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2 years ago

What pressure (in atm) is required to contain 0.034 moles of oxygen gas in a 5.1 L container at a temperature of 25.00C

Chemistry

1 answer:

kherson [118]2 years ago

5 0

Answer:

0.16 atm

Explanation:

Let's look at the formula PV=nRT. First of all, T should be in kelvin, so let's add 273. Now solve for P so we get P = nRT/V.

Now let's plug these in...

P = (0.034 mol)(0.08206 L atm mol^-1 K^-1)(298 K)/5.1 L

After solving and rounding to sig figs we get 0.16 atm

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101.325 kPa = 760 mmHg. So, 25.0 kPa is

gogolik [260]

101.325 kPa = 760 mmHg.
So, 25.0 kPa is 187.515 mmHg

5 0

3 years ago

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.

8 0

3 years ago

Which is a change or signal in the environment that can make an organism

nadya68 [22]

Answer:

I'm pretty sure its stimulus

7 0

3 years ago

Is BaSO3 a poly atomic covalent

earnstyle [38]

Answer:

no it is a ionic compound

5 0

3 years ago

I need help with this plz help

schepotkina [342]

Answer:

$\rm ^{103}_{\phantom{1}40}Zr$ , zirconium-103.

Explanation:

In a nuclear reaction, both the mass number and atomic number will conserve.

Let $^{A}_{Z}\mathrm{X}$ represent the unknown particle.

The mass number of a particle is the number on the upper-left corner. The atomic number of a particle is the number on its lower-left corner under the mass number. For example, for the particle $^{A}_{Z}\mathrm{X}$ , $A$ is the mass number while $Z$ while $Z$ is the atomic number.

Sum of mass numbers on the left-hand side of the equation:

$\underbrace{239}_{^{239}_{\phantom{2}94}\mathrm{Pu}} + \underbrace{1}_{^{1}_{0}\mathrm{n}} = 240$ .

Note that there are three neutrons on the right-hand side of the equation. Sum of mass numbers on the right-hand side:

$\underbrace{A}_{^{A}_{Z}\mathrm{X}} + \underbrace{134}_{^{134}_{\phantom{2}54}\mathrm{Xe}} + \underbrace{3\times 1}_{3\;^{1}_{0}\mathrm{n}} = A + 137$ .

Mass number conserves. As a result,

$A + 137 = 240$ .

Solve this equation for $A$ :

$A = 103$ .

Among the five choices, the only particle with a mass number of 103 is $\rm ^{103}_{\phantom{1}40}Zr$ . Make sure that atomic number also conserves.

3 0

3 years ago