Nts

Chemistry

1 answer:

nordsb [41]2 years ago

4 0

Answer:

118.2 g H₂O

Explanation:

To find the mass of water, you need to (1) convert moles HNO₃ to moles H₂O (via mole-to-mole ratio from reaction coefficients) and then (2) convert moles H₂O to grams H₂O (via molar mass).

The molar mass of water can be found by combining the element's molar masses times their quantities. The ratios/conversions should be arranged in a way that allows for the cancellation of units (desired unit should be in the numerator).

Molar Mass (H₂O): 2(1 g/mol) + 16 g/mol

Molar Mass (H₂O): 18 g/mol

S + 6 HNO₃ ---> H₂SO₄ + 6 NO₂ + 2 H₂O
^ ^

19.7 moles HNO₃ 2 moles H₂O 18 g
-------------------------- x ----------------------- x --------------------- = 118.2 g H₂O
6 moles HNO₃ 1 mole H₂O

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At −15.0 ∘C , a common temperature for household freezers, what is the maximum mass of sucrose (C12H22O11) you can add to 1.50 k

nlexa [21]

Answer: The maximum mass of sucrose you can add is 4158.95 grams.

Explanation:

This is an example of freezing point depression. The formula for calculating this is the following:

ΔT = Kf . b . i

ΔT is the temperature depression

Kf is the cryoscopic constant that is unique for each solvent

b is the molality of the solution (moles of solute per kg of solvent)

i is the Vant Hoff factor

The freezing point of water is 0°C. ΔT equals inicial temperature - final temperature, so it's 0°- (-15°)= 273K - 258K = 15K

The Kf for water is known to be 1.853 K. Kg /mol

i is the number of particles the molecule is split to when ionized. Because sucrose doesn't ionize, its Vant Hoff factor is 1.

If we clear b from the ecuation:

b = ΔT/ Kf . i

b = 15K/ 1.853 K. Kg/mol . 1

b= 8.1 mol/kg

If we can add 8.1 moles to a kg of water before it freezes, we use cross multiplication to calculate how many we can add to 1.5 kg. The answer is 12.15.

The weight of a mole of sucrose is 342.3 grams. So the weight of 12.15 moles of sucrose is 4158.95g.

3 0

3 years ago

1. Determine the final volume of a 1.5 M HCl solution prepared from 20.0 mL

nadezda [96]

Explanation:

Because molarity is classified as moles of solute per liter of water, dilution of the water may result in a reduction of its concentration.

Therefore, because the amount of moles of solute has to be constant for dilution, you will use the molarity and volume of that same target solution to calculate how many moles of solute will be present in the sample of the stock solution that you dilute.

c = $\frac{n}{v}$