1.2*104 is 12000 it is easy just try ur best
Answer:
2
Explanation:
The subscript on Ammoniumwhich is (NH4) is 2.
Hope this helped!
Answer: The 234.74 grams of sample should be ordered.
Explanation:
Let the gram of 114 Ag to ordered be 
The amount required for the beginning of experiment = 0.0575 g
Time requires to ship the sample = 4.2hour = 252 min(1 hr = 60 min)
Half life of the sample =
= 21 min
![\log[N]=\log[N_o]-\frac{\lambda t}{2.303}](https://tex.z-dn.net/?f=%5Clog%5BN%5D%3D%5Clog%5BN_o%5D-%5Cfrac%7B%5Clambda%20t%7D%7B2.303%7D)
![\log[0.0575 g]=\log[N_o]-\frac{0.033 min^{-1}\times 252 min}{2.303}](https://tex.z-dn.net/?f=%5Clog%5B0.0575%20g%5D%3D%5Clog%5BN_o%5D-%5Cfrac%7B0.033%20min%5E%7B-1%7D%5Ctimes%20252%20min%7D%7B2.303%7D)
The 234.74 grams of sample should be ordered.
Answer:
pH =3.8
Explanation:
Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:
HA + H₂O ⇄ H₃O⁺ + A⁻ with Ka = [ H₃O⁺] x [A⁻]/ [HA]
The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.
In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:
HA H₃O⁺ A⁻
Initial, M 0.40 0 0
Change , M -x +x +x
Equilibrium, M 0.40 - x x x
Lets express these concentrations in terms of the equilibrium constant:
Ka = x² / (0.40 - x )
Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,
7.3 x 10⁻⁶ = x² / 0.40 ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³
[H₃O⁺] = 1.71 x 10⁻³
Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.
pH = - log ( 1.71 x 10⁻³ ) = 3.8
Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.
