Question

calculate mass in gram, no. of moles,no.of molecules,no. of atom, no. of electron of 5.6 l co2 at stp ​

Answer 1

Answer:

Explanation:

Explanation:

1) 5.6L of co2 at stp

at stp p is 1atm t is 273 k

v given 5.6l

pv=nRT

1×5.6=273×0.0821×n

r=0.0821

n=0.25mol of CO2 molecule

2) 15 g CO

no of moles of co molecule =mass/molar mass

=15/28=0.5357mol of CO molecules

3) 4g H2

no of moles =mass/molar

molar mass of H2 is 2g/mol

4g/2g/mol=2mol of H2 molecule

4)20 gN2O4

molar mass of N2O4 is =92 g/mol

no of moles =mss/molar mass

=20g/92g/mol

=0.217mol of N2O4 MOLECULE

NOTE:: 1MOL = 6.02×10^23 WHICH IS AVOGADRO CONSTANT

SO IF YOU WANT YOU CAN MULTIPLY ANSWER WITH AVOGADRO NUMBER

Answer 2

Explanation:

Given: Volume = 5.6 L, Molar mass of CO₂ = 44 g

1) Gram molar mass of 22.4 CO₂ at STP = 44 g

   mass of 5.6 L of CO₂ at STP = 5.6 × 44/22.4

                                                    = 11 g

2) Number of moles:

  1 mole of CO₂ = 44 g

  x moles of CO₂= 11 g

  x = 11/44 = 0.25 moles

3) Number of molecules/atoms

44 g has 6.022 × 10²³ number of atoms/molecules

11 g will have =  6.022 × 10²³ × 11/44  

                        = 1.50 × 10²³ number of atoms/molecules

4) Number of electrons = atomic number of CO₂ × number of molecules

                                        = 22 × 1.50 × 10²³

                                        = 33 × 10²³