Answer is: volume of helium is 244.72 liters.
m(He) = 43.7 g.
n(He) = m(He) ÷ M(He).
n(He) = 43.7 g ÷ 4 g/mol.
n(He) = 10.925 mol.
V(He) = n(He) · n(He).
V(He) = 10.925 mol · 22.4 L/mol.
V(He) = 244.72 L.
Vm - molar volume at STP.
n - amount of substance.
Answer is: a lower freezing point has solution of K₂SO₄.
Change in freezing
point from pure solvent to solution: ΔT =i · Kf · b.<span>
Kf - molal freezing-point depression constant for water is 1.86°C/m.
b - molality, moles of solute per
kilogram of solvent.
i - </span>Van't
Hoff factor.<span>
b(K</span>₂SO₄<span>) = 0.35 m.
</span>b(KCl) = 0.5 m.
i(K₂SO₄) = 3.
i(KCl) = 2.
ΔT(K₂SO₄) = 3 · 0.35 m · 1.86°C/m.
ΔT(K₂SO₄) = 1.953°C.
ΔT(KCl) = 2 · 0.5 m · 1.86°C/m.
ΔT(KCl) = 1.86°C.
<u>Answer:</u> The new concentration of lemonade is 3.90 M
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
.....(1)
Molarity of lemonade solution = 2.66 M
Volume of solution = 473 mL
Putting values in equation 1, we get:
Now, calculating the new concentration of lemonade by using equation 1:
Moles of lemonade = 1.26 moles
Volume of solution = (473 - 150) mL = 323 mL
Putting values in equation 1, we get:
Hence, the new concentration of lemonade is 3.90 M
<span>It takes 3 breaths to get to 1.2 l. One breath is then (1.2 l) / 3 breaths = .4l/breath.
To get to 3.0 l we need the difference from 1.2 l.
3.0-1.2 = 1.8 l.
Divide the difference by liters/breath (.4) to get how many needed breaths.
(1.8 l)/(.4 l/breath) = 4.5 breaths to get the balloon to 3.0 l.
In total there were 3 breaths+ 4.5 breaths = 7.5breaths to get to 3.0 l.
To find the total moles multiply 7.5breaths by .060 moles/breath
7.5 breaths*.060moles/breath = .45moles</span>
Answer:
Option A
Explanation:
Number of millimoles of Na3PO4 = 1 × 100 = 100
Number of millimoles of AgNO3 = 1 × 100 = 100
When 1 mole of Na3PO4 is dissociated we get 3 moles of sodium ions and 1 mole of phosphate ion
When 1 mole of AgNO3 is dissociated, we get 1 mole of Ag+ and 1 mole of NO3-
As Ag+ concentration is negligible, the dissociated Ag+ ion must have form the precipitate with phosphate ion and as number of moles of Ag+ and phosphate ion are same, therefore the concentration of phosphate ion must be negligible
Here as 100 millimoles of Na3PO4 is there, we get 300 millimoles of Na+ and 100 millimoles of PO43-
And as 100 millimoles of AgNO3 is there, we get 100 millimoles of Ag+ and 100 millimoles of NO3-
∴ Increasing order of concentration will be PO43- < NO3- < Na+
