Answer:
The vapor pressure for the first solution at 20°C is 17.48 Torr.
The vapor pressure for the second solution at 20°C is 17.49 Torr.
Explanation:
1) mass of solute that is glucose = 10 g
Moles of glucose ,![n_1=\frac{10 g}{180 g/mol}=0.05555 mol](https://tex.z-dn.net/?f=n_1%3D%5Cfrac%7B10%20g%7D%7B180%20g%2Fmol%7D%3D0.05555%20mol)
Mass of water = m
Volume of water = V = 1L = 1000 mL
Density of water = 1 g/ml
![m=d\times V=1 g/mL\times 1000 mL=1000 g](https://tex.z-dn.net/?f=m%3Dd%5Ctimes%20V%3D1%20g%2FmL%5Ctimes%201000%20mL%3D1000%20g)
Moles of water = ![n_2=\frac{1000 g}{18 g/mol}=55.55 mol](https://tex.z-dn.net/?f=n_2%3D%5Cfrac%7B1000%20g%7D%7B18%20g%2Fmol%7D%3D55.55%20mol)
Vapor pressure of the solution = ![p](https://tex.z-dn.net/?f=p)
Vapor pressure of the pure solvent that is water = ![p_o=17.5 Torr](https://tex.z-dn.net/?f=p_o%3D17.5%20Torr)
Mole fraction of solute= ![\chi_1=\frac{n_1}{n_1+n_2}](https://tex.z-dn.net/?f=%5Cchi_1%3D%5Cfrac%7Bn_1%7D%7Bn_1%2Bn_2%7D)
![\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}](https://tex.z-dn.net/?f=%5Cfrac%7Bp_o-p%7D%7Bp_o%7D%3D%5Cfrac%7Bn_1%7D%7Bn_1%2Bn_2%7D)
![\frac{17.5 Torr-p}{17.5 Torr}=\frac{0.05555 mol}{0.05555 mol+55.55 mol}](https://tex.z-dn.net/?f=%5Cfrac%7B17.5%20Torr-p%7D%7B17.5%20Torr%7D%3D%5Cfrac%7B0.05555%20mol%7D%7B0.05555%20mol%2B55.55%20mol%7D)
![p=17.48 Torr](https://tex.z-dn.net/?f=p%3D17.48%20Torr)
The vapor pressure for the first solution at 20°C is 17.48 Torr.
2 ) mass of solute that is sucrose= 10 g
Moles of sucrose ,![n_1=\frac{10 g}{342 g/mol}=0.02924 mol](https://tex.z-dn.net/?f=n_1%3D%5Cfrac%7B10%20g%7D%7B342%20g%2Fmol%7D%3D0.02924%20mol)
Mass of water = m
Volume of water = V = 1L = 1000 mL
Density of water = 1 g/ml
![m=d\times V=1 g/mL\times 1000 mL=1000 g](https://tex.z-dn.net/?f=m%3Dd%5Ctimes%20V%3D1%20g%2FmL%5Ctimes%201000%20mL%3D1000%20g)
Moles of water = ![n_2=\frac{1000 g}{18 g/mol}=55.55 mol](https://tex.z-dn.net/?f=n_2%3D%5Cfrac%7B1000%20g%7D%7B18%20g%2Fmol%7D%3D55.55%20mol)
Vapor pressure of the solution = ![p](https://tex.z-dn.net/?f=p)
Vapor pressure of the pure solvent that is water = ![p_o=17.5 Torr](https://tex.z-dn.net/?f=p_o%3D17.5%20Torr)
Mole fraction of solute= ![\chi_1=\frac{n_1}{n_1+n_2}](https://tex.z-dn.net/?f=%5Cchi_1%3D%5Cfrac%7Bn_1%7D%7Bn_1%2Bn_2%7D)
![\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}](https://tex.z-dn.net/?f=%5Cfrac%7Bp_o-p%7D%7Bp_o%7D%3D%5Cfrac%7Bn_1%7D%7Bn_1%2Bn_2%7D)
![\frac{17.5 Torr-p}{17.5 Torr}=\frac{0.02924 mol}{0.02924 mol+55.55 mol}](https://tex.z-dn.net/?f=%5Cfrac%7B17.5%20Torr-p%7D%7B17.5%20Torr%7D%3D%5Cfrac%7B0.02924%20mol%7D%7B0.02924%20mol%2B55.55%20mol%7D)
![p=17.49 Torr](https://tex.z-dn.net/?f=p%3D17.49%20Torr)
The vapor pressure for the second solution at 20°C is 17.49 Torr.
Answer:
ΔHr = -275 kj
Explanation:
It is possible to obtain the net change in enthalpy for the formation of one mole of lead(II) sulfate from lead, lead(IV) oxide, and sulfuric acid using the reactions:
(1) H₂SO₄(l) → SO₃(g) + H₂O (l) ΔH=+113kJ
(2) Pb(s) + PbO₂(s) + 2SO₃(g) → 2PbSO₄(s) ΔH=−775kJ
If you sum (1) + ¹/₂(2) you will obtain:
H₂SO₄(l) + ¹/₂Pb(s) + ¹/₂PbO₂(s) → PbSO₄(s) + H₂O(l)
Using Hess's law, the net change in enthalpy for this reaction could be obtained as:
ΔHr = ΔH(1) + ¹/₂ΔH(2)
ΔHr = +113kJ + ¹/₂ -775kJ
ΔHr = -275 kJ
Answer:
When you breathe in, or inhale, your diaphragm contracts and moves downward. This increases the space in your chest cavity, and your lungs expand into it. The muscles between your ribs also help enlarge the chest cavity. They contract to pull your rib cage both upward and outward when you inhale.
Explanation:
^^^
Answer:
22.2 g of MgF₂
Explanation:
We star from the reaction:
NaF → sodium fluoride
Mg → Magnesium
2NaF + Mg → MgF₂ + 2Na
2 moles of sodium fluoride react to 1 mol of Mg in order to produce 1 mol of magnesium fluoride and 2 moles of sodium.
We convert mass of the reactant to moles:
30 g . 1mol / 41.98 g = 0.715 mol
We assume that Mg is in excess
As ratio is 2:1, if we have 0.715 moles of NaF we may produce the half of moles, of MgF₂
0.715 mol /2 = 0.357 moles of MgF₂
We convert moles to mass: 0.357 mol . 62.30g /mol = 22.2 g
The limiting reactant in the reaction would be the compound that would be consumed completely in the reaction. The balanced chemical reaction for this problem is written as:
<span>MnO2 + 4HCl → MnCl2 + 2H2O + Cl2
From the reaction and the amounts present of the reactant, the limiting reactant would be HCl or hydrochloric acid. Ten moles of HCl only need 2.5 moles of MnO2. Hope this helps.</span>