A 6-column table with 2 rows. The first column labeled number of washers has entries 1, 2. The second column labeled initial vel
2 answers:
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Answer:
Explanation: For the given reaction:
Concentrations of the ions are not given so we need to think about another way to calculate Kc.
We can calculate the free energy change using the standard cell potential as:
can be calculated using standard reduction potentials.
Standard reduction potential for zinc is -0.76 V and for silver, it is +0.78 V.
=
Reduction takes place at anode and oxidation at cathode. As silver is reduced, it is cathode. Zinc is oxidized and so it is anode.
= 0.78 V - (-0.76 V)
= 0.78 V + 0.76 V
= 1.54 V
Value of n is two as two moles of electrons are transferred in the cell reaction F is Faraday constant and its value is 96485 C/mol of electron .
= -297173.8 J
Now we can calculate Kc using the formula:
T = 25+273 = 298 K
R =
--297173.8 = -(8.314*298)lnKc
297173.8 = 2477.572*lnKc
lnKc = 119.946
<u>Answer:</u> The amount of heat absorbed is 141.004 kJ.
<u>Explanation:</u>
In order to calculate the amount of heat released while converting given amount of steam (gaseous state) to ice (solid state), few processes are involved:
(1):
(2):
(3):
(4):
Calculating the heat absorbed for the process having the same temperature:
......(i)
where,
q is the amount of heat absorbed, m is the mass of sample and is the enthalpy of fusion or vaporization
Calculating the heat released for the process having different temperature:
......(ii)
where,
= specific heat of solid or liquid
are final and initial temperatures respectively
- <u>For process 1:</u>
We are given:
Putting values in equation (i), we get:
- <u>For process 2:</u>
We are given:
Putting values in equation (i), we get:
- <u>For process 3:</u>
We are given:
Putting values in equation (i), we get:
- <u>For process 4:</u>
We are given:
Putting values in equation (i), we get:
Calculating the total amount of heat released:
(Conversion factor: 1 kJ = 1000J)
Hence, the amount of heat absorbed is 141.004 kJ.
Answer:
Explanation:
The half-life of Na-24 (15 h) is the time it takes for half of it to decay.
After one half-life, half (50 %) of the original amount will remain.
After a second half-life, half of that amount (25 %) will remain, and so on.
We can construct a table as follows:
No. of Fraction Mass
<u>half-lives t/da Remaining Remaining/g </u>
0 0 1 18
1 15 ½ 9.0
2 30 ¼ 4.5
3 45 ⅛ 2.2
4 60 ⅟₁₆ 1.1
5 75 ⅟₃₂ 0.56
6 90 ⅟₆₄ 0.28
We see that remain after five half-lives (75 h).