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Vsevolod [243]
3 years ago
14

Suppose that a laboratory analysis of white powder showed 42.59% Na, 12.02% C, and 44.99% oxygen. Would you report that the comp

ound is sodium oxalate or sodium carbonate? (Use 43.38% Na, 11.33% C, and 45.29% O for sodium carbonate, and 34.31% Na, 17.93% C, and 47.76% O for sodium oxalate.)
Chemistry
1 answer:
butalik [34]3 years ago
8 0

Answer:

The white powder is sodium carbonate

Explanation:

If you supposed an ammount of 100 grams of the white powder,  it means there are 42.59 grams of Na, 12.02 grams of C and 44.99 grams of O. If you divided every compound by its molar mass, you will know the moles of every compound. (According to moles=mass (grams)/ molar mass (grams/mole)

C=12.01 g/mole, Na=23 g/mole, O=16 g/mole

So:

C=12.02/12.01= 1.0 moles, Na=42.59/ 23= 1.85 moles, O= 44.99/16= 2.81 moles

Knowing this, we can stimated that the white powder has 1 mol of C and approximately 2 and 3 moles of N and O, respectively

Chemical formula for sodium carbonate is Na2CO3 and the formula for sodium oxalate is Na2C2O4. So the white powder is Na2CO3

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Convert 1000mg=__g 1L=__mL 160cm=__mm 1.4km=__m 109g=__kg 250m=__km 80cm=__m 75mL=__L 5.6m=__cm 6.5g=__mg 170.4m=__cm 564Dg=__g
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Answer:

1000mg= 1g

1L= 1000 mL

160cm = 1600mm

1.4km= 1400m

109 g = 0.109kg

250m= 0.250 km

80cm= 0.8 m

75mL= 0.075L

5.6m= 560 cm

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564 Dg = 5640 g

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600 L=  0.6 KL

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Explanation:

1 mg = 1x10⁻³ g

1 g = 1000 mg

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1 Dg = 10 g

1 dg = 100 mg

1 L = 1000 mL

1 L = 1x10⁻³ KL

1 mL = 1x10⁻³ L

1 km = 1000 m

1 km = 1x10⁶ mm

1 m = 1x10⁻³ km

1 cm = 1x10⁻² m

1 cm = 10 mm

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3 years ago
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