Answer:
The correct answer is: X is nitrogen dioxide, and Y is a metal oxide
Explanation:
Combustion of compound of containing nitrogen and metal will give nitrogen dioxide and metal oxide as product. During combustion reaction a compound reacts with oxygen in order to yield oxides of elements present in the compound.
The general equation is given as:
Hence, the correct answer is :X is nitrogen dioxide, and Y is a metal oxide.
Answer:
a. 0.5 mol
b. 1.5 mol
c. 0.67
Explanation:
Fe3+ + SCN- -----> [FeSCN]2+
a. The ratio of the product to Fe3+ is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of Fe3+ was used. Leaving 0.5 mol remaining at equilibrium
b. The ratio of the product to SCN= is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of SCN- was used. Leaving 1.5 mol remaining at equilibrium
c. KC = 0.5/(0.5*1.5) = 0.67
Answer:
0.185moles
Explanation:
Given parameters:
Volume of O₂ = 49.8L
Unknown:
Number of moles of sucrose required = ?
Solution:
We can assume that the reaction takes place at standard temperature and pressure.
From this, we can find the number of moles of oxygen that reacted and extrapolate to that of sucrose.
Chemical equation;
C₁₂H₂₂0₁₁ + 120₂ → 12CO₂ + 11H₂0
Number moles = 
at STP
Number of moles of oxygen gas = 
= 2.22moles
12 moles of oxygen gas combines with 1 mole of sucrose
2.22 moles of oxygen gas will combine with 
= 0.185moles
Answer:Effect of Catalysts on the Activation Energy. Catalysts provide a new reaction pathway in which a lower Activation energy is offered. A catalyst increases the rate of a reaction by lowering the activation energy so that more reactant molecules collide with enough energy to surmount the smaller energy barrier.
Explanation:
Your answer is in this
Answer:
(a) oxygen
(b) 154g (to 3sf)
(c) 79.9% (to 3sf)
Explanation:
mass (g) = moles × Mr/Ar
note: eqn means chemical equation
(a)
moles of P = 84.1 ÷ 30.973 = 2.7152 moles
moles of O2 = 85÷2(16) = 2.65625 moles
Assuming all the moles of P is used up,
moles of O2 / moles of phosphorus = 5/4 (according to balanced chemical eqn)
moles of O2 required = 5/4 × 2.7152moles = 3.394 moles (more than supplied which is 2.65625moles)
therefore there is insufficient moles of O2 and the limiting reactant is oxygen.
(b)
moles of P2O5 produced
= 2/5 (according to eqn) × 2.7152
= 1.08608moles
mass of P2O5 produced
= 1.08608 × [ 2(30.973) + 5(16) ]
= 154.164g
= approx. 154g to 3 sig. fig.
(c)
% yield = actual/theoretical yield × 100%
= 123/154 × 100%
= 79.870%
= approx. 79.9% (to 3sf)
