Answer:
10.28 mol
Explanation:
S + 2O = SO2
(atm x L) ÷ (0.0821 x K)
(3.45 x 45.6) ÷ (0.0821 x 373)
=5.13726
Then round it to significant figures
=5.14
5.14 mol SO2 x (2 mol O ÷ 1 mol SO2)
=10.28 mol O
There are four main types of air pollution sources:
<span>mobile sources – such as cars, buses, planes, trucks, and trainsstationary sources – such as power plants, oil refineries, industrial facilities, and factoriesarea sources – such as agricultural areas, cities, and wood burning fireplaces<span>natural sources – such as wind-blown dust, wildfires, and volcanoes........hope i helped</span></span>
Answer:
They both include a question,procedure and conclusion.
Answer:
The percent by mass of copper in the mixture was 32%
Explanation:
The ammount of HNO₃ used is:
mol HNO₃ = volume * concentration
mol HNO₃ = 0.015 l * 15.8 mol/l
mol HNO₃ = 0.237 mol
According to the reaction, 4 mol HNO₃ will react with 1 mol Cu and produce 1 mol Cu²⁺. Since we have 0.237 mol HNO₃, the amount of Cu that could react would be (0.237 mol HNO₃ * 1 mol Cu / 4 mol HNO₃) 0.06 mol. This reaction would produce 0.060 mol Cu²⁺, however, only 0.010 mol Cu²⁺ were obtained, indicating that only 0.010 mol Cu were present in the mixture. This means that the acid was in excess, so we can assume that all copper present in the mixture has reacted.
Since 0.010 mol of Cu²⁺ were produced, the amount of Cu was 0.01 mol.
1 mol of Cu has a mass of 63.5 g, then 0.01 mol has a mass of:
0.01 mol Cu * 63.5 g / 1 mol = 0.635 g.
Since this amount was present in 2.00 g mixture, the amount of copper in 100 g of the mixture will be:
100 g(mixture) * 0.635 g Cu / 2 g(mixture) = 32 g
Then, the percent by mass of Cu (which is the mass of Cu in 100 g mixture) is 32%
Answer:
c.
Explanation:
it is c.-1 because oxidation number of cl is-1
