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forsale [732]
4 years ago
8

What is the molarity of a aqueous solution of potassium hydroxide (KOH, 56.106 g/mol) if the molality of the solution is 5.2 mol

/kg
Chemistry
1 answer:
PIT_PIT [208]4 years ago
3 0

Answer:

The molarity is 8.52M

Explanation:

molality = moles of solute/kg of solvent = 5.2mol/kg

moles of solute = 5.2mole

mass of solute = moles of solute × MW = 5.2 × 56.106 = 291.8g

mass of solvent = 1kg = 1000g

mass of solution = mass of solute + mass of solvent = 291.8g + 1000g = 1291.8g

Density of KOH = 2120g/L

Volume of solution = mass/density = 1291.8/2120 = 0.61L

Molarity = number of moles of solute/volume of solution = 5.2/0.61 = 8.52M

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An electric range burner weighing 699.0 grams is turned off after reaching a temperature of 482.0°C, and is allowed to cool down
jasenka [17]

Answer:

0.42 J/gºC

Explanation:

We'll begin by calculating the heat energy used to heat up the water. This can be obtained as follow:

Mass (M) of water = 560 g

Initial temperature (T₁) = 22.7 °C

Final temperature (T₂) = 80.3 °C.

Specific heat capacity (C) of water = 4.18 J/gºC

Heat (Q) absorbed =?

Q = MC(T₂ – T₁)

Q = 560 × 4.18 (80.3 – 22.7)

Q = 2340.8 × 57.6

Q = 134830.08 J

Finally, we shall determine the specific heat capacity of the burner. This can be obtained as follow:

Mass (M) of burner = 699 g

Initial temperature (T₁) = 482.0°C

Final temperature (T₂) = 22.7 °C

Heat (Q) evolved = – 134830.08 J

Specific heat capacity (C) of the burner =?

Q = MC(T₂ – T₁)

–134830.08 = 699 × C (22.7 – 482.0)

–134830.08 = 699 × C × –459.3

–134830.08 = –321050.7 × C

Divide both side by –321050.7

C = –134830.08 / –321050.7

C = 0.42 J/gºC

Therefore, the specific heat capacity of the burner is 0.42 J/gºC

8 0
3 years ago
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BlackZzzverrR [31]
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Explanation:
the two-stage cooling method<span> is a </span><span>Food Code </span>counselled<span> procedure for cooling food in restaurants and foodservice </span>institutions<span>. </span>within the<span> two-stage cooling </span>methodology<span>, food is</span><span> cooled from 140° F (60° C) to 70° F (21° C) </span>among 2<span> hours and to 41° F (5° C) or lower </span>among<span> four hours. Use of this cooling </span>methodology<span> ensures that food is cooled quickly and safely and has no harmful effects.</span>
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Answer:

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