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krok68 [10]
3 years ago
9

Two bottles are connected with a valve. In bottle A, there are 2.4 L of nitrogen gas at 1.8 atm. In bottle B, there are 4.8 L of

nitrogen gas at 1.8 atm.
After opening the valve, what is the pressure of nitrogen gas?
Please answer in atmospheres.
I will give brainliest please answer quickly!
My
Chemistry
1 answer:
ycow [4]3 years ago
4 0

Answer:

1.8 atm

Explanation:

For bottle "A"

P1 = 1.8 atm V1 = 2.4L, P2 = ?, V2 = 2.4 + 4.8 L = 7.2 L,

Thus using P1•V1 = P2•V2, we have;

P2 = (1.8 × 2.4)/7.2 = 0.6 atm

For bottle "B"

P1 = 1.8 atm, V1 = 4.8L, P2 = ?, V2 = 7.2L,

P2 = (1.8 × 4.8)/7.2 = 1.2 atm

We will now add the individual pressures of each bottle;

Total pressure = P(A) + P(B)

Total pressure = 0.6 + 1.2

Total pressure = 1.8 atm.

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Answer:

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Explanation:

<em>Molecular Structure of Each Answer</em>

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C: OH-

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8 0
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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?
pav-90 [236]
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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