Answer:
15 mL of the solute
Explanation:
From the question given above, the following data were obtained:
Solution = 50 mL
Solvent = 35 mL
Solute =?
Solution is simply defined as:
Solution = solute + solvent
With the above formula, we can easily obtain the solute in the solution as follow:
Solution = 50 mL
Solvent = 35 mL
Solute =?
Solution = solute + solvent.
50 = solute + 35
Collect like terms
50 – 35 = solute
15 = solute
Solute = 15 mL
Therefore, 15 mL of the solute is required.
Answer:
Arrhenius acid & Bronsted-Lowry acid
Explanation:
Answer:
Alpha particle
Explanation:
An alpha particle is a helium nucleus, 2 protons and 2 neutrons, loss of an alpha particle give a new element with an atomic number 2 less than the original isotope and an atomic mass that is lower by about 4 amu.
The solution for this problem is:
Get into moles first. .0560 grams over 540.8 grams per mole = 1.04 x l0^-4 moles
Sr3(As04)2 = 3 Sr++(aq) plus 2 As04^-3(aq)
Ksp = (Sr++)^3(As04^-3)^2
(Sr++) = 3 X 1.04 x l0^-4= 3.11 x l0^-4
(As04^-3) = 2 x 1.04 x l0^-4= 2.07 x l0^-4
Ksp = (1.04 x l0^-4)^3 (2.07 x l0^-4)^2 which equals 4.82 x 10^-20
The molar concentration of the KI_3 solution is 0.251 mol/L.
<em>Step 1</em>. Write the <em>balanced chemical equation</em>
I_3^(-) + 2S_2O_3^(2-) → 3I^(-) + S_4O_6^(2-)
<em>Step 2</em>. Calculate the <em>moles of S_2O_3^(2-)</em>
Moles of S_2O_3^(2-)
= 27.9 mL S_2O_3^(2-) ×[0.270 mmol S_2O_3^(2-)/(1 mL S_2O_3^(2-)]
= 7.533 mmol S_2O_3^(2-)
<em>Step 3</em>. Calculate the <em>moles of I_3^(-)
</em>
Moles of I_3^(-) = 7.533 mmol S_2O_3^(2-)))) × [1 mmol I_3^(-)/(2 mmol S_2O_3^(2-)] = 3.766 mmol I_3^(-)
<em>Step 4</em>. Calculate the <em>molar concentration of the I_3^(-)
</em>
<em>c</em> = "moles"/"litres" = 3.766 mmol/15.0 mL = 0.251 mol/L
