Answer:
5.17.
Explanation:
<em>∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.
</em>
[OH⁻] = 1.5 x 10⁻⁹ M.
∴ [H₃O⁺] = 10⁻¹⁴/[OH⁻] = 10⁻¹⁴/(1.5 x 10⁻⁹ M) = 6.66 × 10⁻⁶ M.
∵ pH = - log[H₃O⁺]
<em>∴ pH = - log(6.66 × 10⁻⁶ M) = 5.17.
</em>
Answer:
Zn =⇒ Zn+2(0.10) + 2e- (anode)
Zn+2(?M) + 2e- === Zn(s) (cathode)
Zn + Zn+2(?M) ===⇒ Zn+2(0.10) + Zn
E = E^o -0.0592 log Q; in this case E^o is zero.
E = - 0.0592 /n logQ where n is the number of electrons transferred, in this
case n = 2
23 mV x 1 volt/1000mv = 0.023 Volts
0.023 = -0.0592 / 2 log(0.10) / [Zn+2]
0.023 = -0.0296 { log 0.10 – log [Zn+2] }
0.023 = -0.0296{ -1 - log[Zn+2] }
0.023 = +0.0296 + 0.0296log[Zn+2]
-0.0066 = 0.0296log[Zn+2]
-0.22= log[Zn+2]
[Zn+2] = 10^-0.22 = 0.603 Molar
Answer:
Explanation:
A sigma bond is formed when a hybrid orbital sp overlaps with another hybrid orbital or with s- or p- orbital
Ethylene has a structural formula of H - C≡ C- H
At the ground state; we have :
C | ⇅ | |↑ | ↑
2s 2p
At the excited state, we have:
C | ↑ | |↑ | ↑ | ↑
the hybrid orbital
C = | ↑ | ↑ |
2sp
<span>The density of the solution =1.05 g/ml.
</span><span>The total mass of the resulting solution is = 398.7 g (CaCl2 + water)
</span>
Find moles of CaCl2 and water.
Molar mass of CaCl2 = 110 (approx.)
Moles of CaCl2 = 23.7 / 110 = 0.22
so, moles of Cl- ion = 2 x 0.22 = 0.44 (because each molecule of CaCl2 will give two Cl- ions)
Moles of water = 375 / 18 = 20.83
Now, Mole fraction of CaCl2 = (moles of CaCl2) / (total moles)
total moles = moles of Cl- ions + moles of Ca2+ ions + moles of water
= 0.44 + 0.22 + 20.83
=21.49
So, mole fraction = 0.44 / (21.49) = 0.02
Guess what !!! density is not used. No need
C. A central body is the center of the universe.
