If you add 39 mL of distilled water to 48 mL of a 15 % (m/v)...

If you add 39 mL of distilled water to 48 mL of a 15 % (m/v) NaOH solution, what is the new concentration in % (m/v)?

A 21.8 g sample of water contains how many water molecules?

An unknown hydrocarbon compound was analyzed for hydrogen by elemental analysis and results show that it contains 15.88 % H. What is the empirical formula?

Determine the empirical formula of the compound with the following composition by mass: 92.26 percent C and 7.74 percent H.

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Expert Solution

If you add 39 mL of distilled water to 48 mL of a 15 % (m/v) NaOH solution, what is the new concentration in % (m/v)?

15% NaOH solution means, 15 g NaOH present in 100mL solution.

So, in 48 mL, amount of NaOH = (15/100) * 7.2 g

39 mL water was added to the solution, so final volume is 87 mL.

Thus, 7.2 g NaOH is present in 87 mL of water.

So in 100 mL, amount of NaOH is = (7.2/87)*100 = 8.27 g

So the new concentration is 8.27% (m/v)

A 21.8 g sample of water contains how many water molecules?

moles of water present = 21.8/18 = 1.211

Number of molecules of water present = 1.211 * 6.023 * 10²³ = 7.354 * 10²³

An unknown hydrocarbon compound was analyzed for hydrogen by elemental analysis and results show that it contains 15.88 % H. What is the empirical formula?

Some additional data required to calculate % of carbon. Otherwise,

Determine the empirical formula of the compound with the following composition by mass: 92.26 percent C and 7.74 percent H.

C = 92.26% = 7.688 mol%

H = 7.74% = 7.678 mol%

So, molar ratio of C and H is 1:1

Thus, emperical formula of the compound = C₁H₁

queen_honey_blossom answered 1 year ago

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