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Nesterboy [21]
3 years ago
9

How many grams of KBr are present in 250.0mL of a 1.20M solution

Chemistry
1 answer:
castortr0y [4]3 years ago
4 0
Molar mass KBr = 119 g/mol

Volume in liters: 250.0mL / 1000 => 0.25 L

n = M x V

n = 1.20 x 0,25 => 0.3 moles of KBr

Therefore:

1 mole KBr ----------- 119 g
0.3 moles KBr -------- ??

0.3 x 119 / 1 => 35.7 g of KBr 
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50 mL of 2.2 M HCl is combined with 50 mL of 2.0 M NaOH in a coffee-cup
GaryK [48]

Answer:

-56.1kJ/mol

Explanation:

The reaction between HCl and NaOH is:

NaOH + HCl → NaCl + H₂O + ΔH

<em>Where ΔH is heat change in  the reaction.</em>

<em />

As the temperature of the solution increases, the heat is released and ΔH < 0

The heat released in the reaction is obtained using coffe-cup calorimeter equation:

Q = C×m×ΔT

<em>Where Q is heat</em>

<em>C is specific heat of the solution (4.184J/g°C)</em>

<em>m is mass of solution: Assuming density = 1g/mL, 100mL of solution = 100g</em>

<em>And ΔT is change in temperature (13.4°C)</em>

<em />

Replacing:

Q = C×m×ΔT

Q = -4.184J/g°C×100g×13.4

Q = -5606.6J

Now, in the reaction you have:

<em>Moles HCl:</em>

0.050L * (2.2mol/L) = 0.11 moles

<em>Moles NaOH:</em>

0.050L * (2.0mol/L) = 0.1 moles

That means the moles of reaction are 0.1 moles, and heat change in the chemical reaction is:

5606.6J / 0.1 mol = 56066J =

<h3>-56.1kJ/mol</h3>

<em />

5 0
3 years ago
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