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3 years ago

How many magnesium atoms are there in 1 molecule

Chemistry

2 answers:

jasenka [17]3 years ago

4 0

It is avogrado number. One molecue of magnesium has 6.023 x 10^23 atoms

kirill [66]3 years ago

3 0

If you have one atom of Mg you just have one. You could have many atoms as there are atoms of Mg in the universe. Maybe you mean proton=12, neutrons= 12, electron= 12.

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A solution is prepared by mixing 2.88 mol of acetone with 1.45 mol of cyclohexane at 30°C. 2nd attempt Calculate the χacetone an

Volgvan

First c<span>alculate the mole fraction of each substance:
acetone: 2,88 mol </span>÷ (2,88 mol + 1,45 mol) = 0,665.
cyclohexane: 1,45 ÷ (2,88 mol + 1,45 mol) = 0,335.
Raoult's Law:
P(total) = P(acetone) · χ(acetone) + P(cyclohexane) · χ(cyclohexane).
P(total) = 229,5 torr · 0,665 + 97,6 torr · 0,335.
P(total) = 185,3 torr.
χ for acetone: 229,5 torr · 0,665 ÷ 185,3 torr = 0,823.
χ for cyclohexane: 97,6 torr · 0,335 ÷ 185,3 torr = 0,177.

8 0

3 years ago

How many molecules are in 0.33 moles of CaCO3?

Nesterboy [21]

Answer: hope this is correct

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3 years ago

The energy of any one-electron species in its nth state (n = principal quantum number) is given by E = –BZ2 /n2 where Z is the c

Ivahew [28]

Explanation:

(a) The given data is as follows.

B = $2.180 \times 10^{-18} J$

Z = 4 for Be

Now, for the first excited state $n_{f}$ = 2; and $n_{i} = \infinity$ if it is ionized.

Therefore, ionization energy will be calculated as follows.

I.E = $\frac{-Bz^{2}}{\infinity^{2}} - (\frac{-2.180 \times 10^{-18} J /times (4)^{2}}{(2)^{2}})$

= $8.72 \times 10^{-18} J$

Converting this energy into kJ/mol as follows.

$8.72 \times 10^{-18} J \times 6.02 \times 10^{23} mol$

= 5249 kJ/mol

Therefore, the ionization energy of the $Be^{3+}$ ion in its first excited state in kilojoules per mole is 5249 kJ/mol.

(b) Change in ionization energy is as follows.

$\Delta E = -Bz^{2}(\frac{1}{(4)^{2}} - {1}{(2)^{2}}) = \frac{hc}{\lambda}$

$\frac{hc}{\lambda} = 0.1875 \times 2.180 \times 10^{-18} J \times (4)^{2}$

$\lambda = \frac{6.626 \times 10^{-34} \times 2.998 \times 10^{8} m/s}{0.1875 \times 2.180 \times 10^{-18} J \times 16}$

= $303.7 \times 10^{-10} m$

or, = $303.7^{o}A$

Therefore, wavelength of light given off from the $Be^{3+}$ ion by electrons dropping from the fourth (n = 4) to the second (n = 2) energy levels $303.7^{o}A$ .

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3 years ago

Balanced equation for the combination of magnesium and oxygen to form magnesium oxide

Digiron [165]

2Mg(s) + O2(g) --> 2MgO(s)

8 0

3 years ago

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Vitek1552 [10]

The less dense areas in a sound wave is called a rarefaction or rarefactions if there are multiple.

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3 years ago

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