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.
<span>Avogadro's number
represents the number of units in one mole of any substance. This has the value
of 6.022 x 10^23 units / mole. This number can be used to convert the number of
atoms or molecules into number of moles. We calculate as follows:
0.180 mol Br2 ( </span>6.022 x 10^23 molecules / mole ) = 1.084x10^23 molecules Br2
One: looks to be correct for both answers. Certainly the first one is. The second depends on your other choices. But military use is one.
Two: is correct. Pd has (in this case) an atomic mass of 114 and its number is 46
Three: Even with my slop numbers, 4.98 is the answer (although I get 4.99 but again, my numbers are pretty sloppy).
Four: Slop numbers say 78.3, but 78 is the right answer.
Five: Slop numbers agree with Al2S3. I think that's D
They are all correct. Very Fine Work.
Answer:
333.7 g.
Explanation:
- The depression in freezing point of water (ΔTf) due to adding a solute to it is given by: <em>ΔTf = Kf.m.</em>
Where, ΔTf is the depression in water freezing point (ΔTf = 20.0°C).
Kf is the molal freezing point depression constant of the solvent (Kf = 1.86 °C/m).
m is the molality of the solution.
<em>∴ m = ΔTf/Kf</em> = (20.0°C)/(1.86 °C/m) = <em>10.75 m.</em>
molaity (m) is the no. of moles of solute per kg of the solvent.
∵ m = (no. of moles of antifreeze C₂H₄(OH)₂)/(mass of water (kg))
∴ no. of moles of antifreeze C₂H₄(OH)₂ = (m)(mass of water (kg)) = (10.75 m)(0.5 kg) = 5.376 mol.
∵ no. of moles = mass/molar mass.
<em>∴ mass of antifreeze C₂H₄(OH)₂ = no. of moles x molar mass </em>= (5.376 mol)(62.07 g/mol) =<em> 333.7 g.</em>
The number of grams of Ag2SO4 that could be formed is 31.8 grams
<u><em> calculation</em></u>
Balanced equation is as below
2 AgNO3 (aq) + H2SO4(aq) → Ag2SO4 (s) +2 HNO3 (aq)
- Find the moles of each reactant by use of mole= mass/molar mass formula
that is moles of AgNO3= 34.7 g / 169.87 g/mol= 0.204 moles
moles of H2SO4 = 28.6 g/98 g/mol =0.292 moles
- use the mole ratio to determine the moles of Ag2SO4
that is;
- the mole ratio of AgNo3 : Ag2SO4 is 2:1 therefore the moles of Ag2SO4= 0.204 x1/2=0.102 moles
- The moles ratio of H2SO4 : Ag2SO4 is 1:1 therefore the moles of Ag2SO4 = 0.292 moles
- AgNO3 is the limiting reagent therefore the moles of Ag2SO4 = 0.102 moles
<h3> finally find the mass of Ag2SO4 by use of mass=mole x molar mass formula</h3>
that is 0.102 moles x 311.8 g/mol= 31.8 grams
