ 44 g
X = 89.6 L of O₂