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disa [49]
2 years ago
12

A piece of iron is heated to 120'C then placed in a calorimeter containing 50g of water. The water temperature is raised from 30

°C to 40°C. Assuming the specific heat of water is 4. 18 J/8'C and the specific heat of
Iron is 0. 44 J/g C, what is the mass of the piece of iron?

118. 8

59. 45

39. 6

752
Chemistry
1 answer:
Lana71 [14]2 years ago
3 0

Answer:

Mass of iron = 59.375 gm

Explanation:

Calories ( or joules) are added to the water by the hot steel so at the endpoint they are BOTH at 40 C

 The water gains:

     4.18  j/g-C  * 50 * (40-30 C) = 2090 j

The steel gave up 2090 j going from 120 to 40 C

2090 = .44 j/g-C  * m * (120-40)    solve fro m = 59.375 gm

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Make an observation about the average height of the beanbag for each mass dropped. How does it compare with your calculated kine
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For each of the following substituents, indicate whether it withdraws electrons inductively, donates electrons by hyperconjugati
aleksandrvk [35]

Answer:

Br- Withdraws electrons inductively

       Donates electrons by resonance

CH2CH3 - Donates electrons by hyperconjugation

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OCH3 -  Withdraws electrons inductively

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Explanation:

A chemical moiety may withdraw or donate electrons by resonance or inductive effect.

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Alkyl groups donate electrons by hyperconjugation involving hydrogen atoms.

-NHCH3  and contain species that have lone pair of electrons which can be donated by resonance. Also, the nitrogen and oxygen atoms are very electron withdrawing making the carbon atom to have a -I inductive effect.

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3 0
3 years ago
What is the freezing point of a solution of 465 g of sucrose c12h22o11 dissolved in 575 ml of water?
Amanda [17]

Answer:

The freezing point of the solution is - 4.39 °C.

Explanation:

We can solve this problem using the relation:

<em>ΔTf = (Kf)(m),</em>

where, ΔTf is the depression in the freezing point.

Kf is the molal freezing point depression constant of water = -1.86 °C/m,

density of water = 1 g/mL.

<em>So, the mass of 575 mL is 575 g = 0.575 kg.</em>

m is the molality of the solution (m = moles of solute / kg of solvent = (465 g / 342.3 g/mol)/(0.575 kg) = 2.36 m.

<em>∴ ΔTf = (Kf)(m</em>) = (-1.86 °C/m)(2.36 m) = <em>- 4.39 °C.</em>

<em>∵ The freezing point if water is 0.0 °C and it is depressed by - 4.39 °C.</em>

<em>∴ The freezing point of the solution is - 4.39 °C.</em>

6 0
2 years ago
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