Explanation:
Metals are elements that ionized by loss of electrons.
Ionic and molecular compounds are usually non-metals.
Properties of metals:
- Metals have free mobile electrons and the metallic bonding ensures that.
- They are usually electropositive and freely looses their electrons.
- None of the metal is soluble without a chemical change occurring.
- They are ductile and malleable.
- Metals are good conductor or heat and electricity in their free uncombined state.
- They are lustrous.
B. The specific property of metals accountable for their unusual electrical conductivity is due to the presence of free mobile electrons in their lattices.
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Answer:
B
Explanation:
Lets see here, Gravity is pulling you down but your body is repelling to gravity. Meaning that gravity is pulling down your weight and the scale is reading it.
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Answer:
Crystalline solids, or crystals, have distinctive internal structures that in turn lead to distinctive flat surfaces, or faces. The faces intersect at angles that are characteristic of the substance. When exposed to x-rays, each structure also produces a distinctive pattern that can be used to identify the material.
Explanation:
The idea here is that you need to figure out how many moles of magnesium chloride,
MgCl
2
, you need to have in the target solution, then use this value to determine what volume of the stock solution would contain this many moles.
As you know, molarity is defined as the number of moles of solute, which in your case is magnesium chloride, divided by liters of solution.
c
=
n
V
So, how many moles of magnesium chloride must be present in the target solution?
c
=
n
V
⇒
n
=
c
⋅
V
n
=
0.158 M
⋅
250.0
⋅
10
−
3
L
=
0.0395 moles MgCl
2
Now determine what volume of the target solution would contain this many moles of magnesium chloride
c
=
n
V
⇒
V
=
n
c
V
=
0.0395
moles
3.15
moles
L
=
0.01254 L
Rounded to three sig figs and expressed in mililiters, the volume will be
V
=
12.5 mL
So, to prepare your target solution, use a
12.5-mL
sample of the stock solution and add enough water to make the volume of the total solution equal to
250.0 mL
.
This is equivalent to diluting the
12.5-mL
sample of the stock solution by a dilution factor of
20
.
