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masya89 [10]
3 years ago
8

Why do ionic compounds make good conductors when dissolved in water?

Physics
1 answer:
sveticcg [70]3 years ago
5 0

Answer: The correct answer is B- The ions are free to move in water and carry an electric charge.

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Calculate the unit cell edge length for an 79 wt% Ag- 21 wt% Pd alloy. All of the palladium is in solid solution, and the crysta
ankoles [38]

Answer:

The edge length is 0.4036 nm

Solution:

As per the question:

Density of Ag, \rho = 10.49 g/cm^{3}

Density of Pd, \rho = 12.02 g/cm^{3}

Atomic weight of Ag, A = 107.87 g/mol

Atomic weight of Pd, A' = 106.4 g/mol

Now,

The average density, \rho_{a} = \frac{n A_{avg}} {V_{c}\times N_{A}}

where

V_{c} = a^{3}  = Volume of crystal lattice

a = edge length

n = 4 = no. of atoms in FCC

Therefore,

\rho_{a} = = \frac{n A_{avg}} {V_{c}\times N_{A}}

Therefore, the length of the unit cell is given as:

a = (\frac{nA_{avg}}{\rho_{a}\times N_{a}})^{1/3}            (1)

Average atomic weight is given as:

A_{avg} = \frac{100}{\frac{C_{Ag}}{A_{Ag}} + \frac{C_{Pd}}{A_{Pd}}}

where

C_{Ag} = 79 %

A_{Ag} = 107

C_{Pd} = 21%

A_{Pd} = 106

Therefore,

A_{avg} = \frac{100}{\frac{79}{107} + \frac{21}{106}} = 106.78

In the similar way, average density is given as:

\rho_{a} = \frac{100}{\frac{C_{Ag}}{\rho_{Ag}} + \frac{C_{Pd}}{\rho_{Pd}}}

\rho_{a} = \frac{100}{\frac{79}{10.49} + \frac{21}{12.02}} = 10.78 g/cm^{3}

Therefore, edge length is given by eqn (1) as:

a = (\frac{4\times 106.78}{10.78\times 6.023 X 10^23})^{1/3} = 4.036\times 10^{- 8} cm = 0.4036\times 10^{- 9} m = 0.4036 nm

5 0
3 years ago
What type of energy does a stove use to convert into heat
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4 0
3 years ago
Read 2 more answers
Help with this please.
gregori [183]
Rutherford's experiment<span> utilized positively charged alpha particles (He with a +2 charge) which were deflected by the dense inner mass (nucleus). The conclusion that could be formed from this result was that </span>atoms<span> had an inner core which contained most of the mass of an </span>atom<span> and was positively charged.</span>
8 0
4 years ago
A skier moving at 4.75 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220
disa [49]
First we need to find the acceleration of the skier on the rough patch of snow.
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force, \mu m g. For Newton's second law, the resultant of the forces acting on the skier must be equal to ma (mass per acceleration), so we can write:
ma=-\mu m g
Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:
a=-(0.220)(9.81 m/s^2)=-2.16 m/s^2

Now we can use the following relationship to find the distance covered by the skier before stopping, S:
2aS=v_f^2-v_i^2
where v_f=0 is the final speed of the skier and v_i=4.75 m/s is the initial speed. Substituting numbers, we find:
S=- \frac{v_i^2}{2a}=- \frac{(4.75 m/s)^2}{2(-2.16 m/s^2)}=5.23 m
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3 years ago
In which of the following is positive work done by a person on a suitcase
MrRa [10]

pendajo sMh so stupid

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4 years ago
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