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Zepler [3.9K]
3 years ago
5

Use the periodic table to determine the molar mass of each of the following elements. Use the correct number of significant figu

res.
Potassium (K):
g/mol
Osmium (Os):
g/mol
Manganese (Mn):
g/mol
Chemistry
2 answers:
Mademuasel [1]3 years ago
8 0

Answer:

Potassium (K):   39.10 g/mol

Osmium (Os):   190.23 g/mol

Manganese (Mn):   54.94 g/mol

Explanation:

Just got it right on Edge 2021

Andrews [41]3 years ago
7 0

Answer:

Potassium (K):

⇒ 39.10 g/mol

Osmium (Os):

⇒ 190.23 g/mol

Manganese (Mn):

⇒ 54.94 g/mol

Explanation:

I Just Did The Assignment

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How many are molecules ( or formula) in each sample?
andre [41]

Answer:

  • 4.010 \times 10^{25} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}
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<u>Explanation</u>:

<u>Number of molecules for 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}</u>

\text { Firstly molar mass is calculated of } \mathrm{NaHCO}_{3}:

Atomic mass of Na + H + C + 3(O)  = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.00 g/mol

\text { Number of molecules of } \mathrm{NaHCO}_{3} \text { in } 55.93 \text { kg are as follows: }

55.93 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} \mathrm{NaHCO}_{3}}{84.00 \mathrm{gm} \mathrm{NaHCO}_{3}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number }\right)

=4.010 \times 10^{26} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}

<u>Number of molecules for for \left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}</u>

\text { Firstly molar mass is calculated of } \mathrm{Na}_{3} \mathrm{PO}_{4}

= Atomic mass of 3(Na) + P + 4(O)

= 3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol

459 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} N a_{3} P O_{4}}{163.94 \mathrm{gm} N a_{3} P O_{4}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number) } / 1 \mathrm{mol}\right.

=16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}

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3 years ago
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Answer:

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Instructions:
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Answer:

See explanation

Explanation:

A. Constitutional or structural isomers have the same molecular formula but different structural formulas.

B. Conformational isomers are compounds having the same atom to atom connectivity but differ by rotation about one or more single bonds.

C. Stereo isomers are compounds having the same molecular mass and atom to atom connectivity but different arrangement of atoms and groups in space.

I. Enantiomers are stereo isomers (optical isomers particularly) that are non-superimposable mirror images of each other.

II. Diasteromers are optical isomers that are not mirror images of each other.

Both diasteromers and enantiomers are types of optical isomers which in turn is one of the types of stereo isomers.

Stereo isomers differ from conformational isomers in that the arrangement of atoms in stereo isomers is permanent while conformational isomers results from free rotations in molecules about single bonds.

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<span>The molar mass is 169.09

304.3g/169.09g = 1.799mol which rounds to 1.800 mol</span>
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C. 1.92g/dm^3 (3 s.f.)

Explanation:

Please see attached picture for full solution.

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