Answer:
6.68 X 10^-11
Explanation:
From the second Ka, you can calculate pKa = -log (Ka2) = 6.187
The pH at the second equivalence point (8.181) will be the average of pKa2 and pKa3. So,
8.181 = (6.187 + pKa3) / 2
Solving gives pKa3 = 10.175, and Ka3 = 10^-pKa3 = 6.68 X 10^-11
Atomic mass Hg = 200.59 u.m.a
200.59 g --------------- 6.02x10²³ atoms
( mass Hg ) ----------- 1.20 x10²² atoms
mass Hg = ( 1.20x10²² ) x 200.59 / 6.02x10²³
mass Hg = 2.407x10²⁴ / 6.02x10²³
= 3.998 g of Hg
hope this helps!
a. 34 mL; b. 110 mL
a. A tablet containing 150 Mg(OH)₂
Mg(OH)₂ + 2HCl ⟶ MgCl₂ + 2H₂O
<em>Moles of Mg(OH)₂</em> = 150 mg Mg(OH)₂ × [1 mmol Mg(OH)₂/58.32 mg Mg(OH)₂
= 2.572 mmol Mg(OH)₂
<em>Moles of HCl</em> = 2.572 mmol Mg(OH)₂ × [2 mmol HCl/1 mmol Mg(OH)₂]
= 5.144 mmol HCl
Volume of HCl = 5.144 mmol HCl × (1 mmol HCl/0.15 mmol HCl) = 34 mL HCl
b. A tablet containing 850 mg CaCO₃
CaCO₃ + 2HCl ⟶ CaCl₂ + CO₂ + H₂O
<em>Moles of CaCO₃</em> = 850 mg CaCO₃ × [1 mmol CaCO₃/100.09 mg CaCO₃
= 8.492 mmol CaCO₃
<em>Moles of HCl</em> = 8.492 mmol CaCO₃ × [2 mmol HCl/1 mmol CaCO₃]
= 16.98 mmol HCl
Volume of HCl = 16.98 mmol HCl × (1 mL HCl/0.15 mmol HCl) = 110 mL HCl
Answer : The value of rate of reaction is 
Explanation :
Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.
The given chemical equation is:
Rate law expression for the reaction is:
![\text{Rate}=k[NO]^a[O_2]^b](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BNO%5D%5Ea%5BO_2%5D%5Eb)
As per question,
a = order with respect to 
= 2
b = order with respect to 
= 1
Thus, the rate law becomes:
![\text{Rate}=k[NO]^2[O_2]^1](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BNO%5D%5E2%5BO_2%5D%5E1)
Now, calculating the value of rate of reaction by using the rate law expression.
Given :
k = rate constant = 
[NO] = concentration of NO = 
![[O_2]](https://tex.z-dn.net/?f=%5BO_2%5D)
= concentration of = 
Now put all the given values in the above expression, we get:
Hence, the value of rate of reaction is
Answer:
See Explanation
Explanation:
Let us consider the first two reactions, the initial concentration of CO was held constant and the concentration of Hbn was doubled.
2.68 * 10^-3/1.34 * 10^-3 = 6.24 * 10^-4/3.12 * 10^-4
2^1 = 2^1
The rate of reaction is first order with respect to Hbn
Let us consider the third and fourth reactions. The concentration of Hbn is held constant and that of CO was tripled.
1.5 * 10^-3/5 * 10^-4 = 1.872 * 10^-3/6.24 * 10^-4
3^1 = 3^1
The reaction is also first order with respect to CO
b) The overall order of reaction is 1 + 1=2
c) The rate equation is;
Rate = k [CO] [Hbn]
d) 3.12 * 10^-4 = k [5 * 10^-4] [1.34 * 10^-3]
k = 3.12 * 10^-4 /[5 * 10^-4] [1.34 * 10^-3]
k = 3.12 * 10^-4/6.7 * 10^-7
k = 4.7 * 10^2 mmol-1 L s-1
e) The reaction occurs in one step because;
1) The rate law agrees with the experimental data.
2) The sum of the order of reaction of each specie in the rate law gives the overall order of reaction.
