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givi [52]
3 years ago
5

A sample of krypton gas with a volume of 10.00 L has a temp of 303 K and exerts

Chemistry
1 answer:
marshall27 [118]3 years ago
7 0

<u>Answer:</u> The final volume will be 14.85 L.

<u>Explanation:</u>

To calculate the final volume when temperature increases, we use Charles' Law.

This law states that volume is directly proportional to the temperature of the gas if number of moles and pressure remains constant.

V\propto T\\\\\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 = Initial volume and temperature

V_2\text{ and }T_2 = Final volume and temperature

We are given:

V_1=10L\\&#10;T_1=303K\\&#10;V_2=?L\\&#10;T_2=450K

Putting values in above equation, we get:

\frac{10L}{303K}=\frac{V_2}{450K}

V_2=14.85L

Hence, the final volume of the gas is 14.85L

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What does Hess's law state can be done in order to be able to react solid magnesium with oxygen gas safely (that is, without exp
aleksley [76]

Answer:

C. The reaction can be broken down and performed in steps

Explanation:

Hess's Law of Constant Heat Summation states that irrespective of the number of steps followed in a reaction, the total enthalpy change for the reaction is the sum of all enthalpy changes corresponding to all the steps in the overall reaction. The implication of this law is that  the change of enthalpy in a chemical reaction is independent of the pathway between the initial and final states of the system.

To obtain MgO safely without exposing magnesium to flame, the reaction sequence shown in the image attached may be carried out. Since the enthalpy of the overall reaction is independent of the pathway between the initial and final states of the system, the sum of the enthalpy of each step yields the enthalpy of formation of MgO.

3 0
2 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
A student has 100. mL of 0.400 M CuSO4 (aq) and is asked to make 100. mL of 0.150 M CuSO4 (aq) for a spectrophotometry experimen
GenaCL600 [577]

Answer:

We have to take 37.5 mL of a 0.400 M solution

Explanation:

Step 1: Data given

Stock volume = 100 mL  = 0.100L

Stock concentration 0.400 M

Volume of solution he wants to make = 100 mL = 0.100L

Concentration of solution he wants to make = 0.150 M

Step 2: Calculate the volume of 0.400 M CuSO4 needed

C1*V1 = C2*V2

⇒with C1 = the stock concentration = 0.400M

⇒with V1 = the volume of the stock = TO BE DETERMINED

⇒with C2 = the concentration of the solution he wants to make = 0.150 M

⇒with V2 = the volume of the solution made = 0.100 L

0.400 M * V1 = 0.150M * 0.100L

V1 = (0.150M*0.100L) / 0.400 M

V1 = 0.0375 L = 37.5 mL

We have to take 37.5 mL of a 0.400 M solution

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