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skad [1K]
2 years ago
12

What is the boiling point of a solution made by mixing 75.0g ZnCl2 with 375.0 grams of water? (Kb for water is 0.512 C/m)

Chemistry
1 answer:
monitta2 years ago
5 0
The problem can be solved using the following formula:

ΔTb = i Kb <em>m</em>

i = moles particles/moles solute
Kb = 0.512 °C/m
m = molality =  moles solute/kg solvent

First we can solve for the molality of the solution:

75.0 g ZnCl₂ / 136.286 g/mol = 0.550 mol ZnCl₂

m = 0.550 mol/0.375 kg
m = 1.468 mol/kg

We can now solve for the change in temperature of the boiling point:

ΔTb = i Kb m
ΔTb = (3 mol particles/1 mol ZnCl₂) (0.512 °C/m) (1.468 m)
ΔTb = 2.25 °C

The boiling point of a solution is the initial boiling point plus the change in boiling point:

BP = 100 °C + 2.25 °C
BP = 102.25 °C

The solution will have a boiling point of 102.25 °C.
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3 Apply Many scientific words, such as matter, also have everyday meanings. Use context clues to write your own definition for e
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Answer:

Explanation:

Use context clues to write your own definition for each meaning of the word matter.

Example sentence

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Matter:

__________

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Matter:

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3 0
1 year ago
Please help me I need this
klio [65]
It would be:
1.D
2.A
3.B
4.C

7 0
3 years ago
How many atoms are in this formula? 4 O2 A 2
Effectus [21]

Answer:

8 atoms as 4*2=8

Explanation:

4 0
3 years ago
Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)
Brums [2.3K]

Answer:1) Volume of AgNO_3 required is 55.98 mL.

2) 0.62577 grams of Ag_3PO_4 is produced.

Explanation:

3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)

1) Molarity of AgNO_3,M_1=0.225 M

Volume of AgNO_3.V_1=?

Molarity of Na_3PO_4,M_2=0.135 M

Volume of Na_3PO_4,V_2=31.1 mL=0.0311 L

Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}

\text{number of moles }Na_3PO_4=M_2\times V_2=0.135 mol/L\times 0.0311 L=0.0041985 moles

According to reaction, 1 mole of Na_3PO_4 reacts with 3 mole of AgNO_3, then, 0.0041985 moles of Na_3PO_4 will react with:

\frac{3}{1}\times 0.0041985 moles of AgNO_3 that is 0.0125955 moles.

M_1=0.225 M=\frac{\text{number of moles of }AgNO_3}{V_1}

V_1=\frac{0.0125955 moles}{0.225 M}=0.05598 L=55.98 mL

Volume of AgNO_3 required is 55.98 mL.

2)

Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}

Number of moles of AgNO_3=0.195\times 0.023 L=0.004485 moles

According to reaction, 3 moles of AgNO_3 gives 1 mole of Ag_3PO_4, then 0.004485 moles of AgNO_3 will give:\frac{1}{3}\times 0.004485 moles of Ag_3PO_4 that is 0.001495 moles.

Mass of Ag_3PO_4 =

Moles of Ag_3PO_4 × Molar Mass of Ag_3PO_4

= 0.001495 moles × 418.58 g/mol = 0.62577 g

0.62577 grams of Ag_3PO_4 is produced.

7 0
3 years ago
How many grams are in 4.07x10^15 molecules of calcium hydroxide
maksim [4K]

There are 5.01 × 10-⁷grams in 4.07 x 10¹⁵molecules of calcium hydroxide.

HOW TO CALCULATE MASS:

The mass of a substance can be calculated by multiplying the number of moles in the substance by its molecular mass.

However, given that the number of molecules in calcium hydroxide is 4.07 x 10¹⁵molecules, we need to calculate the number of moles in Ca(OH)2 as follows:

no. of moles of Ca(OH)2 = 4.07 x 10¹⁵ ÷ 6.02 × 10²³

no. of moles = 0.676 × 10-⁸

no. of moles = 6.76 × 10-⁹ moles.

Molar mass of Ca(OH)2 = 74.093 g/mol

Mass of Ca(OH)2 = 74.093 × 6.76 × 10-⁹ moles

Mass = 5.01 × 10-⁷grams.

Therefore, there are 5.01 × 10-⁷grams in 4.07 x 10¹⁵molecules of calcium hydroxide.

Learn more about how to calculate mass at: brainly.com/question/8101390?referrer=searchResults

5 0
2 years ago
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