CH
3
C≡CH
2mole
HCl
A
CH
3
C(Cl)
2
CH
3
Heat
aq.KOH
B
CH
3
COCH
3
Answer:
B
Explanation:
first we apply electricity and by the help of circuits it will get toasted which means it will get heated so the answer is B
A.atomic mass(a)=16
atomic mass(b)=18
b.a and b are isotopes
c.(a)=2,6
(b)=2,6
Answer:
The mass of radon that decompose = 63. 4 g
Explanation:
R.R = P.E/(2ᵇ/ⁿ)
Where R.R = radioactive remain, P.E = parent element, b = Time, n = half life.
Where P.E = 100 g , b = 5.55 days, n = 3.823 days.
∴ R.R = 100/
R.R = 100/
R.R = 100/2.73
R.R = 36.63 g.
The mass of radon that decompose = Initial mass of radon - Remaining mass of radon after radioactivity.
Mass of radon that decompose = 100 - 36.63
= 63.37 ≈ 63.4 g
The mass of radon that decompose = 63. 4 g
Answer:
pH = 12.43
Explanation:
<em>...is titrating 212.7 mL of a 0.6800 M solution of hydrazoic acid (HN3) with a 0.2900 M solution of KOH. The p Ka of hydrazoic acid is 4.72. Calculate the pH of the acid solution after the chemist has added 571.6 mL of the KOH solution to it</em>.
To solve this question we need to know that hidrazoic acid reacts with KOH as follows:
HN3 + KOH → KN3 + H2O
<em>Moles KOH:</em>
0.5716L * (0.2900mol /L) =0.1658 moles of KOH
<em>Moles HN3:</em>
0.2127L * (0.6800mol/L) = 0.1446 moles HN3
As the reaction is 1:1, the KOH is in excess. The moles in excess of KOH are:
0.1658 moles - 0.1446 moles =
0.0212 mol KOH
In 212.7mL + 571.6mL = 784.3mL = 0.7843L
The molarity of KOH = [OH-] is:
0.0212 mol KOH / 0.7843L = 0.027M = [OH-]
The pOH is defined as -log [OH-]
pOH = -log 0.027M
pOH = 1.57
pH = 14 - pOH
pH = 12.43
