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hichkok12 [17]
3 years ago
15

A 15.6 grams of ethanol absorb 868 J as it is heated. The initial temperature is 21.5 degrees Celsius. What is the final tempera

ture if the specific heat of ethanol is 2.41 Joules/ grams Celsius?
Chemistry
1 answer:
77julia77 [94]3 years ago
3 0
You have to use the equation q=mcΔT and solve for T(final).  
T(final)=(q/mc)+T(initial)
q=the amount of energy absorbed or released (in this case 868J)
m=the mass of the sample (in this case 15.6g)
c= the specific heat capacity of the substance (in this case 2.41 J/g°C)
T(initial)=the initial temperature of the sample (in this case 21.5°C)

When you plug everything in, you should get 44.6°C.
Therefore the final temperature of ethanol is 44.6°C

I hope this helps.  Let me know if anything is unclear.
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1) How many moles are in 4.0x10^24 atoms?
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Answer:

<h2>6.64 moles</h2>

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 \times  {10}^{24} }{6.02 \times  {10}^{23} }   \\  = 6.644518...

We have the final answer as

<h3>6.64 moles</h3>

Hope this helps you

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The formation of SO3 from SO2 and O2 is an intermediate step in the manufacture of sulfuric acid, and it is also responsible for
jeka57 [31]

Answer:

118.22 atm

Explanation:

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)      

KP = 0.13 = \frac{p(SO_{3})^{2}}{p(SO_{2})^{2}p(O_{2})}

Where p(SO₃) is the partial pressure of SO₃, p(SO₂) is the partial pressure of SO₂ and p(O₂) is the partial pressure of O₂.

  • With 2.00 mol SO₂ and 2.00 mol O₂ if there was a 100% yield of SO₃, then 2 moles of SO₃ would be produced and 1.00 mol of O₂ would remain.
  • With a 71.0% yield, there are only 2*0.71 = 1.42 mol SO₃, the moles of SO₂ that didn't react would be 2 - 1.42 = 0.58; and the moles of O₂ that didn't react would be 2 - 1.42/2 = 1.29.

The total number of moles is 1.42 + 0.58 + 1.29 = 3.29. With that value we can calculate the molar fraction (X) of each component:

  • XSO₂ = 0.58/3.29 = 0.176
  • XO₂ = 1.29/3.29 = 0.392
  • XSO₃ = 1.42/3.29 = 0.432

The partial pressure of each gas is equal to the total pressure (PT) multiplied by the molar fraction of each component.

  • p(SO₂) = 0.176 * PT
  • p(O₂) = 0.392 * PT
  • p(SO₃) = 0.432 * PT

Rewriting KP and solving for PT:

\frac{p(SO_{3})^{2}}{p(SO_{2})^{2}p(O_{2})}=0.13\\\frac{(0.432*P_{T})^{2}}{(0.176*P_{T})^{2}(0.392*P_{T})} =0.13\\\frac{0.1866*P_{T}^{2}}{0.0121*P_{T}^{3}} =0.13\\\frac{15.369}{P_{T}}=0.13\\P_{T}=118.22 atm

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