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Salsk061 [2.6K]
3 years ago
6

A liquid's freezing point is -38°C and it's boiling point is 357°C. What is the number of Kelvin between the boiling point and t

he freezing point of the liquid?
Chemistry
1 answer:
laiz [17]3 years ago
7 0
The Kelvin temperature K =℃ + 273.15. ℃ means centigrade temperature. So the centigrade temperature number between freezing point and boiling point is equal to the number of Kelvin. So the answer is 357-(-38)=395.
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Revolution means to rotate around a body and a year is how long it takes earth to orbit around the sun. Hope this helps ;)
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3 years ago
Given the balanced equation representing a reaction:Zn(s) +H2SO2(aq) ==> ZnSO4(aq) + H2(g)
miskamm [114]
Zn + H₂SO₄ = ZnSO₄ + H₂

answer (3) <span>single replacement

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3 years ago
How many moles are in 4.818 x 1024 chloride ions
erastovalidia [21]

Answer:

<h3>The answer is 8.00 moles</h3>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L}  \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{4.818 \times  {10}^{24} }{6.02 \times  {10}^{23} }  \\  = 8.00332225...

We have the final answer as

<h3>8.00 moles</h3>

Hope this helps you

4 0
3 years ago
When a 0.235-g sample of benzoic acid is combusted in a bomb calorimeter, the temperature rises 1.643 ∘C . When a 0.275-g sample
andrew-mc [135]

Answer : The heat of combustion per mole of caffeine at constant volume is 76197.18 kJ/mole

Explanation :

First we have to calculate the specific heat calorimeter.

Formula used :

Q=m\times c\times \Delta T

where,

Q = heat of combustion of benzoic acid = 26.38 kJ/g = 26380 J/g

m = mass of benzoic acid = 0.235 g

c = specific heat of calorimeter = ?

\Delta T = change in temperature = 1.643^oC

Now put all the given value in the above formula, we get:

26380J/g=0.235g\times c\times 1.643^oC

c=68323.38J/^oC

Thus, the specific heat of calorimeter is 68323.38J/^oC

Now we have to calculate the heat of combustion of caffeine.

Formula used :

Q=c\times \Delta T

where,

Q = heat of combustion of caffeine = ?

c = specific heat of calorimeter = 68323.38J/^oC

\Delta T = change in temperature = 1.584^oC

Now put all the given value in the above formula, we get:

Q=68323.38J/^oC\times 1.584^oC

Q=108224.23J=108.2kJ

Now we have to calculate the moles of caffeine.

\text{Moles of caffeine}=\frac{\text{Mass of caffeine}}{\text{Molar mass of caffeine}}

Mass of caffeine = 0.275 g

Molar mass of caffeine = 194.19 g/mole

\text{Moles of caffeine}=\frac{0.275g}{194.19g/mole}=0.00142mol

Now we have to calculate the heat of combustion per mole of caffeine at constant volume.

\text{Heat of combustion per mole of caffeine}=\frac{108.2kJ}{0.00142mol}=76197.18kJ/mole

Therefore, the heat of combustion per mole of caffeine at constant volume is 76197.18 kJ/mole

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Answer:

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3 years ago
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