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Jet001 [13]
3 years ago
6

IM DUMMY STUPID AND I NEED HELP. True or False. A force is a push or pull exerted on an object.

Chemistry
2 answers:
trasher [3.6K]3 years ago
4 0

its true cuz of newtons laws?????? ok free points

boyakko [2]3 years ago
3 0

Answer: true

Explanation:

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What quantity of aluminium is deposited when a current of 10A is passed through a solution of aluminium salt for 1930s? (All=27,
Arlecino [84]

Answer:

AL*3+ + 3e- =AL.

(10A×1930)÷96500=0.2mole e-

0.2÷3×27=1.8g(AL)

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3 years ago
Lens coating are made of A.teflon B.bakelite C.melamine D.PVC
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Answer:

C. melamine

Explanation:

a thin coating that eliminates reflections and glare from the front and back surfaces of your lenses.

5 0
3 years ago
Corrosion may be regarded as the destruction of metal by:
NeX [460]
All of the above I think it might not be right
4 0
2 years ago
In a physical change, the _____ does not change.
Ksivusya [100]

Physical changes occur when objects or substances undergo a change that does not change their chemical composition. This contrasts with the concept of chemical change in which the composition of a substance changes or one or more substances combine or break up to form new substances.

7 0
3 years ago
Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit
ser-zykov [4K]

<u>Answer:</u>

<u>For a:</u> The edge length of the unit cell is 314 pm

<u>For b:</u> The radius of the molybdenum atom is 135.9 pm

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the edge length for given density of metal, we use the equation:

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density = 10.28g/cm^3

Z = number of atom in unit cell = 2  (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

N_{A} = Avogadro's number = 6.022\times 10^{23}

a = edge length of unit cell =?

Putting values in above equation, we get:

10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm

Conversion factor used:  1cm=10^{10}pm  

Hence, the edge length of the unit cell is 314 pm

  • <u>For b:</u>

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

R=\frac{\sqrt{3}a}{4}

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

R=\frac{\sqrt{3}\times 314}{4}=135.9pm

Hence, the radius of the molybdenum atom is 135.9 pm

4 0
3 years ago
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