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omeli [17]
3 years ago
8

You need to prepare 2.00 L of 0.100 M Na2CO3 solution. The best procedure is to weigh out: ___________

Chemistry
2 answers:
lyudmila [28]3 years ago
7 0
It would be C I believe
jeka943 years ago
4 0

Answer:

The answer to your question is letter C

Explanation:

Data

Volume = 2 L

Molarity = 0.100 M

Molecular weight Na₂CO₃ = (2 x 23) + (1 x 12) + (3 x 16)

                                           = 46 + 12 + 48

                                           = 106 g

Process

1.- Calculate the grams of Na₂CO₃ needed

                         106 g ----------------  1 mol

                           x      ----------------  0.1 moles

                           x = (0.1 x 106) / 1

                           x = 10.6 g

2.- Calculate the grams of Na₂CO₃ needed for 2 liters of solution

                           10.6 g -------------- 1 liter

                            x        --------------  2 liters

                           x = (10.6 x 2) / 1

                           x = 21.2 grams of Na₂CO₃                        

                     

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4 years ago
Which statement describes the trend in first ionization energy for elements on the periodic table?
snow_tiger [21]
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3 0
4 years ago
Classify the chemical reaction shown here:
lesantik [10]

Answer:

Single-Replacement Reaction

General Formulas and Concepts:

<u>Chemistry - Reactions</u>

  • Synthesis Reactions: A + B → AB
  • Decomposition Reactions: AB → A + B
  • Single-Replacement Reactions: A + BC → AB + C
  • Double-Replacement Reactions: AB + CD → AD + BC

Explanation:

<u>Step 1: Define RxN</u>

Mg + H₂SO₄ → MgSO₄ + H₂

<u>Step 2: Identify RxN</u>

Mg + H₂SO₄ → MgSO₄ + H₂

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7 0
3 years ago
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kvv77 [185]

Answer:

Ammonia is the richest source of nitrogen on a mass percentage basis because it has 82.35% of nitrogen by mass.

Explanation:

Percentage of element in compound :

=\frac{\text{number of atoms}\times text{Atomic mass}}{\text{molar mas of compound}}\times 100

(a) Urea, (NH_2)_2CO

Molar mass of urea = 60 g/mol

Atomic mass of nitrogen = 14 g/mol

Number of nitrogen atoms = 2

N\%=\frac{2\times 14 g/mol}{60 g/mol}\times 100=46.67\%

(b) Ammonium nitrate, NH_4NO_3

Molar mass of ammonium nitrate = 80 g/mol

Atomic mass of nitrogen = 14 g/mol

Number of nitrogen atoms = 2

N\%=\frac{2\times 14 g/mol}{80 g/mol}\times 100=35.00\%

(c) Nitric oxide, NO

Molar mass of nitric oxide  = 30 g/mol

Atomic mass of nitrogen = 14 g/mol

Number of nitrogen atoms = 1

N\%=\frac{1\times 14 g/mol}{30 g/mol}\times 100=46.67\%

(d) Ammonia, NH_3

Molar mass of ammona = 17 g/mol

Atomic mass of nitrogen = 14 g/mol

Number of nitrogen atoms = 1

N\%=\frac{1\times 14 g/mol}{17 g/mol}\times 100=82.35\%​

Ammonia is the richest source of nitrogen on a mass percentage basis because it has 82.35% of nitrogen by mass.

4 0
4 years ago
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