3 years ago

How much water must be added to 40.0 g of CaCl2 to produce a solution that is 35.0 wt% CaCl2?

Chemistry

1 answer:

elixir [45]3 years ago

3 0

Answer:

no te entiendo

Explanation:

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A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat in 0.200 L of solution

Dmitrij [34]

Answer:

The change in internal energy is - 1.19 kJ

Explanation:

Step 1: Data given

Heat released = 3.5 kJ

Volume calorimeter = 0.200 L

Heat release results in a 7.32 °C

Temperature rise for the next experiment = 2.49 °C

Step 2: Calculate Ccalorimeter

Qcal = ccal * ΔT ⇒ 3.50 kJ = Ccal *7.32 °C

Ccal = 3.50 kJ /7.32 °C = 0.478 kJ/°C

Step 3: Calculate energy released

Qcal = 0.478 kJ/°C *2.49 °C = 1.19 kJ

Step 4: Calculate change in internal energy

ΔU = Q + W W = 0 (no expansion)

Qreac = -Qcal = - 1.19 kJ

ΔU = - 1.19 kJ

The change in internal energy is - 1.19 kJ

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evablogger [386]

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