Mechanism for oxidation of alkene by KMnO₄ is provided in the attached image:
Answer:- 0.800 moles of the gas were collected.
Solution:- Volume, temperature and pressure is given for the gas and asks to calculate the moles of the gas.
It is an ideal gas law based problem. Ideal gas law equation is used to solve this. The equation is:
PV=nRT
Since it asks to calculate the moles that is n, so let's rearrange this for n:
V = 19.4 L
T = 17 + 273 = 290 K
P = 746 mmHg
we need to convert the pressure from mmHg to atm and for this we divide by 760 since, 1 atm = 760 mmHg
P = 0.982 atm
R = 
Let's plug in the values in the equation to get the moles.
n = 0.800 moles
So, 0.800 moles of the gas were collected.
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
Answer:
a. 7.8*10¹⁴ He⁺⁺ nuclei/s
b. 4000s
c. 7.7*10⁸s
Explanation:
I = 0.250mA = 2.5 * 10⁻³A
Q = 1.0C
1 e- contains 1.60 * 10⁻¹⁹C
But He⁺⁺ Carrie's 2 charge = 2 * 1.60*10⁻¹⁹C = 3.20*10⁻¹⁹C
(A).
No. Of charge per second = current passing through / charge
1 He⁺⁺ = 2.50 * 10⁻⁴ / 3.2*10⁻¹⁹C
1 He⁺⁺ = 7.8 * 10¹⁴ He⁺⁺ nuclei
(B).
I = Q / t
From this equation, we can determine the time it takes to transfer 1.0C
I = 1.0 / 2.5*10⁻⁴ = 4000s
(C).
Time it takes for 1 mol of He⁺⁺ to strike the target =?
Using Avogadro's ratio,
1.0 mole of He = (6.02 * 10²³ ions/mol ) * (1 / 7.81*10¹⁴ He ions)
Note : ions cancel out leaving the value of the answer in mols.
1.0 mol of He = 7.7 * 10⁸s
The volume of a gas that is required yo react with 4.03 g mg at STP is 1856 ml
calculation/
- calculate the moles of Mg used
moles=mass/molar mass
moles of Mg is therefore=4.03 g/ 24.3 g/mol=0.1658 moles
- by use of mole ratio of Mg:O2 from the equation which is 2:1
the moles 02=0.1679 x1/20.0829 moles
- at STP 1 mole of a gas= 22.4 l
0.0895 moles=? L
- =0.0895 moles x22.4 l/ 1 mole=1.8570 L
into Ml = 1.8570 x1000=1856 ml approximately to 1860
