Answer : The rate of effusion of sulfur dioxide gas is 52 mL/s.
Solution :
According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.
or,
..........(1)
where,
= rate of effusion of nitrogen gas = 
= rate of effusion of sulfur dioxide gas = ?
= molar mass of nitrogen gas = 28 g/mole
= molar mass of sulfur dioxide gas = 64 g/mole
Now put all the given values in the above formula 1, we get:
Therefore, the rate of effusion of sulfur dioxide gas is 52 mL/s.
Answer:
88.46%
Explanation:
Percentage yield is actual/theoretical * 100
138/156 * 100 = 88.4615385
From google but i can explain further if needed. <span> The </span>balanced<span> equation for the reaction of interest contains the stoichiometric ratios of the reactants and products; these ratios </span>can<span> be used as </span>conversion factors<span> for mole-to-mole </span>conversions<span>.</span>
Answer is: 4,4 grams <span>of carbon dioxide gas would be produced.
</span>Chemical reaction: CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O.
m(CaCO₃) = 10 g.
n(CaCO₃) = 10 g ÷ 100 g/mol.
n(CaCO₃) = 0,1 mol.
From chemical reaction: n(CaCO₃) : n(CO₂) = 1 : 1.
n(CO₂) = 0,1 mol.
m(CO₂) = n(CO₂) · M(CO₂).
m(CO₂) = 0,1 mol· 44 g/mol.
m(CO₂) = 4,4 g.
Answer:
4.42 × 10⁻³⁷ m
Explanation:
Step 1: Given and required data
- Mass of the body (m): 1 kg
- Velocity of the body (v): 1500 m/s
- Planck's constant (h): 6.63 × 10⁻³⁴ J.s
Step 2: Calculate the de Broglie wavelenght (λ) of the body
We will use de Broglie's equation.
λ = h / m × v
λ = (6.63 × 10⁻³⁴ J.s) / 1 kg × (1500 m/s) = 4.42 × 10⁻³⁷ m
