Ammonium is NH₄⁺ and Carbonate is CO₃⁻² => Ammonium Carbonate is (NH₄)₂CO₃
<span>Answer: option B. 3.07 g
Explanation:
1) given reaction:
S(s) + O₂ (g) → SO(g)
2) Balanced chemical equation:
</span><span>2S(s) + O₂ (g) → 2SO(g)
3) Theoretical mole ratios:
2 mol S : 1 mol O₂ : 2 mol SO
3) number of moles of 4.5 liter SO₂ at</span><span> 300°C and 101 kPa
use the ideal gas equation:
pV = nRT
with V = 4.5 liter
p = 101 kPa
T = 300 + 273.15 K = 573.15 K
R = 8.314 liter×kPa / (mol×K)
=> n = pV / (RT) =
n = [101 kPa × 4.5 liter] / [8.314 (liter×kPa) / (mol×K) × 573.15 K ]
n = 0.0954 mol SO
4) proportion with the theoretical ratio S / SO
2 mol S x
-------------- = ----------------------
2 mol SO 0.0954 mol SO
=> x = 0.0954 mol S.
5) Convert mol of S to grams by using atomic mass of S = 32.065 g/mol
mass = number of moles × atomic mass
mass = 0.0954 mol × 32.065 g/mol = 3.059 g of S
6) Therefore the answer is the option B. 3.07 g
</span>
Answer:C11 -----^boron+positron
Explanation:the above reaction is nuclear reaction.
Positron emission causes the decrease in atomic number by one.
As in above example carbon-11is converted into boron isotop which has atomic number of 5.
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%
The the forth layer known as the thermosphere can reach as high as 2000 degrees Celsius.