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3 years ago

Calculate the number of moles found in 3.045x1024 atoms of helium. PLS HELP

Chemistry

1 answer:

Kisachek [45]3 years ago

4 0

Explanation:

so for this u have to use this equation where

Moles = number of particle/6.02×10^23

= 3.045 × 10^24/6.02×10^23

= 5.0581

write it to 3 S.F so 5.06 moles

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Answer : The initial temperature of the limestone is $1.05\times 10^{2}^oC$ .

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Mass of limestone = 62.6 g

Mass of water = 75 g

Final temperature of limestone = $51.9^oC$

Final temperature of water = $51.9^oC$

Initial temperature of water = $23.1^oC$

The specific heat capacity of limestone = $0.921J/g^oC$

The specific heat capacity of water = $4.186J/g^oC$

The formula used :

q = m × c × ΔT

where,

q = heat required

m = mass of an element

c = heat capacity

ΔT = change in temperature

According to this, the energy as heat lost by the system is equal to the gained by the surroundings.

Now the above formula converted and we get

$q_{system}=-q_{surrounding}$

$m_{system}\times c_{system}\times (T_{final}-T_{initial})_{system}= -[m_{surrounding}\times c_{surrounding}\times (T_{final}-T_{initial})_{surrounding}]$

$m_{limestone}\times c_{limestone}\times (T_{final}-T_{initial})_{limestone}= -[m_{water}\times c_{water}\times (T_{final}-T_{initial})_{water}]$

Now put all the given values in this formula, we get

$62.6g\times 0.921J/g^{0}C \times (51.9^{0}C-T_{\text{ initial limestone}})= -[75g\times 4.186J/g^{0}C \times (51.9^{0}C-23.1^oC)]$

By rearranging the terms, we get the value of initial temperature of limestone.

$T_{\text{ initial limestone}}=104.926^{0}C=1.05\times 10^{2}^oC$

Therefore, the initial temperature of the limestone is $1.05\times 10^{2}^oC$ .

8 0

3 years ago

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