Question

6. Write the molecular equation,complete ionic equation,net ionic equation for the mixing of carbonic acid and strontium hydroxide. 7. Calculate the wavelength and frequency of light emitted when the electron transitions from n= 3 to n-1. 8. Define and describe the periodic trends: atomic radii, ionization energy and electron affinity 9. Write out the periodic table box for carbon, as well as the electron configuration and orbital diagram for carbon. Draw the atom. 10. What volume of 0.150 M oxalic acid is required to neutralize 15.00mL of 0.300 M sodium hydroxide. Please write balanced equation with phase labels and use unit dimensional analysis to show the solution. 11. Write a balanced equation and calculate how many grams of barium sulfate are produced when 6.00 g barium chloride and 8.00 g copper(II) sulfate are mixed. Calc mass of excess reactant.

Answer 1

6) The balanced molecular equation is--

H₂CO₃(aq) + Sr(OH)₂ ------> SrCO₃ + 2H₂O(l)

Total ionic equation will be--

2H⁺ (aq) + CO₃^2-(aq) + Sr²⁺(aq) + 2OH^- (aq)----> Sr²⁺(aq) +  CO₃^2-(aq)  +  2H₂O(l)

Net ionic equation will be--

2H⁺ (aq) + 2OH^- (aq)----> 2H₂O(l)

=> H⁺ (aq) + OH^- (aq)----> H₂O(l)

10) The balanced reaction will be--

H₂C₂O₄ (aq) + 2NaOH(aq) ----> Na₂C₂O₄ +2 H₂O (l)

Here we see that 1 mole of H₂C₂O₄ require 2 mole of NaOH to produce 1 mole of Na₂C₂O₄ and 2 mole of H₂O

moles of NaOH = C x V = 0.300 M x 15.00 mL = 0.300 mol/L x 0.015 L = 0.0045 mol NaOH

Now,

0.0045 mol NaOH x ( 1 mol H₂C₂O₄ / 2 mol NaOH) = 0.00225 mol H₂C₂O₄

Now,

volume of oxalic acid (H₂C₂O₄ ) can be calculated as---

C = n /V

=> V = n / C = 0.00225 mol / 0.150 M = 0.00225 mol / 0.150 mol /L = 0.015 L = 15 mL

11) The balanced reaction will be--

BaCl₂(aq) + CuSO₄(aq) ----> BaSO₄ (s) + CuCl₂(aq)

Now,

6.00 g BaCl₂ x ( 1 mol BaCl₂ / 208.23 g/mol) x ( 1 mol BaSO₄ / 1 mol BaCl₂ ) = 0.0288 mol BaSO₄

and 8.00 g CuSO₄ x ( 1 mol CuSO₄ / 159.609 g/mol) x ( 1 mol BaSO₄ / 1 mol CuSO₄ ) = 0.0501 g

0.0288 mol BaSO₄  will be the  the correct answer for the amount of BaSO₄ that can be formed. There is only enough CuSO₄ to make this amoun and no more BaSO₄ can be formed because there is not any sulfate left.

Now,

0.0288 mol BaSO₄ x ( 233.38 g BaSO₄ / 1 mol BaSO₄) = 6.72 g BaSO₄