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3 years ago

Calculate the final temperature when 50.0 mL of water at 65.0 °C are added to 25 mL of water at 25.0 °C.

Chemistry

1 answer:

jok3333 [9.3K]3 years ago

7 0

Answer:

x=51.66

Explanation:

1 ml = 1 gram

(mass) (Δt) (Cp) = (mass) (Δt) (Cp)

Substituting values into the above, we then have:

(25)(25-x)(4.184)=(50)(x-65)(4.18)

Solve for x

x= 51.6667

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2. (2 pts) How would you prepare 1.5 liters of 2 M KCI (MW=74.55 g/mol)

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Answer:

Dissolve 226 g of KCl in enough water to make 1.5 L of solution

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1. Calculate the moles of KCl needed

$n = \text{1.5 L} \times \dfrac{\text{2 mol}}{\text{1 L}}= \text{3.0 mol}$

2. Calculate the mass of KCl

$m = \text{3.0 mol} \times \dfrac{\text{74.55 g}}{\text{1 mol}}= \text{224 g}$

3. Prepare the solution

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3 years ago

How much heat is released when 75 g of octane is burned completely if the enthalpy of combustion is -5,500 kJ/mol CsH18? C8H18 +

aalyn [17]

Answer : The correct option is, (D) 3600 kJ

Explanation :

Mass of octane = 75 g

Molar mass of octane = 114.23 g/mole

Enthalpy of combustion = -5500 kJ/mol

First we have to calculate the moles of octane.

$\text{ Moles of octane}=\frac{\text{ Mass of octane}}{\text{ Molar mass of octane}}=\frac{75g}{114.23g/mole}=0.656moles$

Now we have to calculate the heat released in the reaction.

As, 1 mole of octane released heat = -5500 kJ

So, 0.656 mole of octane released heat = 0.656 × (-5500 kJ)

= -3608 kJ

≈ -3600 kJ

Therefore, the heat released in the reaction is 3600 kJ

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3 years ago