Answer: Theoretical Yield = 0.2952 g
Percentage Yield = 75.3%
Explanation:
Calculation of limiting reactant:
n-trans-cinnamic acid moles = (142mg/1000) / 148.16 = 9.584*10⁻⁴ mol
pyridium tribromide moles = (412mg/1000) / 319.82= 1.288*10⁻³ mol
- n-trans-cinnamic acid is the limiting reactant
The molar ratio according to the equation mentioned is equals to 1:1
The brominated product moles is also = 9.584*10⁻⁴ mol
Theoretical yield = (9.584*10⁻⁴ mol) * (Mr of brominated product)
= (9.584*10⁻⁴ mol) * (307.97) = 0.2952 g
Percentage Yield is : Actual Yield / Theoretical Yield = 0.2223/0.2952
= 75.3%
Answer:
We can say, there are 4 mol of oxygen with (4 . NA) = 2.40ₓ10²⁴ atoms
Or just, 4 atoms oxygen.
Explanation:
CH₄ + 2O₂ → CO₂ + 2H₂O
1 mol of methane reacts with 2 mol of oxygen to produce 1 mol of carbon dioxide and 2 mol of water.
We count in 1 mol of CO₂, we have 2 mol of oxygen
In 2 mol of water, we have 2 mol of oxygen.
We can say, there are 4 mol of oxygen with (4 . NA) = 2.40ₓ10²⁴ atoms
Or just, 4 atoms oxygen.
Answer:
Mass = 17.8 g
Explanation:
Given data:
Number of atoms of Ca = 2.68 × 10²³
Mass in gram = ?
Solution:
Avogadro number:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ atoms
2.68 × 10²³ atoms × 1 mole /6.022 × 10²³ atoms
0.445 mol
Mass in gram;
Mass = number of moles × molar mass
Mass = 0.445 mol × 40 g/mol
Mass = 17.8 g
In the following chemical equation, which are the reactants?
2HF + Mgo MgF2 + 2H20
>
A. 2HF + MgF2
B. 2H+ + Mgo
C. MgF2 + H20
D. MgO + H20
A
B
D
In the following chemical equation, which are the reactants?
2HF + Mgo MgF2 + 2H20
>
A. 2HF + MgF2
B. 2H+ + Mgo
C. MgF2 + H20
D. MgO + H20
A
B
D
