type answer only. only ONLY answer if you know. EXERCISE 10 (continued) NAME ELM

Question

type answer only.

only ONLY answer if you know.

EXERCISE 10 (continued) NAME ELMAN 6. An organic compound is analyzed and found to be carbon 51.90%, hydrogen 9.80% and chlorine 38.30. What is the empirical formula of this compound? C s1.90 5140 COTOUUD (male ) - 4012521D EXE H 9.80 9 809 H 9.809 ) To aslom 26.32 38-309 (38.30 more = LOR 3.50 4.575 1.0 moldoet aan binnen broodheid so dobite - 7. A sample of oxygen gas, 0, weighs 28.4 g. How many molecules of O, and how many atoms of O are present in this sample? molecules of i atoms of o 8. A mixture of sand and salt is found to be 48 percent NaCl by mass. How many moles of NaCl are in 74 g of this mixture? 9. What is the mass of 2.6 X 102 molecules of ammonia, NH? 10. A water solution of sulfuric acid has a density of 1.67 g/mL and is 75 percent H So, by mass. How many moles of H50, are contained in 400. mL of this solution? -306 - shift

Explanation / Answer

Answer of (8)

48% of 74g is 35.52g

so there are 35.52 grams of salt.

The molar mass of NaCl is 58.443 grams per mole,

giving, 35.52 g/(58.443 grams per mole) = 0.608 moles of salt.

Answer of (9)  

Avogadro's number 6.022*10^23 particles of a substance equals one mole of NH3

Here, (2.6*10^23)/(6.022*10^23)

we get 0.432, and that is the no. of moles in the substance

The molar mass of NH3 is 17g/mol

To find the mass of 0.432 moles of NH3 using the molar mass of NH3, it gives

0.432mol*17g/mol

We get, 7.3 grams of NH3.

Answer of (7)

Here, the weigh of O2 gas is 28.4g.

The molar mass of O2 is 32 g/mol

Hence, 28.4g/(32 g/mol)= 0.8875 moles

the Avogadro's number is 6.022*10^23

the molecules of O2 is 0.8875 moles*(6.022*10^23)=5.3*10^23 molecules of O2

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