Answer:
82.28g
Explanation:
Given parameters:
Number of moles of hydrogen gas = 7.26 moles
Unknown:
Amount of ammonia produced = ?
Solution:
We have to write the balanced equation first.
N₂ + 3H₂ → 2NH₃
Now, we work from the known to the unknown;
3 moles of H₂ will produce 2 moles of NH₃
7.26 mole of H₂ will produce 
= 4.84 moles of NH₃
Molar mass of NH₃ = 14 + 3(1) = 17g/mol
Mass of NH₃ = number of moles x molar mass = 4.84 x 17 = 82.28g
Answer:
Choices 2 and 4
Explanation:
HCL is formed witthe the elements Hydrogen and Chlorine. Whatever is on the left side of the equation must match up with the right side of the equation.
2. H + Cl --> HCl (we have one hydrogen and one chlorine)
4. 2HCl --> H2 + Cl2 (since the two is distributed to both the H and the Cl, H has 2 and Cl has two on the left side AND on the right side of the equation)
Answer:
1. 2NaN₃(s) → 2Na(s) + 3N₂(g)
2. 14.5 g NaN₃
Explanation:
The answer is incomplete, as it is missing the required values to solve the problem. An internet search shows me these values for this question. Keep in mind that if your values are different your result will be different as well, but the solving methodology won't change.
" The airbags that protect people in car crashes are inflated by the extremely rapid decomposition of sodium azide, which produces large volumes of nitrogen gas. 1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid sodium azide (NaN₃) into solid sodium and gaseous dinitrogen. 2. Suppose 71.0 L of dinitrogen gas are produced by this reaction, at a temperature of 16.0 °C and pressure of exactly 1 atm. Calculate the mass of sodium azide that must have reacted. Round your answer to 3 significant digits. "
1. The <u>reaction that takes place is</u>:
- 2NaN₃(s) → 2Na(s) + 3N₂(g)
2. We use PV=nRT to <u>calculate the moles of N₂ that were produced</u>.
P = 1 atm
V = 71.0 L
n = ?
T = 16.0 °C ⇒ 16.0 + 273.16 = 289.16 K
- 1 atm * 71.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 289.16 K
Now we <u>convert N₂ moles to NaN₃ moles</u>:
- 0.334 mol N₂ *
= 0.223 mol NaN₃
Finally we <u>convert NaN₃ moles to grams</u>, using its molar mass:
- 0.223 mol NaN₃ * 65 g/mol = 14.5 g NaN₃
Answer:
camphor sublimates salt is soluble in water while sand does not sublime and does not dissolve in water you first heat the mixture in a beaker covered with a watch glass camphor will then accumulate on the watch glass then you dissolve the remaining mixture of sand and salt salt will dissolve forming a salt solution then you filter using a filter paper and a beaker the residue on the filter paper is sand while the filtrate is salt solution you then heat the salt solution so that it can evaporate leaving salt particles thus you will have obtained salt sand and camphor
Answer:
At the burner temp. and pressure, 18.85 litres of air is needed to completely combust each gram of propane
Explanation:
The combustion stoichiometry is as follows:
C₃H₈ + 5O₂ = 4 H₂O + 3CO₂ The molecular weights (g/mol) are:
MW 44 5x32 4x18 3x44
So each gram of propane is 1/44 = 0.02272 mol propane
and will need 5 x 0.02272 = 0.1136 mol oxygen
At 0.21 mol fraction oxygen in air, 0.1136 / 0.21 = 0.54 mol air is needed to burn the propane.
At the low pressure in the burner we can use the Ideal Gas Law
PV=nRT, or V = nRT/P
P = 1.1 x 101325 Pa = 111457 Pa
T = 195°C + 273 = 468 K
R = 8.314
and we calculated n = number of moles air = 0.54 mol
So V m³ = 0.54 x 8.314 x 468 / 111457 = 0.0188 m³ = 18.85 litres air.
