Answer 1) : The density of the hot air inside the balloon can be found out by using ideal gas equation;
PV = nRT;
As n is number of moles and in gases, number of moles along with mass per mole is equal to the density of the gas.
If the moles in the gas are more the density will be more.
here, density (ρ) = mass (m) / volume (V); substituting in the ideal gas equation we get,
ρ = mP / RT
Answer 2) ρ (hot air) = ρ (cold air) X 
Here according to the formula because T(hot air) >T(cold air),
So, the density of hot air greater than the density of cold air.
The relationship between the ρ (h) = ρ(c) X
Answer:
7.46 g
Explanation:
From the balanced equation, 2 moles of Mg is required for 2 moles of MgO.
The mole ratio is 1:1
mole = mass/molar mass
mole of 4.50 g Mg = 4.50/24.3 = 0.185 mole
0.185 mole Mg will tiled 0.185 MgO
Hence, theoretical yield of MgO in g
mass = mole x molar mass
0.185 x 40.3 = 7.46 g
Using the combined gas law, where PV/T = constant, we first solve for PV/T for the initial conditions: (4.50 atm)(36.0 mL)/(10.0 + 273.15 K) = 0.57213.
Remember to use absolute temperature.
For the final conditions: (3.50 atm)(85.0 mL)/T = 297.5/T
Since these must equal, 0.57213 = 297.5/T
T = 519.98 K
Subtracting 273.15 gives 246.83 degC.
ΔH(reaction) = ΔH(formation of products) - ΔH(formation of reactants)
ΔH(reaction) = ( 1*ΔH(Pb(s)) + 1*ΔH(CO2(g)) ) - ( 1*ΔH(PbO(s)) + 1*ΔH(CO(g)) )
ΔH(reaction) = ( 0 + -393.5 ) - ( ΔH(PbO(s)) + -110.5 )
ΔH(reaction) = -283 - ΔH(PbO(s))
-131.4 = -283 -ΔH(PbO(s))
ΔH(PbO(s)) = -151.6 kJ
So, the best answer is A.
Answer:
NADH and FADH
Explanation:
An oxidizing agent is a chemical substance which has one among the three listed properties.
1. It can accept electrons.
2. It can remove hydrogen.
3. It can add oxygen to the reactant.
NAD+ and FAD+ serve as an oxidizing agent because they not only accept a pair of electron but also remove hydrogen. In this molecule, they have actually taken hydrogen from carbon where OH functional group was present and oxidizing that particular alcohol group to ketone. Please see the image attached.
