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4 years ago

How many grams of Nacl would you need to prepare 200 ml of a 5 M SOLUTION. ?

Chemistry

1 answer:

vazorg [7]4 years ago

8 0

Answer:

58.44 g of NaCl are needed.

Explanation:

Given data:

Mass of NaCl needed = ?

Volume of solution = 200 mL (200/1000 =0.2 L)

Molarity of solution = 5 M

Solution:

We will solve this problem through molarity formula.

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

Now we will put the values.

5 M = moles of solute / 0.2 L

Moles of solute = 5 mol/L × 0.2 L

Moles of solute = 1 mol

Mass of sodium chloride:

Mass = number of moles × molar mass

Mass = 1 mol × 58.44 g/mol

Mass = 58.44 g

Thus, 58.44 g of NaCl needed.

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3 years ago

Which formula represents a saturated hydrocarbon

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During the naval battles of the South Pacific in World War II, the U.S. Navy produced smoke screens by spraying titanium tetrach

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Explanation:

The reaction given is;

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From the reaction;

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b. How many moles of HCl are formed when 8.44 moles of TiCl4 react?

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3 years ago

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<u><em>Hydroxylation </em></u>

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3 years ago

How many molecules are in 0.55 moles of Cu(NO3)2?

xeze [42]

<h3>Answer:</h3>

3.3 × 10²³ molecules Cu(NO₃)₂

<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>

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

0.55 mol Cu(NO₃)₂

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

Avogadro's Number

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 0.55 \ mol \ Cu(NO_3)_2(\frac{6.022 \cdot 10^{23} \ molecules \ Cu(NO_3)_2}{1 \ mol \ Cu(NO_3)_2})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 3.3121 \cdot 10^{23} \ molecules \ Cu(NO_3)_2$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

3.3121 × 10²³ molecules Cu(NO₃)₂ ≈ 3.3 × 10²³ molecules Cu(NO₃)₂

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3 years ago