Answer:
B
Explanation:
In precipitation reaction, there is the emergence of a solid settling out of the solution. It can be visualized as a case in which a particular soluble solid becomes insoluble due to its reaction with another substance otherwise tagged as the precipitating reagent
It is thus reagent that draws it out of the solution by making it insoluble in its former solution. Hence it acts like it is calling out the substance by using a precipitating reagent
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
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Explanation:
As a neutral lithium atom contains 3 protons and its elemental charge is given as 
. Hence, we will calculate its number of moles as follows.
Moles = 
= 
= 100 mol
According to mole concept, there are 
atoms present in 1 mole. So, in 100 mol we will calculate the number of atoms as follows.
No. of atoms = 
= 
atoms
Since, it is given that charge on 1 atom is as follows.
= 
Therefore, charge present on atoms will be calculated as follows.
Thus, we can conclude that a positive charge of is in 0.7 kg of lithium.
