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 manner in which objects will move is determined by the answer to this question. Unbalanced forces will cause objects to change their state of motion
Explanation:
Since these two forces are of equal magnitude and in opposite directions, they balance each other.
C is the answer because animals and plants need water.
Hey there!:
V1 = 3.05 L
V2 = 3.00 L
P1 = 724 mmHg
P2 = to be calculated
T1 = 298 K
T2 = 273 K
Therefore:
P1*V1 / T1 = P2*V2 / T2
P2 = ( P1*V1 / T1 ) * T2 / V2
P2 = 724 * 3.05 * 273 / 298 * 3.00
P2 = 602838.6 / 894
P2 = 674.31 mmHg
1 atm ----------- 760 mmHg
atm ------------- 674.31 mHg
= 674.31 * 1 / 760
= 0.887 atm
Hope this helps!
Answer:
<h2>The answer is 3.16 mL</h2>
Explanation:
The volume of a substance when given the density and mass can be found by using the formula
From the question
mass = 43 g
density = 13.6 g/mL
The volume is
We have the final answer as
<h3>3.16 mL</h3>
Hope this helps you
