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marissa [1.9K]
3 years ago
11

A sample of hexane (C6H14) has a mass of 0.580 g. The sample is burned in a bomb calorimeter that has a mass of 1.900 kg and a s

pecific heat of 3.21 J/giK. What amount of heat is produced during the combustion of hexane if the temperature of the calorimeter increases by 4.542 K?
A. 8.46 kJ
B. 16.1 kJ
C. 27.7 kJ
D. 47.8 kJ
Chemistry
1 answer:
MatroZZZ [7]3 years ago
6 0

Answer:

27.7 KJ

Explanation:

Q= mC dT

m= 1900 g+0.580 g= 1900.58 g

Q= (1900.58 g * 3.21 KJ/ gK* 4.542 K)

Q=27710 J= 27.7 KJ

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Orlov [11]

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I think the right answer is c/ number of atomic orbitals

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Two elements in the same period have the same number of _____ _____ in their electron clouds.
shutvik [7]

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outer electrons

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If 2,035 cal of heat is added to a 500.0 g sample of water at 35.0°C, what is the final
Lynna [10]

Answer:

39.1 °C

Explanation:

Recall the equation for specific heat:

q=mc \Delta T\\

Where q is the heat, m is the mass, c is the specific heat of the substance (in this case water), and delta T is the change in temperature.

You should know that the specific heat of water is 1 cal/g/C.

Using the information in the question:

2035=500(1)(T-35)\\2035=500T-17500\\500T=19535\\T=39.07

The final temperature is about 39.1 °C.

5 0
3 years ago
A sample of gas occupies 300 mL at 100K. What is its volume when the
iris [78.8K]

If the temperature of the sample of gas increases to the given value, the volume also increases to 600mL.

<h3>What is Charles's law?</h3>

Charles's law states that "the volume occupied by a definite quantity of gas is directly proportional to its absolute temperature.

It is expressed as;

V₁/T₁ = V₂/T₂

Given the data in the question;

  • Initial temperature of gas T₁ = 100K
  • Initial volume of gas V₁ = 300mL
  • Final temperature T₂ = 200K
  • Final volume V₂ = ?

V₁/T₁ = V₂/T₂

V₂ = V₁T₂ / T₁

V₂ = ( 300mL × 200K ) / 100K

V₂ = 60000mLK / 100K

V₂ = 600mL

Therefore, if the temperature of the sample of gas increases to the given value, the volume also increases to 600mL.

Learn more about Charles's law here: brainly.com/question/12835309

#SPJ1

3 0
2 years ago
A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base
tamaranim1 [39]

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ =  0.00375  mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

HN_3 + OH- ---> N^-_{3} + H_2O

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction =  0.025 +  0.0133 = 0.0383  L

Concentration of HN_3 = \dfrac{0.001755}{0.0383} = 0.0458 M

Concentration of N^{-}_3 = \dfrac{ 0.001995 }{0.0383} = 0.0521 M

GIven that :

Ka = 1.9 x 10^{-5}

Thus; it's pKa = 4.72

pH =4.72 +  log(\dfrac{ \ 0.0521}{0.0458})

pH =4.72 + log(1.1376)

pH =4.72 + 0.05598

pH =4.77598

pH ≅ 4.80

3 0
3 years ago
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