Answer:
b < c < a < d
Explanation:
The weak acid with the lowest pKa will be the most acidic. In the other way, the conjugate base which the acid is weak will be strong.
The weak base with the lowest pKb will be the most basic. And the conjugate base of the weak base will be a strong acid.
Ka Acetic acid = 1.8x10-5
Ka HCN = 1.9x10-10
Kb pyridine = 1.7x10-9
Kb NH3 = 1.8x10-5
NH4Br is the conjugate base of a weak base. That means is a strong acid.
NH4Br has the lowest pH
NaBr is the conjugate base of a strong acid, HBr. That means NaBr is neutral
The most basic between the conjugate base of the acetic acid, NaCH3CO2 and KCN is KCN because the acetic acid is the stronger acid regard to HCN.
The rank is:
NH4Br < NaBr < NaCH3CO2 < KCN
b < c < a < dWhat is true of an earthquake that causes major damage to buildings in an area?
It has a high magnitude.
It has a high frequency.
It has a low frequency.
It has a low magnitude.
The correct option is :
=》It has a high magnitude.
the damage caused by an earthquake is proportional to its magnitude, as much the magnitude is, that much damage will be caused by it.
The term used to indicate and earthquake that causes major damage to buildings in an area is called a high magnitude earthquake. Hence, option a is correct.
What is high magnitude earthquake?The most typical way to gauge an earthquake's size is by its magnitude. No matter where you are or how violent the shaking is, it is the same number since it represents the size of the earthquake's source.
The USGS no longer uses the outmoded Richter scale to determine the magnitude of major, teleseismic earthquakes. Several magnitude scales measure various aspects of the earthquake, but the Richter scale measures the biggest wobble (amplitude) on the recording.
Currently, the USGS uses the Moment Magnitude scale to report earthquake magnitudes, however many different magnitudes can be calculated for comparison and research. Therefore, option a is correct.
Find more on earthquakes:
https://brainly.com/question/29500066
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Write the balanced equation for the equilibrium reaction for the dissociation ofsilver chloride in water, and write the K expression for this reaction. Then create an ICE chart. Since we know the equilibrium concentration of the silver ion, we can solve for Ksp.Does it agree with the literature value
Answer:
See explanation
Explanation:
Hello there!
In this case, since the the concentrations are not given, and not even the Ksp, we can solve this problem by setting up the chemical equation, the equilibrium constant expression and the ICE table only:
[tex]AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)[/tex]
Next, the equilibrium expression according to the produced aqueous species as the solid silver chloride is not involved in there:
[tex]Ksp=[Ag^+][Cl^-][/tex]
And therefore, the ICE table, in which x stands for the molar solubility of the silver chloride:
[tex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)[/tex]
I - 0 0
C - +x +x
E - x x
Which leads to the following modified equilibrium expression:
[tex]Ksp=x^2[/tex]
Unfortunately, values were not given, and they cannot be arbitrarily assigned or assumed.
Regards!
The Swedish chemist Karl Wilhellm was the first to produce chlorine in the lab
2NaCl + 2H2SO4 + MnO2 -----> Na2SO4 + MnSO4 + H2O + Cl2
If Dr. Wilhellm started with 50.0 g of each reactant, which reactant is the limiting reactant?
Answer:
Explanation:
Remark
Interesting que8stion. You have to figure out how many mols are present in each reactant. Since all periodic tables are different, I'm going to use rounded numbers. If it is too close, I will go further.
NaCl
Na = 23
Cl = 35.5
1 mol = 58.5 grams
given = 50.0 grams
Mols for the reaction = 50/58.5 = 0.855
H2SO4
H2 = 2*1 2
S = 1 * 32 32
O4 = 4*16 64
1 mol = 98 grams
mols present = 50/98 = 0.510
MnO2
Mn = 1 * 55 = 55
O2 = 2*16 = 32
1 mol = 87 grams
mols available = 50/87 = 0.5747
Discussion
Na Cl and H2SO4 both require 2 moles for every mol of Cl2 produces.
H2SO4 has 0.51 mols available for a reaction
NaCl has 0.855 moles available for a reaction
MnO2 has 0.575 moles available for a reaction.
Given those numbers 0.510 mols of H2SO4 will only produce 0.255 mols of chlorine and the rest will be reduced in a similar manner. H2SO4 is the limiting reagent (reactant).
In other words only 0.510 moles of NaCl will be used and 0.855 - 0.510 moles will be left over on the reactants side.
only 0.575 moles of MnO2 will be used and 0.065 moles will be left over.
The oddity in the result shows up because the balance numbers in the equation give a ratio of 2 to 1 for H2SO4 and NaCl The 2 belongs to the reactants and the 1 for the chlorine.
