Answer:
Theoretical moles of V are 1.6 moles
Explanation:
The theoretical yield of a reaction is defined as the amount of product you would make if all of the limiting reactant was converted into product.
In the reaction:
V2O5(s) + 5Ca(i) → 2V(i) + 5CaO(s)
Based on the reaction, 1 mol of V2O5 needs 5 moles of Ca for a complete reaction. As there are just 4 moles, <em>limiting reactant is Ca.</em> As there are produced 2 moles of V per 5mol of Ca, Theoretical moles of V are:
4 moles of Ca × (2mol V / 5Ca) = <em>1.6 moles of V</em>
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I hope it helps!
Answer:
Conductivity meter
Explanation:
All of the other choices measure pH and can tell you when the pH changes, which is how you know the neutralization reaction has occurred. A conductivity meter does not measure pH.
Answer:
Explanation:
Let the number of half lives be x
<u>Solve this equation to find the value of x:</u>
- 125*(1/2)ˣ = 3.90625
- (0.5)ˣ = 3.90625 / 125
- (0.5)ˣ = 0.03125
- log (0.5)ˣ = log 0.03125
- x = log 0.03125 / log 0.5
- x = 5
Answer:
Explanation:
Combustion reaction is given below,
C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)
Provided that such a combustion has a normal enthalpy,
ΔH°rxn = -1270 kJ/mol
That would be 1 mol reacting to release of ethanol,
⇒ -1270 kJ of heat
Now,
0.383 Ethanol mol responds to release or unlock,
(c) Determine the final temperature of the air in the room after the combustion.
Given that :
specific heat c = 1.005 J/(g. °C)
m = 5.56 ×10⁴ g
Using the relation:
q = mcΔT
- 486.34 = 5.56 ×10⁴ × 1.005 × ΔT
ΔT= (486.34 × 1000 )/5.56×10⁴ × 1.005
ΔT= 836.88 °C
ΔT= T₂ - T₁
T₂ = ΔT + T₁
T₂ = 836.88 °C + 21.7°C
T₂ = 858.58 °C
Therefore, the final temperature of the air in the room after combustion is 858.58 °C
