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IrinaVladis [17]
3 years ago
13

What is the molarity of a 300.0 mL solution containing 25.0 g of NaCl?

Chemistry
1 answer:
Aleks04 [339]3 years ago
8 0

Answer:

1,43 M

Explanation:

molair mass NaCl = 58,44 g/mol

You need to calculate how many moles 25 g is:

25 g / 58,44 g/mol = 0,4278 mol

0,4278 mol / 0,3 L = 1,43 M

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Potassium chlorate decomposes into potassium chloride and oxygen gas. How many grams of oxygen are produced when 1.06 grams of p
kifflom [539]

The amount of oxygen that are produced when 1.06 grams of potassium chlorate decompose completely is 0.64 grams.

<h3>What is the relation between mass & moles?</h3>

Relation between the mass and moles of any substance will be represented as:

  • n = W/M, where
  • W = given mass
  • M = molar mass

Moles of potassium chlorate = 1.66g / 122.5g/mol = 0.0135mole

Given chemical reaction is:

2KClO₃ → 2KCl + 3O₂

From the stoichiometry of the reaction, it is clear that:

2 moles of KClO₃ = produces 3 moles of O₂

0.0135 moles of KClO₃ = produces (3/2)(0.0135)=0.02 moles of O₂

Mass of oxygen = (0.02mol)(32g/mol) = 0.64 g

Hence produced mass of oxygen is 0.64 grams.

To now more about mass & moles, visit the below link:
brainly.com/question/18983376

#SPJ1

5 0
2 years ago
Which gas is used to take out blueprint​
tiny-mole [99]

The blueprint process

The best known is a process using ammonium ferric citrate and potassium . The paper is impregnated with a solution of ammonium ferric citr

7 0
3 years ago
A) fluorite<br><br> B) orthoclase<br><br> C) apatite<br><br> D) gypsum
Sidana [21]
C




I’m pretty sure it’s c
5 0
2 years ago
What is the electron (subshell) configurations for 15P
prisoha [69]
58 ce is the correct answer!
6 0
3 years ago
Read 2 more answers
Of the following solutions, which has the greatest buffering capacity?
Goshia [24]

Answer:

d. 0.121 M HC2H3O2 and 0.116 M NaC2H3O2

Explanation:

Hello,

In this case, since the pH variation is analyzed via the Henderson-Hasselbach equation:

pH=pKa+log(\frac{[Base]}{[Acid]} )

We can infer that the nearer to 1 the ratio of of the concentration of the base to the concentration of the acid the better the buffering capacity. In such a way, since the sodium acetate is acting as the base and the acetic acid as the acid, we have:

a. \frac{[Base]}{[Acid]}=\frac{0.497M}{0.365M}=1.36

b. \frac{[Base]}{[Acid]}=\frac{0.217M}{0.521M}=0.417

c. \frac{[Base]}{[Acid]}=\frac{0.713M}{0.821M}=0.868

d. \frac{[Base]}{[Acid]}=\frac{0.116M}{0.121M}=0.959

Therefore, the d. solution has the best buffering capacity.

Regards.

4 0
3 years ago
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