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2 years ago

Given 6.23 mol Al find the atoms of Al

Chemistry

1 answer:

swat322 years ago

3 0

Answer:

The answer is

<h3>3.75 × 10²⁴ atoms of Al</h3>

Explanation:

To find the number of atoms of Al given it's number of moles we use the formula

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question

n = 6.23 mol

We have

N = 6.23 × 6.02 × 10²³

We have the final answer as

<h3>3.75 × 10²⁴ atoms of Al</h3>

Hope this helps you

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2 years ago

1. 100gm of a 55% (M/M) nitric acid solution is to be diluted to 20% (M/M) nitric acid.

stealth61 [152]

Answer:

The volume of water to be added is 0.175 liters of water

Explanation:

The given concentration of the nitric acid = 55% (M/M)

The mass of the nitric acid solution = 100 gm

The concentration solution is to diluted to = 20% (M/M)

The 100 g 55%(M/M) nitric acid solution gives 55g nitric acid in 100 g of solution

Therefore, to have 20% (M/M) nitric acid solution with the 55 g nitric acid, we get

Let "x" represent the volume of the resulting solution, we have;

20% of x = 55 g of nitric acid

∴ 20/100 × x = 55 g

x = 55 g × 100/20 = 275 g

The mass of extra water to be added = The mass of the 20%(M/M) solution solution of nitric acid - The current mass of the 55%(M/M) solution of nitric acid

The mass of extra water to be added = 275 g - 100 g = 175 g

Volume = Mass/Density

The density of water ≈ 1 g/ml

∴ The volume of water to be added that gives 175 g of water = 175 g/(1 g/ml) = 175 ml. = 0.175 l

The volume of water to be added = 0.175 liters of water.

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2 years ago

What mass (g) of magnesium nitride (Mg3N2) can be made from the reaction of 1.22 g of magnesium with excess nitrogen? __Mg + __N

Citrus2011 [14]

<h3>Answer:</h3>

1.69 g Mg₃N₂

<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis
Reactions RxN

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Unbalanced] Mg + N₂ → Mg₃N₂

[RxN - Balanced] 3Mg + N₂ → Mg₃N₂

[Given] 1.22 g Mg

[Solve] grams Mg₃N₂

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

[RxN] 3 mol Mg → Mg₃N₂

[PT] Molar Mass of Mg - 24.31 g/mol

[PT] Molar Mass of N - 14.01 g/mol

Molar Mass of Mg₃N₂ - 3(24.31) + 2(14.01) = 100.95 g/mol

<u>Step 3: Stoich</u>

[DA] Set up: $\displaystyle 1.22 \ g \ Mg(\frac{1 \ mol \ Mg}{24.31 \ g \ Mg})(\frac{1 \ mol \ Mg_3N_2}{3 \ mol \ Mg})(\frac{100.95 \ g \ Mg_3N_2}{1 \ mol\ Mg_3N_2})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 1.68873 \ g \ Mg_3N_2$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1.68873 g Mg₃N₂ ≈ 1.69 g Mg₃N₂

8 0

2 years ago