They didnt have a powerfull enough microscope to see them an when they could they were moving to fast unless they were cold
Answer:
305 litres of NO gas will be produced from 916 L of NO₂
Explanation:
Given the balanced equation of the chemical reaction as follows:
3 NO₂ (g) + H₂O( l) —— 2 HNO₃ (l) + NO (g)
Under standard conditions, 3 moles of No₂ will react with 1 mole of water to produce 1 mole of NO gas.
Molar volume of a gas at STP is 22.4 L
Number of moles of NO₂ gas present in 916 L = 916/22.4 = 40.893 moles of NO₂ gas
From the mole ratio of NO₂ to NO in the equation of reaction,
Number of moles of NO that will be produced = 1/3 × 40.893 moles = 13.631 moles of NO gas
Volume of 13.631 moles of NO gas = 13.631 × 22.4
Volume of NO gas produced = 305.334L
Therefore, Volume of NO gas produced from the reaction of 916 L of NO₂ with water = 305 L
P₄ + 10Cl₂ ---> 4PCl₅
stoichiometry of P₄ to PCl₅ is 1:4
number of moles of P₄ reacted - 28.0 g / 124 g/mol = 0.22 mol
Cl₂ is in excess therefore P₄ is the limiting reactant, amount of product formed depends on amount of limiting reactant present
according to molar ratio of 1:4
number of PCl₅ moles formed -0.22 mol x 4 = 0.88 mol
0.88 mol of PCl₅ is formed
Answer:
c is the answer then check it out
Answer:
0.259 kJ/mol ≅ 0.26 kJ/mol.
Explanation:
- To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat absorbed by ice (Q = ??? J).
m is the mass of the ice (m = 100.0 g).
c is the specific heat of water (c of ice = 4.186 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 21.56°C - 25.0°C = -3.44°C).
<em>∵ Q = m.c.ΔT</em>
∴ Q = (100.0 g)(4.186 J/g.°C)(-3.44°C) = -1440 J = -1.44 kJ.
<em>∵ ΔH = Q/n</em>
n = mass/molar mass = (100.0 g)/(18.0 g/mol) = 5.556 mol.
∴ ΔH = (-1.44 kJ)/(5.556 mol) = 0.259 kJ/mol ≅ 0.26 kJ/mol.
