Answers:
See below
Step-by-step explanation:
1. Most food energy
(a) Pringles
Heat from Pringles + heat absorbed by water = 0
m₁ΔH + m₂CΔT = 0
1.984ΔH + 100 × 4.184 × 18 = 0
1.984ΔH + 7530 = 0
ΔH = -7530/1.984 = -3800 J/g
(b) Cheetos
0.884ΔH + 418.4 × 13 = 0
ΔH = -5400/0.884 = -6200 J/g
Cheetos give you more food energy per gram.
(c) Snickers
Food energy = 215 Cal/28 g × 4184 J/1 Cal = 32 000 J/g
The food energy from Cheetos is much less than that from a Snickers bar
2. Experimental uncertainty
The experimental values are almost certainly too low.
Your burning food is heating up the air around it, so much of the heat of combustion is lost to the atmosphere.
3. Percent efficiency
Experimental food energy = 3800 J/g
Actual food energy = 150 Cal/28 g × 4184 J/1 Cal = 22 000 J/g
% Efficiency = Experimental value/Actual value × 100 %
= 3800/22 000 × 100 %
= 17 %
Answer:
The percentage yield is 78.2g
Explanation:
Given, mass of propane = 42.8 g , sufficient O2 percent yield = 61.0 % yield.
Reaction - C3H8(g)+5O2(g)------> 3CO2(g)+4H2O(g)
First we need to calculate the moles of propane
Moles of propane = 
g.mol-1
= 0.971 moles
So, moles of CO2 from the moles of propane
1 mole of C3H8(g) = 3 moles of CO2(g)
So, 0.971 moles of C3H8(g) = ?
= 2.913 moles of CO2
So theoretical yield = 2.913 moles 
44.0 g/mol
= 128.2 g
So, the actual mass of CO2 = percent yield theoretical yield / 100 %
= 61.0 % 128.2 g / 100 %
= 78.2 g
the mass of CO2 that can be produced if the reaction of 42.8 g of propane and sufficient oxygen has a 61.0 % yield is 78.2 g
Melting points and boiling points of molecular compound are usually lower than ionic compounds. This is so as only a small amount of energy is required to overcome the weak intermolecular forces of attraction (Van de Waals forces) thus having lower m.p. and b.p.
Ionic compounds require a large amount of energy to overcome the strong electrostatic forces of attraction between the ions, hence having higher mp and bp
D IS THE ANSWER TO YOUR QUESTION
It's NF3 (one Nitrogen and 3 Flouride)
