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>
The answer to the question is the first opti.on.A
The total number of atoms make up the products :
D) 1 carbon, 4 hydrogen, and 4 oxygen
<h3>Further explanation
</h3>
Complete combustion of Hydrocarbons with Oxygen will produce CO₂ and H₂O compounds.
If O₂ is insufficient there will be incomplete combustion produced by CO and H and O
Hydrocarbon combustion reactions (especially alkanes)
For combustion of methane (CH₄) and two molecules of oxygen (O₂).
The number of atoms make up the products
CO₂ : 1 carbon, 2 oxygen
2H₂O : 4 hydrogen , 2 oxygen
we are given the reaction Cu + 2AgNO3 ---> Cu(NO3)2 + 2Ag. This means for every mole Cu used, there are 2 moles of Ag produced. In this case, given 31.75 g Cu, converting to moles through molar mass and using stoichiometric ratio and the molar mass of Ag, the mass Ag produced is 107.9 grams.
Answer:
B. created by man made dams
Explanation:
I live in Texas
