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lys-0071 [83]
2 years ago
11

The same amount of heat transferred into identical masses of different substances produces different temperature changes. Calcul

ate the final temperature when 1.00 kcal of heat transfers into 1.00 kg of steel, originally at 20.00; c = 0.108 kcal/kg 0C .
29.3 0 C


26.9 0 C


32.3 0 C


27.4 0 C
Chemistry
1 answer:
vivado [14]2 years ago
8 0

The assumption is Originally at 20°C

Now

  • m=1
  • Q=1
  • c=0.108

We need ∆T

  • Q=mc∆T
  • 1=1(0.108)∆T
  • 0.108∆T=1
  • ∆T=1/0.108=9.26°C

Now

  • ∆T=T_f-20
  • 9.26=T_f-20
  • T_f=29.26°C

Option A

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Given the balanced equation representing a reaction: H2 → H + H What occurs during this reaction? 1. Energy is absorbed as bonds
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In the given reaction, the bond between the hydrogen-hydrogen in H_2 are breaking into two hydrogen. That means during the bond breaking, some energy is required or absorbed to break the bonds.

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A student dissolves of 15 g aniline in of a solvent with a density of . The student notices that the volume of the solvent does
VikaD [51]

The given question is incomplete. The complete question is ;

A student dissolves of 15 g aniline in 200 ml of a solvent with a density of 1.05 g/ml. The student notices that the volume of the solvent does not change when the aniline dissolves in it. Calculate the molarity and molality of the student's solution. Be sure each of your answer entries has the correct number of significant digits.

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Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

Molarity=\frac{n\times 1000}{V_s}

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n= moles of solute  

V_s = volume of solution in ml = 200 ml

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Now put all the given values in the formula of molarity, we get

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Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

Molarity=\frac{n\times 1000}{W_s}

where,

n = moles of solute

W_s = weight of solvent in g

{\text {moles of solute}=\frac{\text {given mass}}{\text {molar mass}}=\frac{15g}{93g/mol}=0.16mol

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mass of solvent = mass of solution - mass of solute = (210 - 15) g = 195 g

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Therefore, the molality of solution is 0.82m

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A.

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