More drag is created because the air molecules are not moving out of the way of the airplane
Molar mass CuNO₃ = 125.55 g/mol
number of moles:
4.80 / 125.55 => 0.038 moles of CuNO₃
M = n / V
0.270 = 0.038 / V
V = 0.038 / 0.270
V = 0.1407 L
hope this helps!
Answer:
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Answer:
The number of oxygen in 7.9 × 10⁻¹ moles of CO₂ is 9.5116 × 10²³ atoms of oxygen
Explanation:
The given parameters are;
The number of moles of CO₂ = 7.9 × 10⁻¹ moles
The number of molecules of CO₂ in one mole of CO₂ = 6.02 × 10²³
The number of molecules, , of CO₂ in 7.9 × 10⁻¹ moles is given as follows;
= 7.9 × 10⁻¹ × 6.02 × 10²³ ≈ 4.76 × 10²³ molecules
The number of atoms of oxygen in one molecule of CO₂ = 2 oxygen atoms
Therefore;
The number of oxygen in 7.9 × 10⁻¹ moles of CO₂ = 2 × 4.76 × 10²³ ≈ 9.5116 × 10²³
The number of oxygen in 7.9 × 10⁻¹ moles of CO₂ ≈ 9.5116 × 10²³ atoms of oxygen.
Answer:
oxygen is limiting reactant
Explanation:
Given data:
Mass of phosphorus = 25.0 g
Mass of oxygen = 50.0 g
What is limiting reactant ?
Solution:
Chemical equation:
P₄ + 5O₂ → P₄O₁₀
Number of moles of P₄:
Number of moles = mass/molar mass
Number of moles = 25.0 g/ 123.89 g/mol
Number of moles = 0.20 mol
Number of moles of O₂:
Number of moles = mass/molar mass
Number of moles = 50.0 g/ 32 g/mol
Number of moles = 1.56 mol
now we will compare the moles of reactants with product:
P₄ : P₄O₁₀
1 : 1
0.20 : 0.20
O₂ : P₄O₁₀
5 : 1
1.56 : 1/5×1.56 = 0.312 mol
Less number of moles of product are formed by the oxygen thus it will act as limiting reactant.