Answer:
Percent relative error = 0.191%
Explanation:
Relative error is a measure of accuracy (How closeness is the measure to the accepted value) for a determined data. The formula is:
Percent relative error = |Experimental - Accepted| / Accepted * 100
<em>Where experimental is the average of the data:</em>
(40.220g/mol + 40.654g/mol + 40.314g/mol + 40.165g/mol + 40.554g/mol) / 5 =
40.381g/mol
Replacing using accepted value = 40.304g/mol:
|40.381g/mol - 40.304g/mol| / 40.304g/mol * 100
<h3>Percent relative error = 0.191%</h3>
Electrons are added to the same principal energy level.
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.
Answer:
A high pH value indicates a high concentration of OH- ion
Explanation:
The higher the OH- ion concentration high will be the pH.In simple words if the concentration of OH- ions are increased then the pH of the solution will also increase which means the solution will turns towards basic with increasing its OH- ion concentration.
Let us assume that the OH- concentration of a solution is 10-9 so the pOH of that solution will be 9 and the pH will be 5.
Now the concentration of OH-ion of that solution is increased from 10-9 to 10-8 now the pOH of that solution is 8 and the pH is 6.
1.34 L of HF
Explanation:
We have the following chemical reaction:
Sn (s) + 2 HF (g) → SnF₂ (s) + H₂ (g)
First we calculate the number of moles of SnF₂:
number of moles = mass / molecular weight
number of moles of SnF₂ = 5 / 157 = 0.03 moles
From the chemical reaction we see that 1 mole of SnF₂ are produced from 2 moles of SnF₂. This will mean that 0.03 moles of SnF₂ are produced from 0.06 moles of HF.
Now at standard temperature and pressure (STP) we can use the following formula to calculate the volume of HF:
number of moles = volume / 22.4 (L/mole)
volume of HF = number of moles × 22.4
volume of HF = 0.06 × 22.4 = 1.34 L
Learn more about:
problems with gases at STP
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