Answer: first option is not a termination
∙CH3 + Cl2 → CH3Cl + Cl∙
Explanation:
Since a radical is formed as part of the product it means it's a propagation step and not a termination step, at termination no free radical exist as product
Answer:
the mass of the air in the room is 4.96512 kg ( in 0°C)
Answer:
The answer is Sodium Sulfate = Na2SO4
Explanation:
Molar mass of sulfate = 1 (S) + 4 (O) = 1 (32) + 4 (16) = 32 + 64 = 96
Molar mass of sodium sulfate = 2 (23) + 96 = 46 + 96 = 142
% of Sulfate = (96/142)*100 = 67.6%
Percent mistake in Studen A,
(I) % mistake = (67.6 - 68.6)/67.6 = 1.48
(ii) % mistake = (67.6 - 66.2)/67.6 = 2.07
(iii) % mistake = (67.6 - 67.1)/67.6 = 0.74
For understudy B
(I) % mistake = (67.6 - 66.7)/67.6 = 1.33
(ii) % mistake = (67.6 - 66.6)/67.6 = 1.48
(iii) % mistake = (67.6 - 66.5)/67.6 = 1.63
Sutdent An is some how exact.
Understudy B is exact however not precise.
Answer:
We'll have 82 moles ZnO and 41 moles S
Explanation:
Step 1: data given
Number of moles Zinc (Zn) = 82 moles
Number of moles sulfur oxide (SO2) = 42 moles
Step 2: The balanced equation
2Zn + SO2 → 2ZnO + S
Step 3: Calculate the limiting reactant
For 2 moles Zinc we need 1 mol sulfur oxide to produce 2 moles zinc oxide and 1 mol sulfur
Zinc is the limiting reactant. It will completely be consume (82 moles). Sulfur oxide is in excess. There will react 82/2 = 41 moles
There will remain 42-41 = 1 mol SO2
Step 4: Calculate moles of products
For 2 moles Zinc we need 1 mol sulfur oxide to produce 2 moles zinc oxide and 1 mol sulfur
For 82 moles Zinc we'll have 82 moles of Zinc Oxide (ZnO)
For 82 moles Zinc we'll have 82/2 = 41 moles of sulfur
We'll have 82 moles ZnO and 41 moles S
