<span>(P1/T1) = (P2/T2)
T must be in kelvin first!</span>
N(C)=5,02·10²² atoms
calculation check:
N(C)=(1/12)*6.022*10²³=0.5018*10²³≈5.02·10²²
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>
Answer:
The correct answer is 2, 0M
Explanation:
We calculate the molarity, which is a concentration measure that indicates the moles of solute (in this case NaCl) in 1000ml of solution (1 liter):
250 ml solution----- 0,5 moles of NaCl
1000 ml solution----x= (1000 ml solution x 0,5 moles of NaCl)/250 ml solution
x= 2,0 moles of NaCl --> <em>The solution is 2 M</em>
"X" in the reaction above is acetic acid with a chemical formula CH3COOH. The chemical reaction would be NaOH + CH3COOH = NaCH3COO + H2O. This is a neutralization reaction in which it produces a salt and water. The salt produced is called sodium acetate.
