Question

In: Chemistry

Complete the table: C^6H^12O^6 ---> 17.4 needs mole and number of particles Pb ---> needs mass...

Complete the table: C^6H^12O^6 ---> 17.4 needs mole and number of particles Pb ---> needs mass and moles 9.80x10^21 CF^4 24.0 kg ---> needs moles and numbers of particles C ---> needs mass and number of particles 0.0408 moles

Solutions

Expert Solution

Q1.

C6H12O6 = 17.4 grams

1 mol of C6H12O6 = 180 g/mol

then,

mol = mass/MW = 17.4/180 = 0.09666 moles

no of particles:

1 mol = 6.022*1023 particles

0.09666 mol = (0.09666)(6.022*10^23) = 5.820*10^22 particles

Q2.

9.8*10^21 particles of Pb,

1 mol of Pb = 6.022*10^23 partilces

mol of Pb = particles / Na = (9.8*10^21)/(6.022*10^23) = 0.016273 moles

1 mol of Pb = 207.2 g

then

0.016273 mol of Pb = 0.016273*207.2 = 3.37176 g of Pb

Q3.

24 kg of CF4 = 24000 g of CF4

1 mol of CF4 = 88.0043 g

mol of CF4 = mass/MW = (24000)/(88.0043) = 272.7139 mol

1 mol of CF4 = 6.022*10^23 particles

272.7139 mol --- (272.7139)(6.022*10^23) = 1.64228*10^26 particles of CF4

Q4.

1 mol of C = 12 g

then

0.0408 mol = 12*0.0408 = 0.4896 g of C

1 mol of C = 6.022*10^23 particles

0.0408 mol = (0.0408)(6.022*10^23) = 2.456976*10^22 particles of C


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