Answer:
YIkES One thing u can do i ask your teacher about it or say that u need help with it the cant make fun of u because u need help with your work.
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:
0.402 moles
Explanation:
1 mole NaCl/58.44g Nacl ×23.5g NaCl
grams get cancelled out and you are left with moles
Answer:
The bottom number on each element of the periodic table are called the 4f series or lanthanoids and 5f or actanoids. They are also called inner transition elements.
At atmospheric pressure water boils at 100 degC.
So the water would be gas/vapor/steam.
