The kinetic theory describes the matter of particles in all matter
1.66 M is the concentration of the chemist's working solution.
<h3>What is molarity?</h3>
Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.
In this case, we have a solution of Zn(NO₃)₂.
The chemist wants to prepare a dilute solution of this reactant.
The stock solution of the nitrate has a concentration of 4.93 M, and he wants to prepare 620 mL of a more dilute concentration of the same solution. He adds 210 mL of the stock and completes it with water until it reaches 620 mL.
We want to know the concentration of this diluted solution.
As we are working with the same solution, we can assume that the moles of the stock solution will be conserved in the diluted solution so:
=
(1)
and we also know that:
n = M x 
If we replace this expression in (1) we have:
x
=
x 
Where 1, would be the stock solution and 2, the solution we want to prepare.
So, we already know the concentration and volume used of the stock solution and the desired volume of the diluted one, therefore, all we have to do is replace the given data in (2) and solve for the concentration which is
:
4.93 x 210 = 620 x
= 1.66 M
This is the concentration of the solution prepared.
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Answer:
1 mole of 02=6.022*10^23 02 molecules 1 02 molecules
Explanation:
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Answer:
pH = -log[H⁺] = -log(4.75 x 10⁻⁷) = -(- 6.32 ) = 6.32
Explanation:
Given 4.3 x 10⁻⁷M H₂CO₃ (Ka1 = 4.2 x 10⁻⁷ & Ka2 = 4.8 x 10⁻¹¹)
Note: The Ka2 value for the 2nd ionization step is so small (Ka2 = 4.8 x 10⁻¹¹) It will be assumed all of the hydronium ions (H⁺) come from the 1st ionization step.
1st Ionization step
H₂CO₃ ⇄ H⁺ + HCO₃⁻
C(initial) 4.3 x 10⁻⁷ 0 0
ΔC -x +x +x
C(final) 4.3 x 10⁻⁷ - x x x
Note: the 'x' value in this analysis can not be dropped as the Conc/Ka value is less than 10². In this case he C/Ka ratio* (4.3E-7/4.2E-7 ≈ 1) is far below 10².
So, one sets up the equilibrium equation to be quadric and the x-value can be determined.
Ka1 = [H⁺][HSO⁻]/[H₂SO₃] = (x)(x)/(4.3 x 10⁻ - x) = x²/(4.3 x 10⁻⁷ - x) = 4.2 x 10⁻⁷
=> x² = 4.2 x 10⁻⁷(4.3 x 10⁻⁷ - x)
=> x² + 4.2 x 10⁻⁷x - 1.8 x 10⁻¹³ = 0
a = 1, b = 4.2 x 10⁻⁷, c = - 1.8 x 10⁻¹³
x = b² ± SqrRt(b² - 4(1)(-1.8 x 10⁻¹³ / 2(1) = 4.75 x 10⁻⁷
x = [H⁺] = 4.75 x 10⁻⁷
pH = -log[H⁺] = -log(4.75 x 10⁻⁷) = -(- 6.4 ) = 6.4
______________________
* The Concentration/Ka-value is the simplification test for quadratic equations used in Equilibrium studies. If the C/Ka > 100 then one can simplify the C(final) by dropping the 'x' if used in this type analysis. However, if the C/Ka value is < 100 then the x-value must be retained and the solution is determined using the quadratic equation formula.
for ax² + bx + c = 0
x = b² ± SqrRt(b² - 4ac) / 2a