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Anarel [89]
3 years ago
9

______ molecules have no net electrical charge

Chemistry
1 answer:
Reptile [31]3 years ago
4 0

Answer:

A nonpolar molecule has no separation of charge

Explanation:

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Hot water deep within earth can _______ minerals and transport them someplace else.
Vedmedyk [2.9K]

Answer:

Dissolve

Hope this helps!

Explanation:

5 0
2 years ago
PLEASEEE HELPPPP ASAP!!
Goshia [24]

Atoms are made up of three subatomic particles called protons, neutrons, and electrons.

Protons and neutrons are located in the nucleus.

All protons have a positive charge.

All neutrons have no charge or are neutral.

Electrons orbit around the nucleus and have a negative charge.

8 0
3 years ago
Read 2 more answers
Calculate the unit cell edge length for an 78 wt% Ag- 22 wt% Pd alloy. All of the palladium is in solid solution, and the crysta
BaLLatris [955]

Answer:

The unit cell edge length for the alloy is 0.405 nm

Explanation:

Given;

concentration of Ag, C_{Ag} = 78 wt%

concentration of Pd, C_P_d = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, A_A_g = 107.87 g/mol

atomic weight of iron, A_P_d = 106.4 g/mol

Step 1: determine the average density of the alloy

\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }

\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3

Step 2: determine the average atomic weight of the alloy

A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }

A _{Avg.} = \frac{100}{\frac{78}{107.87} + \frac{22}{106.4} } = 107.54 \ g/mole

Step 3: determine unit cell volume

V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell

Step 4: determine the unit cell edge length

Vc = a³

a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm= 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

7 0
3 years ago
What is the average atomic mass of all of the naturally occurring isotopes of nickel in amu?
My name is Ann [436]
In general chemistry, isotopes are substances that belong to one specific element. So, they all have the same atomic numbers. But they only differ in the mass numbers, or the number of protons and neutrons in the nucleus. In a nutshell, they only differ in the number of neutrons.

For Nickel, there are 5 naturally occurring isotopes. Their identities, masses and relative abundance are listed below

  Isotope                Abundance           Atomic Mass
   Ni-58                    68.0769%              <span>57.9353 amu
   Ni-60                    </span>26.2231%              <span>59.9308 amu
   Ni-61                    </span>1.1399 %               <span>60.9311 amu
   Ni-62                    </span>3.6345%                <span>61.9283 amu
   Ni-64                    </span>0.9256%                <span>63.9280 amu

To determine the average atomic mass of Nickel, the equation would be:
Average atomic mass = </span>∑Abundance×Atomic Mass

Using the equation, the answer would be:
Average atomic mass = 57.9353(68.0769%) + 59.9308(26.2231%) + 60.9311(1.1399%) + 61.9283(3.6345%) + 63.9280(0.9256%)

Average atomic mass = 58.6933 amu
3 0
4 years ago
<img src="https://tex.z-dn.net/?f=what%20%5C%3A%20is%20%5C%3A%20acid%20%5C%3A%20%20%20%5C%3A%20%20%5C%3A%20%7B%3F%7D%20" id="Tex
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5 0
3 years ago
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