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goblinko [34]
3 years ago
13

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

Chemistry
1 answer:
vazorg [7]3 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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Explanation:

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Answer:

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Based on gas law, it is possible to find the ΔH of a reaction using ΔH of half reactions.

Using the reactions:

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<em>(5) </em>CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH° = -890.3 kJ

The sum of 4×(4) + (5) gives:

4C(graphite) + 8H₂(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)

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Now, this reaction - 4×(1) gives:

4CO(g) + 8H₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)

ΔH° = -1189.54kJ - 4×-110.5 = <em>-747.54kJ</em>

<em></em>

Thus <em>ΔH° of the reaction is -747.54kJ</em>

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