Answer:
The 9-v transistor battery is a direct current source, while the power supply from the wall socket is an alternating current source.
Explanation:
A direct current (DC) is one that flows continuously without any minimum and maximum values. This is majorly obtained from generators, batteries or cells etc.
An alternating current (AC) is a type that flows in a sinusoidal manner, having minimum and maximum value with respect to its flowing frequency. An example is the one obtained from the wall electrical source , generators etc.
An alternating current is not mainly used for electrolysis because of its nature. It would cause a switch of the electrodes in every cycle. Therefore, a direct current source is used.
The 9-v transistor battery is a direct current source, while the power supply from the wall is an alternating current source.
Question 1 :
V1/T1 = V2/T2
3.0L/273K = V2/373K
To get the value of Z, cross multiply
3.0L x 373K = 273K x V2
1119 = 273V2
Divide both sides by 273
1119/273 = 273V2/273
4.10L = V2
The new volume is 4.10 liters
Question 2 :
P1/T1 = P2 /T2
P1 = 880 kPA= 880 *10^3 Pa
T1 = 250 K
T2 = 303 K
P2 =?
Substituting for P2
P2 = P1 T2/ T1
P2 = 880 kPa * 303 / 250
P2 = 266,640 kPa/ 250
P2 = 1066.56 kPa.
The new pressure of the gas is 1066.56 kPa
Question 3 :
Given that:
Volume of gas V = 4.80L
(since 1 liter = 1dm3
4.80L = 4.80dm3)
Temperature T = 62°C
Convert Celsius to Kelvin
(62°C + 273 = 335K)
Pressure P = 2.9 atm
Number of moles of gas N = ?
Apply ideal gas equation
pV = nRT
2.9atm x 4.8dm3 = n x (0.0082 atm dm3 K-1 mol-1 x 335K)
13.92 atm dm3 = nx 2.747 atm dm3 mol-1
n = 13.92/2.747
n = 5.08 moles
There are 5.08 moles of gas contained in the sample
Question 4 :
Volume of gas V = 3.47L
(since 1 liter = 1dm3
3.47L = 3.47dm3)
Temperature T = 85.0°C
Convert Celsius to Kelvin
(85.0°C + 273 = 358K)
Pressure P = ?
Number of moles of gas N = 0.100 mole
Apply ideal gas equation
pV = nRT
p x 3.47dm3 = 0.10 x (0.0082 atm dm3 K-1 mol-1 x 358K)
p x 3.47dm3 = 0.29 atm dm3
p = (0.29 atm dm3 / 3.47 dm3)
p = 0.085 atm
If 1 atm = 760 mm Hg
0.085atm = 0.085 x 760
= 64.6 mm Hg
The pressure of the gas is 64.6 mm hg
Explanation:
The given reaction is as follows.
Hence, number of moles of NaOH are as follows.
n = 
= 0.005 mol
After the addition of 25 ml of base, the pH of a solution is 3.62. Hence, moles of NaOH is 25 ml base are as follows.
n = 
= 0.0025 mol
According to ICE table,
Initial: 0.005 mol 0.0025 mol 0 0
Change: -0.0025 mol -0.0025 mol +0.0025 mol
Equibm: 0.0025 mol 0 0.0025 mol
Hence, concentrations of HA and NaA are calculated as follows.
[HA] = 
[NaA] =
![[A^{-}] = [NaA] = \frac{0.0025 mol}{V}](https://tex.z-dn.net/?f=%5BA%5E%7B-%7D%5D%20%3D%20%5BNaA%5D%20%3D%20%5Cfrac%7B0.0025%20mol%7D%7BV%7D)
Now, we will calculate the 
value as follows.
pH = 
![pK_{a} = pH - log \frac{[A^{-}]}{[HA]}](https://tex.z-dn.net/?f=pK_%7Ba%7D%20%3D%20pH%20-%20log%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
= 
= 3.42
Thus, we can conclude that of the weak acid is 3.42.
The balanced chemical reaction describing this decomposition is as follows:
<span>4c3h5n3o9 .............> 6N2 + 12CO2 +10H2O + O2
From the periodic table:
mass of oxygen = 16 grams
mass of nitrogen = 14 grams
mass of hydrogen = 1 gram
mass of carbon = 12 grams
Therefore:
mass of </span><span>C3H5N3O9 = 3(12) + 5(1) + 3(14) + 9(16) = 227 grams
mass of O2 = 2(16) = 32 grams
From the balanced chemical equation:
4(227) = 908 grams of </span>C3H5N3O9 produce 32 grams of O2. Therefore, to know the amount of oxygen produced from 4.5*10^2 grams <span>C3H5N3O9, all we need to do is cross multiplication as follows:
amount of oxygen = (4.5*10^2*32) / (908) = 15.859 grams</span>
Does anyone know the answer?
