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4 years ago

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Chemistry

1 answer:

larisa [96]4 years ago

3 0

Answer:

In FeO iron atom exist as Fe+ cation but in Fe2O3 iron atom exist as Fe3+ cation.

Explanation:

Feo=Fe+ + O-

Fe2O3 =2Fe3+ + 3O2-

The difference in the charge of iron cation deeply affect the compound formulas because in FeO the Fe cation has 1 unit positive charge where in Fe2O3 the Fe cation has 3 units positive charge.

That"s why these two chemical compounds made up of similar atoms differ in their chemical formulas.

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Complete the chart. Atomic Number: 10. 1s

Goshia [24]

2s 2p
3s 3p 3d
4s 4p 4d 4f
5s 5p 5d 5f
6s 6p 6d
7s 7p
hope it helps

7 0

3 years ago

Calcium has a cubic closest packed structure (fcc) as a solid. Assuming that calcium has an atomic radius of 197 pm, calculate t

Paha777 [63]

Answer:

$1.536 g/cm^3$ is the density of solid calcium.

Explanation:

Formula used :

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

where,

$\rho$ = density

Z = number of atom in unit cell

M = atomic mass

$(N_{A})$ = Avogadro's number

a = edge length of unit cell

We have:

r = 197 pm = $197\times 10^{-12} m = 197\times 10^{-12}\times 100 cm$

$r=197\times 10^{-10} m$

$a=r\times 2\sqrt{2}$

$a=197\times 10^{-10} m\times 2\sqrt{2}=557.200\times 10^{-10} cm$

M = 40 g/mol

Z = 4

On substituting all the given values , we will get the value of 'a'.

$\rho =\frac{4\times 40 g/mol}{6.022\times 10^{23} mol^{-1}\times (557.200\times 10^{-10} cm)^{3}}$

$\rho =1.536 g/cm^3$

$1.536 g/cm^3$ is the density of solid calcium.

5 0

3 years ago

How many moles would be in 85.OmL of 0.750M KOH?

Harlamova29_29 [7]

Answer: There are 0.0637 moles present in 85.0 mL of 0.750 M KOH.

Explanation:

Given: Volume = 85.0 mL (1 mL = 0.001 L) = 0.085 L

Molarity = 0.750 M

It is known that molarity is the number of moles of solute present in liter of a solution.

Therefore, moles present in given solution are calculated as follows.

$Molarity = \frac{moles}{Volume (in L)}\\0.750 M = \frac{moles}{0.085 L}\\moles = 0.0637 mol$

Thus, we can conclude that there are 0.0637 moles present in 85.0 mL of 0.750 M KOH.

7 0

3 years ago

JJ Thomson's cathode-ray tube experiment showed which of the following?

Flura [38]

The answer would be D

5 0

3 years ago

A 0.21 m3 drum contains 100L of a mixture ofseveral degreasing solvents in water.The concentraton oftrichloroethylen in the head

LiRa [457]

Answer:

The concentration is 6.42×10^-4 M

Explanation:

Number of moles trichloroethylene = pressure × volume/Henry's constant = 0.00301×0.21/0.00985 = 0.0642 mol

Volume of mixture = 100 L

Concentration of trichloroethylene = number of moles of trichloroethylene/volume of solution = 0.0642/100 = 6.42×10^-4 M

5 0

4 years ago