Awnser: The awnser is cloudy
Explanation: I don't have an explanation I just know it
Colligative properties calculations are used for this type of problem. Calculations are as follows:
ΔT(boiling point) = 101.02 °C - 100.0 °C= 1.02 °C
<span>ΔT(boiling point) = (Kb)m
</span>m = 1.02 °C / 0.512 °C kg / mol
<span>m = 1.99 mol / kg
</span><span>ΔT(freezing point) = (Kf)m
</span>ΔT(freezing point) = 1.86 °C kg / mol (<span>1.99 mol / kg)
</span>ΔT(freezing point) = 3.70 <span>°C
</span>Tf - T = 3.70 <span>°C
T = -3.70 </span><span>°C</span>
Answer:
490.83 oK
Explanation:
First of all, we better agree on the meaning of the word <em>adiabatically</em>. It means without the loss or gain of heat. So nothing is given up to or taken from the environment.
This also assumes that no change has occurred in the pressure.
T1 = 310 oK
T2 = ?
V1 = 12 L
V2 = 19 L
T1/V1 = T2/V2
310 / 12 = x/19 Multiply both sides by 19
310*19 / 12 = T2 Multiply 310 * 19 on the left.
5890 / 12 = T2 Divide by 12
490.83 = T2
The correct option is A.
All the elements in group 1 and hydrogen has only one electron in their outermost shells, thus, they all have similar electron configuration. The elements int the periodic table are grouped in such a way that, all elements in the same group have the same number of electrons in their outermost shells. This is one of the principal reason why hydrogen is grouped with group 1 elements.<span />
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: