Answer:
44 g oxygen are needed.
Explanation:
Given data:
Mass of oxygen needed = ?
Mass of ammonia = 18.2 g
Solution:
Chemical equation:
4NH₃ + 5O₂ → 4NO + 6H₂O
Now we will calculate the number of moles of ammonia:
Number of moles = mass/molar mass
Number of moles = 18.2 g/ 17 g/mol
Number of moles = 1.1 mol
Now we will compare the moles of ammonia with oxygen from balance chemical equation.
NH₃ : O₂
4 : 5
1.1 : 5/4×1.1 = 1.375 mol
Mass of oxygen needed:
Mass = number of moles × molar mass
Mass = 1.375 mol × 32 g/mol
Mass = 44 g
The solubility of carbon dioxide at 400 kPa at room temperature is ;
( B ) 0.61 CO2/L
<u>Given data </u>
pressure of CO₂ = 400 Kpa = 3.95 atm
Kh of CO₂ = 3.3 * 10⁻² mol/L.atm
<h3>Calculate the solubility of carbon dioxide </h3>
Solubility = pressure * Kh value of CO₂
= 3.95 atm * 3.3 * 10⁻² mol / L.atm
= 0.13 mol/l CO₂
= 0.61 CO₂ / L
Hence we can conclude that the solubility of CO₂ at 400 kPa is 0.13 mol/l CO₂.
Learn more about solubility : brainly.com/question/23946616
From the options the closest answer is ( B ) 0.61 CO₂ / L
I would say that you should wear a lab coat, safety goggles, and gloves
when the teacher says so - not everything in a lab is dangerous, so
there is no need to always wear these. But when the teacher says you
should - then you should.
Answer:
25.35%
Explanation:
Again let me restate the the equation of the reaction;
H2O (ℓ) + 2 MnO4 - (aq) + 3 CN- (aq) → 2 MnO2 (s) + 3 CNO- (aq) + 2 OH- (aq)
Amount of potassium permanganate reacted = 10.2/1000 * 0.08035 = 8.1957 * 10^-4 moles
If 2 moles of MnO4 - reacts with 3 moles of CN-
8.1957 * 10^-4 moles of MnO4 - reacts with 8.1957 * 10^-4 * 3/2
= 1.229 * 10^-3 moles of CN-
Mass of CN- reacted = 1.229 * 10^-3 moles of CN- * 26.02 g/mol
= 0.03 g
Hence, percentage of the cyanide = 0.03 g/0.1183 g * 100
= 25.35%
False, the hydrogen atom does not form the basis for all life.
