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lawyer [7]
3 years ago
15

A solution of NaCl(aq)NaCl(aq) is added slowly to a solution of lead nitrate, Pb(NO3)2(aq)Pb(NO3)2(aq) , until no further precip

itation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 13.25 g PbCl2(s)13.25 g PbCl2(s) is obtained from 200.0 mL200.0 mL of the original solution. Calculate the molarity of the Pb(NO3)2(aq)Pb(NO3)2(aq) solution.
Chemistry
1 answer:
Vinvika [58]3 years ago
3 0

Answer:

The molarity of the Pb(NO3)2 solution is 0.24 M

Explanation:

Assuming there was no loss during filtration, we can calculated amount of moles Pb(NO3)2 from the 13.25 g of PbCl2 ussing molar mass of  PbCl2

13.25 g / 278.1 g = 0.04 moles of PbCl2

According balanced equation 2NaCl + Pb(NO3)2 = 2NaNO3 + PbCl2, 1 mol of Pb(NO3)2 produces 1 mol of PbCl2. So, 0.04 moles of Pb(NO3)2 produces PbCl2

Molarity is = moles of solute/volumen of solution (L)=0.04/0.2= 0.24

I hope my answer helps you

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2 years ago
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15 points + brainliest pls help :(
Komok [63]

Answer:

27 g

Explanation:

M(C6H12O6) = 6*12 + 12*1 + 6*16 = 180 g/mol

100 mL = 0.1 L solution

1.5 M = 1.5 mol/L

1.5 mol/L * 0.1 L = 0.15 mol  C6H12O6

0.15 mol * 180 g/1 mol = 27 g C6H12O6

6 0
2 years ago
11cm+11.38cm+500.55cm=
Vika [28.1K]

Answer:522.93 centimeters

Explanation:

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8 0
3 years ago
A 42.0g sample of compound containing only C and H was analyzed. The results showed that the sample contained 36.0g of C and 6.0
user100 [1]

Answer:

What is the empirical formula of the compound?

Explanation:

When the relative masses of elements in a hydrocarbon are given, it is possible to use this information to obtain the empirical formula by dividing the given masses of each element by the relative atomic masses of the element. The lowest ratio is now used to divide through to obtain the empirical formula of the compound.

The empirical formula only shows that ratio of atoms of each element present in the compound. From the information provided, the empirical formula of the compound is CH2. Hence the answer.

5 0
3 years ago
How many grams of Na2So4 would be formed if 0.75 moles of NaOH reacted?
Nata [24]

Answer:

\boxed{\text{53 g }}

Explanation:

You don't give the reaction, but we can get by just by balancing atoms of Na.

We know we will need the partially balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.

M_r:                                 142.04  

             2NaOH + … ⟶ Na₂SO₄ + …  

n/mol:      0.75

1. Use the molar ratio of Na₂SO₄ to NaOH to calculate the moles of NaF.

Moles of Na₂SO₄ = 0.75 mol NaOH × (1 mol Na₂SO₄/2 mol NaOH

= 0.375 mol Na₂SO₄

2. Use the molar mass of Na₂SO₄ to calculate the mass of Na₂SO₄.

Mass of Na₂SO₄ = 0.375 mol Na₂SO₄ × (142.04 g Na₂SO₄/1 mol Na₂SO₄) = 53 g Na₂SO₄

The reaction produces \boxed{\text{53 g }} of Na₂SO₄.

4 0
3 years ago
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