The answer for the following mention bellow.
- <u><em>Therefore the final temperature of the gas is 260 k</em></u>
Explanation:
Given:
Initial pressure (
) = 150.0 kPa
Final pressure (
) = 210.0 kPa
Initial volume (
) = 1.75 L
Final volume (
) = 1.30 L
Initial temperature (
) = -23°C = 250 k
To find:
Final temperature (
)
We know;
According to the ideal gas equation;
P × V = n × R ×T
where;
P represents the pressure of the gas
V represents the volume of the gas
n represents the no of moles of the gas
R represents the universal gas constant
T represents the temperature of the gas
We know;
= constant
×
= 
Where;
(
) represents the initial pressure of the gas
(
) represents the final pressure of the gas
(
) represents the initial volume of the gas
(
) represents the final volume of the gas
(
) represents the initial temperature of the gas
(
) represents the final temperature of the gas
So;
= 
(
) =260 k
<u><em>Therefore the final temperature of the gas is 260 k</em></u>
<u><em></em></u>
From the calculations, the half life of the material is 6.5 days.
<h3>What is radioactivity?</h3>
The term radioactivity has to do with the spontaneous disintegration of a specie.
Uisng the formula;
N=Noe^-kt
N= amount at time t = 0.135 Ci or 4.995 ×10^9 Bq
No = amount initially present = 1.75 x 10^12 Bq
k = rate constant = ?
t = time taken = 55 days
Hence;
4.995 ×10^9 = 1.75 x 10^12e^-55k
4.995 ×10^9/1.75 x 10^12 = e^-55k
2.85 * 10^-3 = e^-55k
ln2.85 * 10^-3 = -55k
k = ln2.85 * 10^-3/-55
k = 0.1066 day-1
Half life = 0.693/ 0.1066 day-1
= 6.5 days
Learn more about radioactivity:brainly.com/question/1770619
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Distillation is the <span>method that can be used to seperate parts of liquid mixture when the entire mixture can pass through a filter.</span>
Answer:
Plants, algae, and a group of bacteria called cyanobacteria are the only organisms capable of performing photosynthesis
Answer: The mass of lead deposited on the cathode of the battery is 1.523 g.
Explanation:
Given: Current = 62.0 A
Time = 23.0 sec
Formula used to calculate charge is as follows.

where,
Q = charge
I = current
t = time
Substitute the values into above formula as follows.

It is known that 1 mole of a substance tends to deposit a charge of 96500 C. Therefore, number of moles obtained by 1426 C of charge is as follows.

The oxidation state of Pb in
is 2. So, moles deposited by Pb is as follows.

It is known that molar mass of lead (Pb) is 207.2 g/mol. Now, mass of lead is calculated as follows.

Thus, we can conclude that the mass of lead deposited on the cathode of the battery is 1.523 g.