Question

A gas at STP which contain 6.023× 10 ^ 23atoms and forms diatomic molecule Will occupy the Volume in litre is​

Answer 1

ɠเѵεɳ :-

• A gas at STP which contain 6.023× 10²³ atoms

ƭσ ƒเɳ∂ :-

• Forms diatomic molecule Will occupy the Volume in litre is

αɳรωεɾ :-

• 11.2 L

εאρℓαɳαƭเσɳ :-

• As we know,

• $\sf 6.02×10²³ \: \: atoms = 1 \: \: mole$

• Number of moles of a diatomic molecule = $\sf \frac{1}{2} \: \: mole$

• Total volume occupied

• = 22.4×0.5

• = $\sf {\underline{ \underline {11.2L}}}$

I hope it helps you ❤️✔️

Answer 2

ɠเѵεɳ :-

ɠเѵεɳ :-A gas at STP which contain 6.023× 10²³ atoms

ɠเѵεɳ :-A gas at STP which contain 6.023× 10²³ atomsƭσ ƒเɳ∂ :-

ɠเѵεɳ :-A gas at STP which contain 6.023× 10²³ atomsƭσ ƒเɳ∂ :-Forms diatomic molecule Will occupy the Volume in litre is

ɠเѵεɳ :-A gas at STP which contain 6.023× 10²³ atomsƭσ ƒเɳ∂ :-Forms diatomic molecule Will occupy the Volume in litre isαɳรωεɾ :-

ɠเѵεɳ :-A gas at STP which contain 6.023× 10²³ atomsƭσ ƒเɳ∂ :-Forms diatomic molecule Will occupy the Volume in litre isαɳรωεɾ :-11.2 L

ɠเѵεɳ :-A gas at STP which contain 6.023× 10²³ atomsƭσ ƒเɳ∂ :-Forms diatomic molecule Will occupy the Volume in litre isαɳรωεɾ :-11.2 Lεאρℓαɳαƭเσɳ :-

ɠเѵεɳ :-A gas at STP which contain 6.023× 10²³ atomsƭσ ƒเɳ∂ :-Forms diatomic molecule Will occupy the Volume in litre isαɳรωεɾ :-11.2 Lεאρℓαɳαƭเσɳ :-As we know,

ɠเѵεɳ :-A gas at STP which contain 6.023× 10²³ atomsƭσ ƒเɳ∂ :-Forms diatomic molecule Will occupy the Volume in litre isαɳรωεɾ :-11.2 Lεאρℓαɳαƭเσɳ :-As we know,\begin{gathered} \sf 6.02×10²³ \: \: atoms = 1 \: \: mole \\ \end{gathered}6.02×10²³atoms=1mole

ɠเѵεɳ :-A gas at STP which contain 6.023× 10²³ atomsƭσ ƒเɳ∂ :-Forms diatomic molecule Will occupy the Volume in litre isαɳรωεɾ :-11.2 Lεאρℓαɳαƭเσɳ :-As we know,\begin{gathered} \sf 6.02×10²³ \: \: atoms = 1 \: \: mole \\ \end{gathered}6.02×10²³atoms=1moleNumber of moles of a diatomic molecule = \begin{gathered} \sf \frac{1}{2} \: \: mole \\ \end{gathered}21mole

ɠเѵεɳ :-A gas at STP which contain 6.023× 10²³ atomsƭσ ƒเɳ∂ :-Forms diatomic molecule Will occupy the Volume in litre isαɳรωεɾ :-11.2 Lεאρℓαɳαƭเσɳ :-As we know,\begin{gathered} \sf 6.02×10²³ \: \: atoms = 1 \: \: mole \\ \end{gathered}6.02×10²³atoms=1moleNumber of moles of a diatomic molecule = \begin{gathered} \sf \frac{1}{2} \: \: mole \\ \end{gathered}21moleTotal volume occupied

ɠเѵεɳ :-A gas at STP which contain 6.023× 10²³ atomsƭσ ƒเɳ∂ :-Forms diatomic molecule Will occupy the Volume in litre isαɳรωεɾ :-11.2 Lεאρℓαɳαƭเσɳ :-As we know,\begin{gathered} \sf 6.02×10²³ \: \: atoms = 1 \: \: mole \\ \end{gathered}6.02×10²³atoms=1moleNumber of moles of a diatomic molecule = \begin{gathered} \sf \frac{1}{2} \: \: mole \\ \end{gathered}21moleTotal volume occupied= 22.4×0.5

ɠเѵεɳ :-A gas at STP which contain 6.023× 10²³ atomsƭσ ƒเɳ∂ :-Forms diatomic molecule Will occupy the Volume in litre isαɳรωεɾ :-11.2 Lεאρℓαɳαƭเσɳ :-As we know,\begin{gathered} \sf 6.02×10²³ \: \: atoms = 1 \: \: mole \\ \end{gathered}6.02×10²³atoms=1moleNumber of moles of a diatomic molecule = \begin{gathered} \sf \frac{1}{2} \: \: mole \\ \end{gathered}21moleTotal volume occupied= 22.4×0.5= \begin{gathered} \sf {\underline{ \underline {11.2L}}} \\ \end{gathered}11.2L

ɠเѵεɳ :-A gas at STP which contain 6.023× 10²³ atomsƭσ ƒเɳ∂ :-Forms diatomic molecule Will occupy the Volume in litre isαɳรωεɾ :-11.2 Lεאρℓαɳαƭเσɳ :-As we know,\begin{gathered} \sf 6.02×10²³ \: \: atoms = 1 \: \: mole \\ \end{gathered}6.02×10²³atoms=1moleNumber of moles of a diatomic molecule = \begin{gathered} \sf \frac{1}{2} \: \: mole \\ \end{gathered}21moleTotal volume occupied= 22.4×0.5= \begin{gathered} \sf {\underline{ \underline {11.2L}}} \\ \end{gathered}11.2LI hope it helps you!!