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Alla [95]
2 years ago
13

How many grams of H3PO4 are in 234 mL of a 2.5 M solution of H3PO4?

Chemistry
1 answer:
motikmotik2 years ago
5 0

Answer:

57 grams of H3PO4

Explanation:

M= moles/ liters

convert mL to L

234 mL x 1L/1000mL = 0.234L

Rearrange the Molarity formula to solve for moles.

moles= MxL

moles= 2.5M x 0.234L

moles= 0.585 mol

Use the molar mass of H3PO4 to get to grams

0.585 mol x 97.994 grams/1 mol = 57.326 grams of H3PO4

round to two sig figs for 57 grams

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Vishwanath had some money. he spent 3 upon 4 part of money to buy goods for his birthday,1 upon 5 part of money give to his sist
lord [1]

Answer:

The correct answer is - 800.

Explanation:

Given:

Total amount = ? or assume x

spend in buying birthday item = 3/4 of x

given to sister = 1/5 of x

remaining to mother = 40

solution:

the remaning amount = x- (3x/4+x/5) = 4=

=> x- 19x/20 = 40

=> x = 20*40

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thus, the correct answer is = 800

8 0
2 years ago
A certain liquid X has a normal boiling point of 108.30 °C and a boiling point elevation constant Kb=1.07 °C kg/mol. A solution
Fynjy0 [20]

Answer:

34,6g of (NH₄)₂SO₄

Explanation:

The boiling-point elevation describes the phenomenon in which the boiling point of a liquid increases with the addition of a compound. The formula is:

ΔT = kb×m

Where ΔT is Tsolution - T solvent; kb is ebullioscopic constant and m is molality of ions in solution.

For the problem:

ΔT = 109,7°C-108,3°C = 1,4°C

kb = 1.07 °C kg/mol

Solving:

m = 1,31 mol/kg

As mass of X = 600g = 0,600kg:

1,31mol/kg×0,600kg = 0,785 moles of ions. As (NH₄)₂SO₄ has three ions:

0,785 moles of ions×\frac{1(NH_{4})_{2}SO_{4}}{3Ions} = 0,262 moles of (NH₄)₂SO₄

As molar mass of (NH₄)₂SO₄ is 132,14g/mol:

0,262 moles of (NH₄)₂SO₄×\frac{132,14g}{1mol} = <em>34,6g of (NH₄)₂SO₄</em>

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I hope it helps!

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