Answer:
64.0 g/mol.
Explanation:
- Thomas Graham found that, at a constant temperature and pressure the rates of effusion of various gases are inversely proportional to the square root of their masses.
<em>∨ ∝ 1/√M.</em>
where, ∨ is the rate of diffusion of the gas.
M is the molar mass of the gas.
<em>∨₁/∨₂ = √(M₂/M₁)</em>
∨₁ is the rate of effusion of the unknown gas.
∨₂ is the rate of effusion of He gas.
M₁ is the molar mass of the unknown gas.
M₂ is the molar mass of He gas (M₂ = 4.0 g/mol).
<em>∨₁/∨₂ = 0.25.</em>
∵ ∨₁/∨₂ = √(M₂/M₁)
∴ (0.25) =√(4.0 g/mol)/(M₁)
<u><em>By squaring the both sides:</em></u>
∴ (0.25)² = (4.0 g/mol)/(M₁)
∴ M₁ = (4.0 g/mol)/(0.25)² = 64.0 g/mol.
Answer:
The answer to your question is 27 g of Al
Explanation:
Data
mass of Al = ?
moles of Al₂O₃ = 0.5
The correct formula for the product is Al₂O₃
Balanced chemical reaction
4Al + 3O₂ ⇒ 2Al₂O₃
Process
1.- Calculate the molar mass of the product
Al₂O₃ = (27 x 2) + (16 x 3)
= 54 + 48
= 102 g
2.- Convert the moles of Al₂O₃ to grams
102 g ---------------- 1 mol
x ---------------- 0.5 moles
x = (0.5 x 102) / 1
x = 51 g of Al₂O₃
3.- Use proportions to calculate the mass of Al
4(27) g of Al --------------- 2(102) g of Al₂O₃
x --------------- 51 g
x = (51 x 4(27)) / 2(102)
x = 5508 / 204
x = 27 g of Al
Answer: 0.374 mol CO2
Explanation:
We can calculate the number of moles of CO, n, required to produce the specified quantity of heat. This is done by dividing the required released heat, q, by the enthalpy of formation per unit mole of the substance, , based on the chemical equation, such that:
Therefore, 0.374 mol of CO2 must be reacted in order to produce 147 kJ of energy?
Answer : The specific heat (J/g-K) of this substance is, 0.780 J/g.K
Explanation :
Molar heat capacity : It is defined as the amount of heat absorbed by one mole of a substance to raise its temperature by one degree Celsius.
1 mole of substance releases heat = 92.1 J/K
As we are given, molar mass of unknown substance is, 118 g/mol that means, the mass of 1 mole of substance is, 118 g.
As, 118 g of substance releases heat = 92.1 J/K
So, 1 g of substance releases heat =
Thus, the specific heat (J/g-K) of this substance is, 0.780 J/g.K
The best example I
can give that is observable but does not contain matter is:
sunlight or simply light
Light does not contain matter because it is a beam of energy
or to be specific made up of packet of photons.
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