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Evgesh-ka [11]
3 years ago
8

How much heat, in calories, is needed to raise the temperature of 125.0 g of Lead (c lead = 0.130 J/g°C) from 17.5°C to 41.1°C?

Chemistry
1 answer:
pav-90 [236]3 years ago
8 0
95.6 cal
are needed.
Explanation:
Use the following equation:
q
=
m
c
Δ
T
,
where:
q
is heat energy,
m
is mass,
c
is specific heat capacity, and
Δ
T
is the change in temperature.
Δ
T
=
T
final
−
T
initial
Known
m
=
125 g
c
Pb
=
0.130
J
g
⋅
∘
C
T
initial
=
17.5
∘
C
T
final
=
42.1
∘
C
Δ
T
=
42.1
∘
C
−
17.5
∘
C
=
24.6
∘
C
Unknown
q
Solution
Plug the known values into the equation and solve.
q
=
(
125
g
)
×
(
0.130
J
g
⋅
∘
C
)
×
(
24.6
∘
C
)
=
400. J

(rounded to three significant figures)
Convert Joules to calories
1 J
=
0.2389 cal
to four significant figures.
400
.
J
×
0.2389
cal
1
J
=
95.6 cal

(rounded to three significant figures)
95.6 cal
are needed.
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GuDViN [60]

Answer:

41.17g

Explanation:

We are given the following parameters for Flourine gas(F2).

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Pressure = 4.00× 10³mmHG

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PV = nRT

Step 1

We find the number of moles of Flourine gas present.

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Converting to Kelvin

= °C + 273k

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= 296k

V = Volume = 5.00L

R = 0.08206L.atm/mol.K

P = Pressure (in atm)

In the question, the pressure is given as 4.00 × 10³mmHg

Converting to atm(atmosphere)

1 mmHg = 0.00131579atm

4.00 × 10³ =

Cross Multiply

4.00 × 10³ × 0.00131579atm

= 5.263159 atm

The formula for number of moles =

n = PV/RT

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The molar mass of Flourine gas =

F2 = 19 × 2

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3 0
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Explanation:

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Answer:

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Explanation:

To answer the question we need to know the following;

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