Answer:
Detail is given below
Explanation:
Atomic radii trend along group:
As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.
As the size of atom increases the ionization energy from top to bottom also decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus.
In A we can see that there is one positive charge and force of attraction is 2.30×10⁻⁸ N and distance is 0.10 nm
In B we can see that negative charge is further away from nucleus because of greater distance thus force of attraction will be less. 0.58×10⁻⁸ N
In C this distance further increases and force also goes in decreasing 0.26×10⁻⁸ N.
Answer:
O₂; KCl; 33.3
Explanation:
We are given the moles of two reactants, so this is a limiting reactant problem.
We know that we will need moles, so, lets assemble all the data in one place.
2KCl + 3O₂ ⟶ 2KClO₃
n/mol: 100.0 100.0
1. Identify the limiting reactant
(a) Calculate the moles of KClO₃ that can be formed from each reactant
(i)From KCl
(ii) From O₂
O₂ is the limiting reactant, because it forms fewer moles of the KClO₃.
KClO₃ is the excess reactant.
2. Moles of KCl left over
(a) Moles of KCl used
(b) Moles of KCl left over
n = 100.0 mol - 66.67 mol = 33.3 mol
The answer is Ka = 1.00x10^-10.
Solution:
When given the pH value of the solution equal to 11, we can compute for pOH as
pOH = 14 - pH = 14 - 11.00 = 3.00
We solve for the concentration of OH- using the equation
[OH-] = 10^-pOH = 10^-3 = x
Considering the sodium salt NaA in water, we have the equation
NaA → Na+ + A-
hence, [A-] = 0.0100 M
Since HA is a weak acid, then A- must be the conjugate base and we can set up an ICE table for the reaction
A- + H2O ⇌ HA + OH-
Initial 0.0100 0 0
Change -x +x +x
Equilibrium 0.0100-x x x
We can now calculate the Kb for A-:
Kb = [HA][OH-] / [A-]
= x<span>²</span> / 0.0100-x
Approximating that x is negligible compared to 0.0100 simplifies the equation to
Kb = (10^-3)² / 0.0100 = 0.000100 = 1.00x10^-4
We can finally calculate the Ka for HA from the Kb, since we know that Kw = Ka*Kb = 1.0 x 10^-14:
Ka = Kw / Kb
= 1.00x10^-14 / 1.00x10^-4
= 1.00x10^-10
<em>Moritz is watching his little brother play in a mud puddle. Moritz notices that when his brother stirs it with a stick, after a few moments the dirt settles back to the bottom of the puddle. His brother is creating a; </em>SUSPENSION
<em>Anjali knows that whole milk has more fat than skim milk. However, the solid fat doesn’t seem to separate from the liquid milk even after it has been in the fridge for a few days. This is evidence that milk is a;</em> COLLOID
Explanation:
We will balance equation which describes the reaction between sulfuric acid and sodium bicarbonate: as follows.
Next we will calculate how many moles of 
are present in 85.00 mL of 1.500 M sulfuric acid.
As, Molarity = 
1.500 M = 
n = 0.1275 mol
Now set up and solve a stoichiometric conversion from moles of to grams of 
. As, the molar mass of is 84.01 g/mol.
= 21.42 g
So unfortunately, 15.00 grams of sodium bicarbonate will "not" be sufficient to completely neutralize the acid. You would need an additional 6.42 grams to complete the task.
