Matematically speaking, maybe because:
The number of substances = number of elements + number of different combinations of those elements
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>
It is water soluble so is also soluble in aqueous solutions of NaOH or NaHCO3.
Answer: 1090°C
Explanation: According to combined gas laws
(P1 × V1) ÷ T1 = (P2 × V2) ÷ T2
where P1 = initial pressure of gas = 80.0 kPa
V1 = initial volume of gas = 10.0 L
T1 = initial temperature of gas = 240 °C = (240 + 273) K = 513 K
P2 = final pressure of gas = 107 kPa
V2 = final volume of gas = 20.0 L
T2 = final temperature of gas
Substituting the values,
(80.0 kPa × 10.0 L) ÷ (513 K) = (107 kPa × 20.0 L) ÷ T2
T2 = 513 K × (107 kPa ÷80.0 kPa) × (20.0 L ÷ 10.0 L)
T2 = 513 K × (1.3375) × (2)
T2 = 1372.275 K
T2 = (1372.275 - 273) °C
T2 = 1099 °C
Answer:
72
Explanation:
The pattern here may be hard to find at first, but it's this: the number in the middle of the triangle = (number at lower left corner of triangle x number at upper vertex of triangle) + (number at upper vertex of triangle x number at lower right corner of triangle).
Thus, for the missing value...
Missing value = (3x8) + (8x6) = 24+48 = 72.
Could you tell me what concept in chemistry relates to this? I'm interested.
Also check out stylesben's answer. Seems like there's several ways of doing this.
