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docker41 [41]
3 years ago
6

A unit of heat energy that was formerly used frequently was the calorie. Look up the definition of the calorie in your textbook

or a handbook and record it here. Calculate the calorimeter constant for your calorimeter in cal/°C .
Chemistry
1 answer:
Svetlanka [38]3 years ago
3 0

Answer and Explanation:

Calorie is the unit of heat energy . There are 2 units with the same name 'calorie' which is widely used.

'The amount of heat energy required to increase the temperature of 1 gram of water by mass by 1^{\circ}C or 1 K is known as small calorie or gram calorie'.

Another one is large calorie which can be defined as :

'The amount of heat energy required to make arise in temperature of water 1 kg by mass by 1^{\circ}C or 1 K is known as large calorie or  kilcalorie and is represented as Cal or kcal'.

After the adoption of SI system, thee units of the metric system cal, C or kilocal are considered deprecated or obsolete with the SI unit for heat energy as 'joule or J'

1 cal = 4.184 J

1C or 1 kilocal = 4184 J

Calorimeter constant:

Calorimeter constant, represented as 'C_{cal}' is used to quantify the heat capacity or the amount of heat of a calorimeter.

It can be calculated by ther given formula:

{\displaystyle C_{cal}}={\frac {\Delta {H}}{\Delta {T}}}}}

where,

{\Delta {T}} = corresponding temperature change

{\Delta {H} = enthalpy change

Its unit is J/K or J/1^{\circ}C[/tex] which can be convertyed to cal/1^{\circ}C[/tex] by dividing the calorimeter constant by 4.184 or 4184 accordingly.

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The ph of a solution prepared by mixing 45.0 ml of 0.183 m koh and 35.0 ml of 0.145 m hcl is ________.
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Answer:

12.6.

Explanation:

  • We should calculate the no. of millimoles of KOH and HCl:

no. of millimoles of KOH = (MV)KOH = (0.183 M)(45.0 mL) = 8.235 mmol.

no. of millimoles of HCl = (MV)HCl = (0.145 M)(35.0 mL) = 5.075 mmol.

  • It is clear that the no. of millimoles of KOH is higher than that of HCl:

So,

[OH⁻] = [(no. of millimoles of KOH) - (no. of millimoles of HCl)] / (V total) = (8.235 mmol - 5.075 mmol) / (80.0 mL) = 0.395 M.

∵ pOH = -log[OH⁻]

∴ pOH = -log(0.395 M) = 1.4.

∵ pH + pOH = 14.

∴ pH = 14 - pOH = 14 - 1.4 = 12.6.

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Explanation:

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