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2 years ago

How much heat would be released by 1.3 kg of water if it's temperature went from 295 k to 274 k

Chemistry

1 answer:

kicyunya [14]2 years ago

4 0

Answer: The energy released will be 114.223 kJ.

Explanation:

To calculate the amount of heat released, we use the formula:

$Q= m\times c\times \Delta T$

Q = heat gained or released

m = mass of the substance = 1.3kg = 1300 g (Conversion factor: 1kg = 1000g)

c = heat capacity of water = 4.184 J/g K

$\Delta T={\text{Change in temperature}}=(274-295)K=-21K$

Putting values in above equation, we get:

$Q=1300g\times 4.184J/gK\times (-21)K$

Q = -114223 Joules = -114.223 kJ (Conversion Factor: 1kJ = 1000J)

Negative sign implies heat released.

Hence, heat released by 1.3 kg of water is 114.223kJ.

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2 years ago

Which statement describes a scientific theory?

vichka [17]

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C) is the best description, although a scientific theory is not necessarily a "fact". They are, however, based on evidence.

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3 years ago

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2 years ago

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A gas of unknown identity diffuses at a rate of 155 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate o

Ostrovityanka [42]

Answer:

19.07 g mol^-1

Explanation:

The computation of the molecular mass of the unknown gas is shown below:

As we know that

$\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }$

where,

Diffusion rate of unknown gas = 155 mL/s

CO_2 diffusion rate = 102 mL/s

CO_2 molar mass = 44 g mol^-1

Unknown gas molercualr mass = M_unknown

Now placing these values to the above formula

$\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}$

After solving this, the molecular mass of the unknown gas is

= 19.07 g mol^-1

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3 years ago