Answer:
1) When 69.9 g heptane is burned it releases 5.6 mol water.
2) C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O.
Explanation:
- Firstly, we should balance the equation of heptane combustion.
- The balanced equation is: <em>C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O.</em>
This means that every 1.0 mole of complete combustion of heptane will release 8 moles of H₂O.
- We need to calculate the no. of moles in 69.9 g of heptane that is burned using the relation: <em>n = mass/molar mass.</em>
n of 69.9 g of heptane = mass/molar mass = (69.9 g)/(100.21 g/mol) = 0.697 mol ≅ 0.7 mol.
<em><u>Using cross multiplication:</u></em>
1.0 mol of heptane releases → 8 moles of water.
0.7 mol of heptane releases → ??? moles of water.
<em>∴ The no. of moles of water that will be released from burning (69.9 g) of water</em> = (0.7 mol)(8.0 mol)/(1.0 mol) = <em>5.6 mol.</em>
<em>∴ When 69.9 g heptane is burned it releases </em><em>5.6</em><em> mol water. </em>
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Answer: a) 211 mm Hg
b) 0.629 grams
Explanation:
According to Dalton's law, the total pressure is the sum of individual pressures.

= total pressure = 750 mmHg
= 124 mm Hg
= 218 mm Hg
= 197 mm Hg
= ?


Thus the partial pressure of the helium gas is 211 mmHg.
b) According to the ideal gas equation:
P = Pressure of the gas = 211 mmHg = 0.28 atm (760mmHg=1atm)
V= Volume of the gas = 13.0 L
T= Temperature of the gas = 282 K
R= Gas constant = 0.0821 atmL/K mol
n= moles of gas= ?
Mass of helium= 
Thus mass of helium gas present in a 13.0-L sample of this mixture at 282 K is 0.629 grams
A, energy stored in chemical bonds of molecules, is correct because chemical energy is the potential energy in the molecule, for example the chemical energy in coal is converted to electrical energy through power plants. The energy was already stored in the coal, just in a different form.
Answer : The energy released by an electron in a mercury atom to produce a photon of this light must be, 
Explanation : Given,
Wavelength = 
conversion used : 
Formula used :

As, 
So, 
where,
= frequency
h = Planck's constant = 
= wavelength = 
c = speed of light = 
Now put all the given values in the above formula, we get:


Therefore, the energy released by an electron in a mercury atom to produce a photon of this light must be, 
The correct answer to this question is letter "B) 8.4 × 10-4 M."
2CO+O2<->2CO2
-x -x +x
It started as 2CO2 so +x, then it decomposed into 2CO and O2, so-x
2CO-oxidized, O2-Oxidized, 2CO2-reduced
the easiest way is to look at the equation, so if we have some reactant and no product
A<->2C+D
-x +2x +x
so if we had product concentration, and no reactant concentration
3A<-> C+D
+3x -x-x
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