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3 years ago

How many calories are absorbed in a process that absorbs 0.128 joules?

Chemistry

1 answer:

earnstyle [38]3 years ago

7 0

Answer:

There are 0.0305 calories in 0.128 joules

Explanation:

Given that,

Heat absorbed, Q = 0.128 J

We need to find the heat energy absorbed in calories.

We know that the relation between joules and calories is as follows :

1 calorie = 4.184 J

1 J = (1/4.184) J

So,

$0.128\ J=\dfrac{0.128 }{4.184}\\\\=0.0305\ cal$

So, there are 0.0305 calories in 0.128 joules

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At 298 K the standard enthalpy of combustion of sucrose is -5645 kJ/mol and the standard reaction Gibbs energy is -5798 kJ/mol.

natka813 [3]

Explanation:

The given data is as follows.

T = 298 K, $\Delta H^{o}$ = -5645 kJ/mol

$\Delta G^{o}$ = -5798 kJ/mol

Relation between $\Delta H$ and $\Delta G$ are as follows.

$\Delta G^{o}$ = $\Delta H^{o} - T \Delta S^{o}$

-5798 kJ/mol = -5645 kJ/mol - $298 \times \Delta S^{o}$

-153 kJ/mol = - $298 \times \Delta S^{o}$

$\Delta S^{o}$ = 0.513 kJ/mol K

Now, temperature is $37^{o}C$ = (37 + 273) K = 310 K

Since, $\Delta G$ = $\Delta H^{o} - T \Delta S^{o}$

= $-5645 kJ/mol - 310 K \times 0.513 kJ/mol K$

= (-5645 kJ/mol - 159.03 kJ/mol)

= -5804.03 kJ/mol

As, change in Gibb's free energy = maximum non-expansion work

$\Delta G = \Delta G_{310 K} - \Delta G_{298 K}$

= -5804.03 kJ/mol - (-5798 kJ/mol)

= -6.03 kJ/mol

Therefore, we can conclude that the additional non-expansion work is -6.03 kJ/mol.

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3 years ago

How many liters of HF can be made from 50.0 g of a 12.0M HF solution

aliina [53]

Answer:

Explanation:

The first thing we calculate are the moles: (50.0 g)/(20.0063 g/mol) =2.5 moles

Then we use the inverse formula for the molarity: (2.5 mol)/(12.0 M) = 0.208 L

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3 years ago

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vagabundo [1.1K]

Answer:

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Explanation:

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