Suppose that a laboratory analysis of white powder showed 42.59% na, 12.02% c, and 44.99% oxygen. would you report that the compound 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.)

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

the answer is the first option

explanation:

just googled it m8. not that hard

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water and alcohols have similar properties because water molecules contain hydroxyl groups that can form hydrogen bonds with other water molecules and with alcohol molecules, and likewise alcohol molecules can form hydrogen bonds with other alcohol molecules as well as with water.