Answer:
THE MASS OF NITROGEN GAS IN THIS CONDITIONS IS 0.0589 g
Explanation:
In an ideal condition
PV = nRT or PV = MRT/ MM where:
M = mass = unknown
MM =molar mass = 28 g/mol
P = pressure = 2 atm
V = volume = 25 mL = 0.025 L
R = gas constant = 0.082 L atm/mol K
T = temperature = 290 K
n = number of moles
The gas in the question is nitrogen gas
Molar mass of nitrogen gas = 14 * 2 = 28 g/mol
Then equating the variables and solving for M, we have
M = PV MM/ RT
M = 2 * 0.025 * 28 / 0.082 * 290
M = 1.4 / 23.78
M = 0.0589 g
The mass of the nitrogen gas at ideal conditions of 2 atm, 25 mL volume and 290 K temperature is 0.0589 g
Answer: If you think about it, B. would be the most reasonable answer with the given factors.
A combustion reaction is a reaction that reacts in the presence of oxygen molecules. Methane will release -3115 kJ/mol of heat.
<h3>What is a combustion reaction?</h3>
A combustion reaction includes the reaction between the chemical reactant and oxygen molecule to produce the product. The combustion reaction between methane and oxygen is given as:
CH₄(g) + 2O₂ (g) → CO₂(g) + 2H₂O (l), ΔH = -890 kJ/mol
The stoichiometry coefficient from the reaction gives 1 mole of methane releases -890 kJ/mol enthalpy.
So, 3.5 moles methane will release = 3.5 × -890 = -3115 kJ/mol
Therefore, -3115 kJ/mol of heat is released.
Learn more about combustion reaction here:
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Molar mass NaCl = 58.44 g/mol
number of moles:
mass NaCl / molar mass
145 / 58.44 => 2.4811 moles of NaCl
Volume = 3.45 L
Therefore :
M = moles / volume in liters:
M = 2.4811 / 3.45
M = 0.719 mol/L⁻¹
hope this helps!
