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dsp73
3 years ago
15

What is the percentage sodium in sodium oxide

Chemistry
2 answers:
Sav [38]3 years ago
5 0
Sodium oxide =-N220 i think
Alekssandra [29.7K]3 years ago
4 0
<span>
Sodium Oxide= Na2O
The formula mass of Na2O is (2x23) + 16 = 62g/mol

% Na= (46/62) x 100 = 74%
% O= (16/62) x 100 = 26%</span><span>
</span>
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Which of the following is true about the mass of atoms?
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2. Answer the following questions about a sample of calcium phosphate:
weqwewe [10]

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a) <u>310.18 g/mol</u>

<u>b) 4.352 moles Ca3(PO4)2</u>

<u>c) 2.6 * 10^24 molecules</u>

<u>d) 5.24 * 10^24 P atoms</u>

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Explanation:

Step 1: Data given

Atomic mass of Ca = 40.08 g/mol

Atomic mass of P = 30.97 g/mol

Atomic mass of O = 16.0 g/mol

Step 2: Calculate molecular weight of Ca3(PO4)2

Molecular weight of Ca3(PO4)2 = 3*atomic mass of Ca + 2* atomic mass of P and 8* atomic mass of O

Molecular weight of Ca3(PO4)2 = 3*40.08 + 2*30.97 + 8*16.0  =<u> 310.18 g/mol</u>

Step 3: Calculate moles of Ca3(PO4)2 in 1350 grams

Moles Ca3(PO4)2 = mass Ca3(PO4)2 /molar mass

Moles Ca3(PO4)2 = 1350 grams / 310.18 g/mol

Moles Ca3(PO4)2 = <u>4.352 moles</u>

Step 4: Calculate molecules in 1350 grams

Molecules = moles * number of Avogadro

Molecules = 4.352 moles * 6.02 * 10^23

Molecules = <u>2.6 *10^24 molecules</u>

<u />

Step 5: Calculate moles Phosphorus

For 1 mol Ca3(PO4)2 we need 2 moles P

For 4.352 moles Ca3(PO4)2 we have 2*4.352 = 8.704 moles

Step 6: Calculate P atoms

Atoms P = 8.704 moles * 6.02*10^23

Atoms P =<u> 5.24 * 10^24 P atoms</u>

<u />

Step 7: Calculate moles Calcium in 1350 grams

For 1 mol Ca3(PO4)2 we have 3 moles Ca

For 4.352 moles we have 3*4.352 = <u>13.056 moles Ca</u>

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<u>Step 8:</u> Calculate mass of 2.1 * 10^25 molecules of Ca3(PO4)2

Moles Ca3(PO4)2 = 2.1 * 10^25 / 6.02 * 10^23

Moles Ca3(PO4)2 = 34.9 moles

Mass Ca3(PO4)2 = 34.9 moles * 310.18 g/mol

Mass Ca3(PO4)2 = 10825.3 grams

4 0
3 years ago
A reaction that is spontaneous at low temperature can become non spontaneous at high temperature if δh is\
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Answer:

Reaction

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A reaction that is spontaneous at low temperature

6 0
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