In the following molecule: SiHFClBr, list the oxidation state of each atom Which

Question

In the following molecule: SiHFClBr, list the oxidation state of each atom Which molecule has a chelating ligand? (choose from above) The chelating ligand is a dentate ligand. Which molecule has cis ligands? (choose from above Which molecule can have no isomers? (choose from above) What does VSEPR stand for? A student reacted 50.0 ml of 1.05 M sodium hydroxide with 50.0 ml of 0.50 M hydrochloric acid in a coffee cup calorimeter. From the temperature vs. time plot, the average Ti was 25.7 degree C. while the Tf of the reaction mixture was 31.2 degree C. Show all your work. What is the enthalpy of neutralization? A student dissolved 2.0 g NaOH in 50 ml of water. Ti = 25.7 degree C and Tf = 35.2 degree C. Calculate the DeltaH_nuel. Show all you work. A student reacted 55 ml of 0.50 M hydrochloric acid diluted to 100 ml with 2.0 g of Sodium Hydroxide. The T_i = 22.3 degree. T_f = 45.3 degree. Show all your work. What is the enthalpy of neutralization? Was Hess' Law validated? Work MUST be shown to receive credit. If concentrated hydrochloric acid is 12.M, how many ml of acid would you use to make 400 ml of 2M hydrochloric acid? How much water would you use? How many grams of NaOH would you use to make 400 ml of 2M? sodium hydroxide solution? How would you figure out how much water to use? Write the conventional equation for the reaction of phosphoric acid and potassium hydroxide Write the net ionic equation for the reaction of phosphoric acid and potassium hydroxide A solution of barium nitrate reacts with a solution of potassium hydroxide to form barium hydroxide octohydrate according to the following equation. Ba(NO_3)2(aq) + 2KOH(aq) rightarrow 2KNO_3(aq) + Ba(OH)_2,8H_2O(s) When 50.0 ml of 0.614 M KOH was added to 75.0 ml of 0.347 M Ba(NO_3)2, 6.51 g of Ba(0H)2.81l20 was isolated. Calculate the number of moles of KOH originally present. Calculate the number of moles of Ba(NO_3)_2 originally present. Identify the limiting reagent Calculate the theoretical yield of Ba(OH)_2.8H_2O in moles. Calculate the theoretical yield of Ba(OH)_2.8H_2O in grams.

Explanation / Answer

1)

Oxidation state of Si in Bromoiodosilane is

Answer 2)

NaOH + HCl à NaCl + H2O

Calculate the heat produced during the neutralisation reaction:

heat produced = total mass x specific heat capacity x change in temperature

q = mtotal x Cg x T

mtotal = mass(NaOH) + mass(HCl) = 50.0 + 50.0 = 100.0 g

mass (NaOH) = (1.05M*0.050L)*40 = 2.1 g

mass (HCl) = (0.5M × 0.05L) *36 = 0.9 g

Here HCl is limiting reagent because number of moles of HCl are less than NAOH,

Moles of HCl= 0.025 and no. of moles of NaOH = 0.0525

By considering reaction stoichiometry number of moles of water formed = 0.025

Cg = specific heat capacity of solutions (water)= 4.18 JoC-1g-1 =

T = Tf - Ti = 31.2 – 25.7 = 5.5 oC

q =(2.1+0.9)× 4.18 x 5.5 = 68.97 J

Molar heat (enthalpy) of neutralisation is the energy liberated per mole of water formed during a neutralisation reaction.

    Hneut is the symbol given to the molar heat of neutralisation.

    Hneut is usually given in units of kJ mol-1

enthalpy change for the reaction = -heat produced

    Hneut will be negative because the reaction is exothermic

Hneut = heat liberated per mole of water = -1 x q ÷ moles of water

Hneut = -1 x 68.97 ÷ 0.025 = -2758.8 J mol-1 = 2.758 KJ mol-1

3)

2 g of NaOH dissolved in 50ml of Water, Ti= 25.7; Tf= 35.2;

Cg = specific heat capacity of solutions (water)= 4.18 JoC-1g-1

q = mtotal x Cg x T

q= 2 * 4.18 * (35.2-25.7) = 79.42 J

    Hneut will be negative because the reaction is exothermic

Hneut = -1 x 79.42 ÷ (2/40) = -1588.4 J mol-1 = 1.588 KJ mol-1

4)

55ml of 0.5M HCl diluted 100ml with 2g of NaOH; Ti= 22.3, Tf= 45.3

q = mtotal x Cg x T

Cg = specific heat capacity of solutions (water)= 4.18 JoC-1g-1

mtotal = [(0.5*0.055L)*36] + 2 = 2.99 g

q = 2.99 * 4.18 * (45.3-22.3) = 287.45J

    Hneut will be negative because the reaction is exothermic

Hneut = -1 x 287.45 ÷ total number of moles of water liberated

             = -1 x 287.45 ÷ 0.275= 1045.27J = 1.045 KJ

5)

Conc of acid is 12M

To make 400ml of 2M HCl

M1V1=M2V2

V1= M2V2/M1 = 66.66 ml

Take 66.66 ml of acid and dilute it upto 400ml with water.

6)

H3PO3(Aq)+3KOH(Aq) --> K3PO3(aq) + 3HOH(l)

Net ionic equation H1+(aq) + OH-1(aq) --> H2O(l)

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