Answer:
1) The temperature inside a diesel engine is high enough for two atmospheric gases nitrogen and oxygen to react and combine to form nitrogen oxides, NOₓ
At very high temperatures of above 1,300 °C, which are temperatures obtainable in the internal combusting chamber of a diesel engine under 100% load, air containing N₂ and O₂ is sucked in, and they combine to produce approximately 85% of the NOₓ pollutants produced from moving engine sources
2) a) N₂O contributes to the Greenhouse effect and therefore global warming
b) NO₂ contributes to the formation of smog and NO₂ combines with water to produce nitric acid which falls as acid rain, and causes irritation in the eyes and the respiratory system airways
c) Nitric oxide gas, NO, has the effect of depleting the Ozone layer which increases global warming and increased penetration of UVB
Explanation:
Answer:
(1) I shifts toward product and II shifts toward reactant.
Explanation:
Increasing the temperature of an endothermic reaction (∆H is positive) shifts the equilibrium position to the right thus favoring product formation.
Increasing the temperature of an exothermic reaction (∆H is negative) shifts the equilibrium position to the left thus favoring the backward reaction.
-130KJ is the standard heat of formation of CuO.
Explanation:
The standard heat of formation or enthalpy change can be calculated by using the formula:
standard heat of formation of reaction = standard enthalpy of formation of product - sum of enthalpy of product formation
Data given:
Cu2O(s) ---> CuO(s) + Cu(s) ∆H° = 11.3 kJ
2 Cu2O(s) + O2(g) ---> 4 CuO(s) ∆H° = -287.9 kJ
CuO + Cu ⇒ Cu2O (-11.3 KJ) ( Formation of Cu2O)
When 1 mole Cu20 undergoes combustion 1/2 moles of oxygen is consumed.
Cu20 + 1/2 02 ⇒ 2CuO (I/2 of 238.7 KJ) or 119.35 KJ
So standard heat of formation of formation of Cu0 as:
Cu + 1/2 02 ⇒ CuO
putting the values in the equation
ΔHf = ΔH1 + ΔH2 (ΔH1 + ΔH2 enthalapy of reactants)
heat of formation = -11.3 + (-119.35)
= - 130.65kJ
-130.65 KJ is the heat of formation of CuO in the given reaction.
To make it easier, assume that we have a total of 100 g of a compound. Hence, we have 58.80g of xenon, 7.166g of oxygen, and 34.04g of fluorine.
Know we will convert each of these masses to moles by using the atomic masses:
58.8/131.3 = 0.45 mole of Xe
7.166/16 = 0.45 mole of O
34.04/19 = 1.79 mole of F
Now, we will divide all the mole numbers by the smallest among them and get the number of atoms in the compound:
Xe = 0.45/0.45 = 1
O = 045/0.45 = 1
F = 1.79/0.45 = 3.98 = 4
So, the empirical formula of the compound XeOF₄
