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sveta [45]
3 years ago
11

How do you convert 642cg to kilograms

Chemistry
2 answers:
Eva8 [605]3 years ago
5 0
0.642 kg..................................................
murzikaleks [220]3 years ago
3 0
La habla espanol yo soy un pedro!
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Calculate the ionization energy, ????????, of the one‑electron ion Be3+. The electron starts in the lowest energy level, ????=1.
spin [16.1K]

Answer:

3.49X10⁻¹⁷ J

Explanation:

The energy of an electron is given by the equation:

E = - (2.18x10^{-18})Z^2(\frac{1}{n^2})

Where Z is the number of protons of the atom and n is the energy level of the electron. For Be, Z = 4.

When n tends to infinity (1/n²) tends to 0, and at this point, the electron has left the atom, so it has ionized.

The ionization energy then is the energy of the electron that left the atom less the energy of the electron in the energy level:

I.E = - (2.18x10⁻¹⁸)x4²x0 - (-(2.18x10⁻¹⁸)x4²x(1/1²))

I.E = 3.49X10⁻¹⁷ J

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3 years ago
Determine the mass of 5.20 moles of c6h12
Gwar [14]
Mass is equal to moles x molar mass, and the molar mass of C6H12 is 84, therefore the mass is 436.8 g, but 437 rounded to correct significant figures
5 0
4 years ago
The iodide ion concentration in a solution may be determined by the precipitation of lead iodide. Pb2 (aq) 2I-(aq) PbI2(s) A stu
MatroZZZ [7]

Answer:

M_{I^-}=0.6841M

Explanation:

Hello.

In this case, since the precipitation of the lead iodide is related to the iodide ion in solution, if we make react lead (II) nitrate with an iodide-containing salt, a possible chemical reaction would be:

Pb(NO_3)_2+2I^-\rightarrow PbI_2+2NO_3^-

In such a way, since 15.71 mL of a 0.5770-M solution of lead (II) nitrate precipitates out lead (II) iodide, we can first compute the moles of lead (II) nitrate in the solution:

n_{Pb(NO_3)_2}=0.5570\frac{molPb(NO_3)_2}{L}*0.01571L=0.01393molPb(NO_3)_2

Next, since there is a 1:2 mole ratio between lead (II) nitrate and iodide ions, we compute the moles of those ions:

n_{I^-}=0.01393molPb(NO_3)_2*\frac{2molI^-}{1molPb(NO_3)_2} =0.02785molI^-

Finally, since the mixing of the two solutions produce a final volume of 40.71 mL (0.04071 L), the resulting concentration (molarity) of the iodide ions in the student's unknown turns out:

M_{I^-}=\frac{0.02785molI^-}{0.04071L}\\\\ M_{I^-}=0.6841M

Best regards!

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3 years ago
Which part of the factory is most similar to the nucleus of a living cell
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The control room is similar to the nucleus, because the nucleus is in charge of cell functions.
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How many grams is 1.3 moles of tin?
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