<h3><u>Answer;</u></h3>
Oxygen
<h3><u>Explanation;</u></h3>
- <em><u>A bunsen burner attains a blue gas flame when there is enough oxygen for complete combustion. </u></em>
- When there is sufficient oxygen, the gas flame appears blue because complete combustion creates enough energy to excite and ionize the gas molecules in the flame.
- In other words; increasing oxygen supply, less black body-radiating soot is produced as a result of a more complete combustion and the reaction creates enough energy to excite and ionize gas molecules in the flame, and a blue flame is formed.
Answer:
Moles of water in 7.1×10²⁵ molecules are 118 mol.
Explanation:
Given data:
Number of molecules of water = 7.1×10²⁵ molecules
Moles of water in 7.1×10²⁵ molecules = ?
Solution;
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.
For example,
1 mole = 6.022 × 10²³ molecules of water
7.1×10²⁵ molecules of water × 1 mol / 6.022 × 10²³ molecules of water
1.18×10² moles of water 0r 118 moles of water
Answer:
39.3%
Explanation:
CaF2 + H2SO4 --> CaSO4 + 2HF
We must first determine the limiting reactant, the limiting reactant is the reactant that yields the least number of moles of products. The question explicitly says that H2SO4 is in excess so CaF2 is the limiting reactant hence:
For CaF2;
Number of moles reacted= mass/molar mass
Molar mass of CaF2= 78.07 g/mol
Number of moles reacted= 11g/78.07 g/mol = 0.14 moles of Calcium flouride
Since 1 mole of calcium fluoride yields two moles of 2 moles hydrogen fluoride
0.14 moles of calcium fluoride will yield 0.14×2= 0.28 moles of hydrogen fluoride
Mass of hydrogen fluoride formed (theoretical yield) = number of moles× molar mass
Molar mass of hydrogen fluoride= 20.01 g/mol
Mass of HF= 0.28 moles × 20.01 g/mol= 5.6 g ( theoretical yield of HF)
Actual yield of HF was given in the question as 2.2g
% yield of HF= actual yield/ theoretical yield ×100
%yield of HF= 2.2/5.6 ×100
% yield of HF= 39.3%
