Answer:
Amount of water
The thermometer
Explanation
In an experiment, there is always a dependent variable and an independent variable. The independent variable is manipulated and its effect on the dependent variable is observed.
The control is that factor in the experiment that must remain constant so that effect of the independent variable on the dependent variable can be observed.
In this case, the independent variable is the amount of sodium chloride while the dependent variable is the temperature at which the solution boils.
The controls must be the amount of water which must be held constant and the same thermometer used to measure the temperature so that the effect of the amount of sodium chloride on the temperature of the solution can be studied.
Answer:
i) The pCa before initiating the titration is 2.6
ii) The pCa is 6.67
Explanation:
please look at the solution in the attached Word file
Answer: Option (b) is the correct answer.
Explanation:
The given reaction is as follows.
In this reaction, potassium ions form bond with sulfate ions and results in the formation of potassium sulfate. Similarly, hydroxide ions combine with hydrogen ions to result in the formation of water.
Thus, we can conclude that it is a double replacement reaction occurs because potassium ions bond with sulfate ions to form a salt.
Answer:
The theoretical yield of
Li
3
N
is
20.9 g
.
Explanation:
Balanced Equation
6Li(s)
+
N
2
(
g
)
→
2Li
3
N(s)
In order to determine the theoretical yield, we must first find the limiting reactant (reagent), which will determine the greatest possible amount of product that can be produced.
Molar Masses
Li
:
6.941 g/mol
N
2
:
(
2
×
14.007
g/mol
)
=
28.014 g/mol
Li
3
N
:
(
3
×
6.941
g/mol Li
)
+
(
1
×
14.007
g/mol N
)
=
34.83 g/mol Li
3
N
Limiting Reactant
Divide the mass of each reactant by its molar mass, then multiply times the mole ratio from the balanced equation with the product on top and the reactant on bottom, then multiply times the molar mass of
Li
3
N
.
Lithium
12.5
g Li
×
1
mol Li
6.941
g Li
×
2
mol Li
3
N
6
mol Li
×
34.83
g Li
3
N
1
mol Li
3
N
=
20.9 g Li
3
N
Nitrogen Gas
34.1
g N
2
×
1
mol N
2
28.014
g N
2
×
2
mol Li
3
N
1
mol N
2
×
34.83
g Li
3
N
1
mol Li
3
N
=
84.8 g Li
3
N
Lithium produces less lithium nitride than nitrogen gas. Therefore, the limiting reactant is lithium, and the theoretical yield of lithium nitride is
20.9 g
.
Explanation:
From the equation;
4 Al + 3 O2 = 2 Al2O3
The mole ratio of Oxygen is to Aluminium hydroxide is 3:2.
Therefore; moles of Al2O3 is
(0.5/3 )× 2 = 0.333 moles
Therefore; The moles of aluminium oxide will be 0.333 moles