The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of 
) =-1275.0
enthalpy of combustion of oxygen(Δ of 
) = zero
enthalpy of combustion of carbon dioxide(Δ of 
) = -393.5
enthalpy of combustion of water(Δ of 
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Are u sure this is the right option? Well, antimony can be decomposed. Including octane.
In which region is the substance in both the solid phase and the liquid phase ?
2
Answer:
the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷
Explanation:
Given the data in the question;
+ 
⇄ 
Formation constant Kf
Kf = / ( [][] ) = 5.0 × 10¹⁰
Now,
[] = 
; ∝₄ = 0.35
so the equilibrium is;
+ 
⇄ + 4H⁺
Given that; 
= { 1 mol reacts with 1 mol }
so at equilibrium, = = x
∴
+ ⇄
x + x 0.010-x
since Kf is high, them x will be small so, 0.010-x is approximately 0.010
so;
Kf = / ( [][] ) = / ( [][] ) = 5.0 × 10¹⁰
⇒ / ( [][] ) = 5.0 × 10¹⁰
⇒ 0.010 / ( [x][ 0.35 × x] ) = 5.0 × 10¹⁰
⇒ 0.010 / 0.35x² = 5.0 × 10¹⁰
⇒ x² = 0.010 / ( 0.35 × 5.0 × 10¹⁰ )
⇒ x² = 0.010 / 1.75 × 10¹⁰
⇒ x² = 0.010 / 1.75 × 10¹⁰
⇒ x² = 5.7142857 × 10⁻¹³
⇒ x = √(5.7142857 × 10⁻¹³)
⇒ x = 7.559 × 10⁻⁷
Therefore, the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷
Answer:
An educated guess based on what you already know.
Explanation:
