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kotegsom [21]
2 years ago
6

A student did an experiment to determine the specific heat capacity of a metal alloy. The student put a sample of the alloy in b

oiling water for several minutes, then quickly transferred the alloy into a calorimeter containing water originally at 25°C. The temperature of the water was monitored over time. The data are given in the graph above.
(a) What is the value of ΔT that the student should use to calculate the value of q, the heat gained by the water?

(b) In terms of what occurs at the particulate level, explain how the temperature of the water increases after the alloy sample is added.

(c) The student claims that thermal equilibrium is reached at time t. Justify the student’s claim. In your justification, include a description of what occurs at the particulate level when the alloy and the water have reached thermal equilibrium.

Chemistry
1 answer:
guapka [62]2 years ago
7 0

Answer:

a) ∆T=T1-T2

b) At the particle level the temperature changes are the result of the added energy causing the particles of water to move more vigorously. Either the particles of solid vibrate more vigorously about their fixed positions or the particles of liquid and gas move about their container more rapidly.

c) The state in which two substances in physical contact do not share any heat energy. The temperature of two substances in thermal equilibrium is said to be the same. Also see thermodynamics.

Explanation:

hope that helped good luck!

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Answer:

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Explanation:

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8 0
2 years ago
A solution is made by dissolving 23.5 grams of glucose (C6H12O6) in 0.245 kilograms of water. If the molal freezing point consta
aalyn [17]

Answer:

- 0.99 °C ≅ - 1.0 °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,

m is the molality of the solution (m = moles of solute / kg of solvent = (23.5 g / 180.156 g/mol)/(0.245 kg) = 0.53 m.

<em>∴ ΔTf = (Kf)(m)</em> = (-1.86 °C/m)(0.53 m) =<em> - 0.99 °C ≅ - 1.0 °C.</em>

4 0
3 years ago
Thomson discovered the what in his cathode ray experiment?
Leona [35]
He used a magnet to find that the negatively charged particles to be smaller then the atom.
He discovered electrons.
3 0
3 years ago
How many atoms are in a sample of 68.7 g copper (Cu)?
dlinn [17]
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8 0
3 years ago
Read 2 more answers
CH3COOH  CH3COO– + H+
Oxana [17]

(a)

pH = 4.77

; (b)

[

H

3

O

+

]

=

1.00

×

10

-4

l

mol/dm

3

; (c)

[

A

-

]

=

0.16 mol⋅dm

-3

Explanation:

(a) pH of aspirin solution

Let's write the chemical equation as

m

m

m

m

m

m

m

m

l

HA

m

+

m

H

2

O

⇌

H

3

O

+

m

+

m

l

A

-

I/mol⋅dm

-3

:

m

m

0.05

m

m

m

m

m

m

m

m

l

0

m

m

m

m

m

l

l

0

C/mol⋅dm

-3

:

m

m

l

-

x

m

m

m

m

m

m

m

m

+

x

m

l

m

m

m

l

+

x

E/mol⋅dm

-3

:

m

0.05 -

l

x

m

m

m

m

m

m

m

l

x

m

m

x

m

m

m

x

K

a

=

[

H

3

O

+

]

[

A

-

]

[

HA

]

=

x

2

0.05 -

l

x

=

3.27

×

10

-4

Check for negligibility

0.05

3.27

×

10

-4

=

153

<

400

∴

x

is not less than 5 % of the initial concentration of

[

HA

]

.

We cannot ignore it in comparison with 0.05, so we must solve a quadratic.

Then

x

2

0.05

−

x

=

3.27

×

10

-4

x

2

=

3.27

×

10

-4

(

0.05

−

x

)

=

1.635

×

10

-5

−

3.27

×

10

-4

x

x

2

+

3.27

×

10

-4

x

−

1.635

×

10

-5

=

0

x

=

1.68

×

10

-5

[

H

3

O

+

]

=

x

l

mol/L

=

1.68

×

10

-5

l

mol/L

pH

=

-log

[

H

3

O

+

]

=

-log

(

1.68

×

10

-5

)

=

4.77

(b)

[

H

3

O

+

]

at pH 4

[

H

3

O

+

]

=

10

-pH

l

mol/L

=

1.00

×

10

-4

l

mol/L

(c) Concentration of

A

-

in the buffer

We can now use the Henderson-Hasselbalch equation to calculate the

[

A

-

]

.

pH

=

p

K

a

+

log

(

[

A

-

]

[

HA

]

)

4.00

=

−

log

(

3.27

×

10

-4

)

+

log

(

[

A

-

]

0.05

)

=

3.49

+

log

(

[

A

-

]

0.05

)

log

(

[

A

-

]

0.05

)

=

4.00 - 3.49

=

0.51

[

A

-

]

0.05

=

10

0.51

=

3.24

[

A

-

]

=

0.05

×

3.24

=

0.16

The concentration of

A

-

in the buffer is 0.16 mol/L.

hope this helps :)

6 0
2 years ago
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