Answer:
![[A_t]=0.037\ M](https://tex.z-dn.net/?f=%5BA_t%5D%3D0.037%5C%20M)
Explanation:
(a)
Given that:
Rate constant, k = 0.0347 min⁻¹
The expression for the half-life is:-
Where, k is rate constant
So,
(b)
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Given that:
The rate constant, k = 
min⁻¹
t = 60 min
Initial concentration = 0.300 M
Final concentration =?
Applying in the above equation, we get that:-
![[A_t]=0.300\times e^{-0.0347\times 60}\ M](https://tex.z-dn.net/?f=%5BA_t%5D%3D0.300%5Ctimes%20e%5E%7B-0.0347%5Ctimes%2060%7D%5C%20M)
![[A_t]=0.3\times \frac{1}{e^{2.082}}\ M](https://tex.z-dn.net/?f=%5BA_t%5D%3D0.3%5Ctimes%20%5Cfrac%7B1%7D%7Be%5E%7B2.082%7D%7D%5C%20M)
<span>1.
</span>The equilibrium expression shows the ratio
between products and reactants. This expression is equal to the concentration
of the products raised to its coefficient divided by the concentration of the
reactants raised to its coefficient. The correct equilibrium expression for the
given reaction is:<span>
<span>H2CO3(aq) + H2O(l)
= H3O+(aq) + HCO3-1(aq)
Kc = [HCO3-1] [H3O+] / [H2O] [H2CO3]</span></span>
Answer:
Element D would have the lowest ionization.
(D) is correct option
Explanation:
Given that,
Number of electron in element A = 17
Number of electron in element B = 3
Number of electron in element C = 18
Number of electron in element D = 11
We know that,
The name of element A is chlorine which have 17 electron.
The name of element B is lithium which have 3 electron.
The name of element C is argon which have 18 electron.
The name of element D is sodium which have 11 electron.
We know that,
The ionization energy of the element decreases from up to down in group and increases from left to right in a period.
We need to find the element would have the lowest ionization
Using periodic table
According to periodic table the lowest Ionization energy of sodium.
Hence, Element D would have the lowest ionization.
(D) is correct option
Answer: Since the reaction is 2Ca(NO3)2 = 2CaO + 4NO2 + O2
1) - given the stoichiometric coefficients, we know that 2 moles of Ca(NO3)2 will produce 4 moles of NO2, hence, 1 mole will produce 2 moles of NO2
2) - 328 g produces 22.4 L, since One mole of any gas at S.T.P. occupies the same volume which is 22.4 L.
Hence, 65.6 g produces (65.6*22.4)/328 = 4.48 L
3) - 328 g produces 112 g CaO. Therefore, 65.6 g produces = (65.6*112)/328 = 22.4 g CaO.
4) - Given the stoichiometric coefficients, we know 5 moles of gaseous products are already being produces ( 4+1) by 2 moles of reactant.
5) - 44.8 L at STP = 2 moles of NO2, since One mole of any gas at S.T.P. occupies the same volume which is 22.4 L.
Hence, to produce 2 moles of NO2, we need 1 mole of reactant = 164 g
Hope this helps :)
Explanation:
