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Lelu [443]
2 years ago
5

A chemist carefully measures the amount of heat needed to raise the temperature of a 894.0g sample of a pure substance from to −

5.8°C to 17.5°C. The experiment shows that 4.90kJ of heat are needed. What can the chemist report for the specific heat capacity of the substance?
Chemistry
1 answer:
astraxan [27]2 years ago
4 0

Answer:

C = 0.2349 J/ (g °C)

Explanation:

Mass, m = 894.0g

Initial Temperature = −5.8°C

Final Temperature =  17.5°C

Temperature change = 17.5°C - (−5.8°C) = 23.3

Heat, H = 4.90kJ = 4900 J

Specific heat capacit, C = ?

The relationship between these quantities is given by the equation;

H = mCΔT

C = H / mΔT

C = 4900 / (894)(23.3)

C = 0.2349 J/ (g °C)

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3 years ago
2. The diagram below shows steps in the exothermic chemical reaction of bromomethane with
vovangra [49]

Answer:

Here's what I get  

Explanation:

1. Balanced equation

HQ⁻ + CH₃-Br ⟶ HQ-CH₃ + Br⁻

(I must use HQ because the Brainly Editor thinks the O makes a forbidden word)

2. Mechanism

HQ⁻ + CH₃-Br ⟶[HQ···CH₃···Br]⁻⟶ HQ-CH₃ + Br⁻

        A                           C                         B

The hydroxide ion attacks the back side of the carbon atom in the bromomethane (A).

At the same time as the Q-H bond starts to form, the C-Br bond starts to break.

At the half-way point, we have a high-energy intermediate (C) with partially formed C-O and C-Br bonds.

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See the diagram below.

 

8 0
3 years ago
If you added 15,000 calories to 2.0 L of water that was at 25.0 degrees C, what temperature would it be at when you finished?
rewona [7]

<u>Answer:</u>

<em>When we finish, the temperature would be 32.5℃</em>

<em></em>

<u>Explanation:</u>

Density of water = mass/volume

So,

Mass of water = Density × Volume

\\\\$=1.0   \times  2.0 L$\\\\$=1.0 \frac{g}{m L} \times 2000 m L$\\\\$\quad=2000 g$

$Q=m \times c \times \Delta T$

where

\Delta T = Final T - Initial T

Q is the heat energy in calories

c is the specific heat capacity (for water 1.0  cal/(g℃))  

m is the mass of water

plugging in the values  

$15000 \mathrm{Cal}=2000 \mathrm{g} \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times \Delta T$

\\$\Delta T=\frac{15000 \mathrm{cal}}{2000 \mathrm{g} \times \frac{1.0 \mathrm{cal}}{g^{\circ} \mathrm{C}}}$\\\\$\Delta T=7.5^{\circ} \mathrm{C}$

Final T = ∆T + Initial T

= 7.5℃ + 25℃ = 32.5℃ (Answer).

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