<span>Pentanol contains five carbons and has a hydroxyl group on the second carbon.
(C5H11OH)</span>
True
explanation:
i looked it up
Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
Note: This question is incomplete and lacks very important data to solve this question. But I have found the similar question which shows the profiles about which question discusses. Using the data from that question, I have solved the question.
a) We need to find the major species from A to F.
Major Species at A:
1. 
Major Species at B:
1. 
2. 
Major Species at C:
1. 
Major Species at D:
1. 
2. 
Major Species at E:
1. 
Major Species at F:
1. 
b) pH calculation:
At Halfway point B:
pH = pK
+ log[
]/[H
]
pH = pK
= 6.35
Similarly, at halfway point D.
At point D,
pH = pK
+ log [H
]/[H2
]
pH = pK
= 10.33
1. The box like figure in the given image is the [BATTERY SOURCE] from where the current drawn into the circuit.
2. A string connecting positive terminal of battery to the bulb is an [ELECTRIC WIRE] through which current flows in the circuit.
3. A bubble like object in the circuit is a [BULB] which lights up when current moves through the circuit.
4. A component connected to the negative terminal of batter source is a [SWITCH].
The answer are uppercase.
I hope this helps c:
Question options:
a) 2.05
b) 0.963
c) 0.955
d) 1.00
Answer:
b) 0.963
Explanation:
H2SO4→ HSO4- + H3O+
HSO4- + H2O ⇌ SO42- + H3O+
Construct ICE table:
HSO4- (aq) + H2O ⇌ SO42- (aq) + H3O+ (aq)
I 0.1 solid & 0 0.1
C -x liquid + x + x
E 0.1 - x are ignored x 0.1 + x
Calculate x
Ka = products/reactants
= ![\frac{[SO42-] [H3O+]}{[HSO4-]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BSO42-%5D%20%5BH3O%2B%5D%7D%7B%5BHSO4-%5D%7D)
0.011 = 
0.011 x (0.1 -x) = o.1x + x^2
0.0011 - 0.011 x - o.1x - x^2 = 0
0.0011 - 0.011 x - x^2 = 0
Use formula to solve for quadratic equation
x =
/ 2a
a = -1, b = -0.111, c = 0.001
Solve for x
x =
/ 2(-1)
x = 0.111 +,-
/ -2
x = 0.111 +,-
/ -2
x = 
x =
, x = 
x =
, x = 
x = - 0.12015 , x = 0.00915
x cannot be negative, so
x = 0.00915 M
Calculate [H3O+]
[H3O+] = 0.1 M + x
[H3O+] = 0.1 M + 0.00915 M
[H3O+] = 0.10915 M
Clculate pH
pH = - log [ H3O+]
pH = - log [ 0.10915]
pH = 0.963