Answer:
Kₐ = 6.7 x 10⁻⁴
Explanation:
First lets write the equilibrium expression, Ka , for the dissociation of hydrofluoric acid:
HF + H₂O ⇄ H₃O⁺ + F⁻
Kₐ = [ H₃O⁺ ] [ F⁻ ] /[ [ HF ]
Since we are given the pH we can calculate the [ H₃O⁺ ] ( pH = - log [ H₃O⁺ ] , and because the acid dissociates into a 1: 1 relation , we will also have [F⁻ ]. The [ HF ] is given in the question so we have all the information that is needed to compute Kₐ.
pH = -log [ H₃O⁺ ]
1.68 = - log [ H₃O⁺ ]
Taking antilog to both sides of this equation:
10^-1.68 = [ H₃O⁺ ] ⇒ 2.1 X 10⁻² M= [ H₃O⁺ ]
[ F⁻ ] = 2.1 X 10⁻² M
Solving for Kₐ :
Kₐ = ( 2.1 X 10⁻² ) x ( 2.1 X 10⁻² ) / 0.65 = 6.7 x 10⁻⁴
(Rounded to two significant figures, the powers of 10 have infinite precision )
Answer:
Freezing point is -2.81°C
Explanation:
34g/342gmol^-1 = 0.0994mol
n = m/mr
Molarity= 0.994/ 0.66 = 1.51M
◇T = -i × m ×Kf
Where ◇T is freezing depression
i= Vant Hoff factor
m = molarity
Kf = freezing content = 1.
860kgmol^-1
◇T =-1 × 1.51 × 1.860 = - 2.81°C
If you only want to balance nuclear reactions, then you should know that number of nucleons are conserved before and after nuclear reaction. Also, charge is conserved as well.
Other things which are conserved in a nuclear reaction are:
Conservation of:
1. Parity
2. Spin
3. angular momentum(vector sum of intrinsic spin and orbital angular momentum)
4. linear momentum
5. Isotopic spin
6. Energy
A: C₆H₁₂O₆ + 6H₂O + 6O₂
6CO₂ + 12H₂O = C₆H₁₂O₆ + 6H₂O + 6O₂
Answer:
2Ba₃(PO₄)₂ +6SiO₂ ⇒ P₄O₁₀ +6BaSiO₃
Explanation:
Equating coefficients, you get ...
aBa₃(PO₄)₂ +bSiO₂ ⇒ cP₄O₁₀ +dBaSiO₃
For Ba: 3a = d
For P: 2a = 4c
For O: 8a +2b = 10c +3d
For Si: b = d
__
Expressing everything in terms of b and c, we get ...
d = b
a = b/3 = 2c
From the second, b = 6c, so we have ...
a = 2c
b = 6c
c = c
d = 6c
And we can write the equation with c=1 as ...
2Ba₃(PO₄)₂ +6SiO₂ ⇒ P₄O₁₀ +6BaSiO₃
