Answer:
3.4 × 10^23 molecules
Explanation:
To find the number of molecules present in C6H14, we multiply the number of moles in the compound by Avagadro's number (6.02 × 10^23 atoms).
number of molecules = number of moles (mol) × 6.02 × 10^23?
Number of molecules = 0.565 × 6.02 × 10^23
3.4 × 10^23 molecules
Number of Atoms in Gold for given mass can be calculated using following formula,
# of Moles = Number of Atoms / 6.022 × 10²³
Or,
Number of Atoms = Moles × 6.022 × 10²³ ------- (1)
Calculating Moles,
As,
Moles = Mass / M.mass
So,
Moles = 4.25 g / 196.96 g/mol
Moles = 0.0215
Putting value of mole in eq.1,
Number of Atoms = 0.0215 × 6.022 × 10²³
Number of Atoms = 1.299 × 10²²
Result:
4.25 g of Gold Nugget contains 1.299 × 10²² Atoms.
Answer : The mole fraction and partial pressure of 
and 
gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.
Explanation : Given,
Moles of 
= 1.79 mole
Moles of 
= 1.20 mole
Moles of = 3.71 mole
Now we have to calculate the mole fraction of and gases.
and,
and,
Thus, the mole fraction of and gases are, 0.267, 0.179 and 0.554 respectively.
Now we have to calculate the partial pressure of and gases.
According to the Raoult's law,
where,
= partial pressure of gas
= total pressure of gas = 5.78 atm
= mole fraction of gas
and,
and,
Thus, the partial pressure of and gases are, 1.54, 1.03 and 3.20 atm respectively.
Here's the balanced equation for given Double displacement reaction ~
The products fored are : Lead Iodide ( PbI2 ) and Potassium Nitrate ( KNO3 )
Answer:
2.5×10⁶ s
Explanation:
From the question given above, the following data were obtained:
Rate constant (K) = 2.8×10¯⁷ s¯¹
Half-life (t½) =?
The half-life of a first order reaction is given by:
Half-life (t½) = 0.693 / Rate constant (K)
t½ = 0.693 / K
With the above formula, we can obtain the half-life of the reaction as follow:
Rate constant (K) = 2.8×10¯⁷ s¯¹
Half-life (t½) =?
t½ = 0.693 / K
t½ = 0.693 / 2.8×10¯⁷
t½ = 2.5×10⁶ s
Therefore, the half-life of the reaction is 2.5×10⁶ s
