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:
726 torr
Explanation:
Generally, atmospheric pressure can be measured using a manometer which is in form of a U-shaped tube. In addition, 1 mm Hg is equivalent to 1 torr. Therefore, 752 torr is equivalent to 752 mm Hg. Therefore, the total pressure will be equivalent to the atmospheric pressure (mm Hg) + the mercury height.
In this case, the mercury height = -26 mm
Thus:
The helium pressure = 752 - 26 = 726 mm Hg
This is also equivalent to 726 torr
Answer:
23.8 L
Explanation:
There is some info missing. I think this is the original question.
<em>Calculate the volume in liters of a 0.0380M potassium iodide solution that contains 150 g of potassium iodide. Be sure your answer has the correct number of significant digits.</em>
<em />
The molar mass of potassium iodide is 166.00 g/mol. The moles corresponding to 150 grams are:
150 g × (1 mol/166.00 g) = 0.904 mol
0.904 moles of potassium iodide are contained in an unknown volume of a 0.0380 mol/L potassium iodide solution. The volume is:
0.904 mol × (1 L/0.0380 mol) = 23.8 L
A functional group is a group of atoms that determines many of the properties of an organic molecule.
For example functional group for alcohol is hydroxyl group (-OH) and for carboxylic acid is <span>carboxyl group (-COOH).
</span>Alpha carbon is <span>the first </span>carbon<span> atom after the carbon that attaches to the functional group.</span>
Answer: 55.84L
Explanation: Please see attachment for explanation.
