-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.
Answer:
1. Volume as STP = 755 L
2. Outside temperature = 255 K
3. Percentage yield = 70.5%
Explanation:
1. At STP, pressure = 101.3 kpa, temperature = 0°C or 273.15 K
Using the general gas equation :
P1V1/T1 = P2V2/T2
P1 = 620 kpa
V1 = 140 L
T1 = 37°C or (273.15 + 37) K = 310.15 K
P2 = 101.3 kpa
V2 = ?
T2 = 273.15 K
V2 = P1V1T2/P2T1
V2 = 620 × 140 × 273.15 / 101.3 × 310.15
V2 = 755 L
2. Using Charles' gas law:
V1/T1 = V2/T2
V1 = 2.5 L
T1 = 290 K
V2 = 2.2 L
T2 = ?
T2 = V2T1/VI
T2 = 2.2 × 290 / 2.5
T2 = 255 K
3. Equation of reaction : 2 Al + 3 CuSO4 ---> Al2 (SO4)3 + 3 Cu
From equation of the reaction, 2 moles of Al produces 3 moles of Cu
Molar mass of Al = 27 g; Molar mass of Cu = 63.5 g
2 moles of Al = 2 × 27 g = 54 g; 3 moles of Cu = 3× 63.5 = 190.5 g
54 g of Al produces 190.5 g of Cu
1.87 g of Al will produce 190.5/54 × 1.87 g of Cu = 6.60 g of Cu
Percentage yield = actual yield /theoretical yield × 100%
Percentage yield = 4.65/6.60 × 100%
Percentage yield = 70.5%
It is an energy source that the world has an abundance in.
<h2>Answer:</h2>
Correct option is B.
B. To relate the type of box material to the warmth of air within the box.
<h2>Explanation:</h2>
Kate gathered three boxes of the same size made of different materials: glass, clear plastic, and aluminum painted black. She placed them on a window sill in the sun for an hour and then measured the warmth of the air in each box. She actually did this to relate the type of box material to the warmth of air within the box.
Answer:
Explanation:
Hello,
In this case, considering that the moles of hydrioiodic acid remain unchanged during the dilution process:
One could apply the following equality in terms of molarity:
Whereas the subscript 1 accounts for the solution before the dilution and 2 after the dilution, therefore, the required volume of 6.00 M acid is:
Best regards.
