110 a) Determine the number of moles of C Hs required to completely react3.0 gra

Question

110 a) Determine the number of moles of C Hs required to completely react3.0 grams of H20. What is the percent yield of H20? (15 points) Given the following balanced chemical equatio met, 103 grams of C,Hs was reacted with 105 grams of O2 to .0130 MNaBr(aq) solution from a 2.50 MNaBr In a particular experiment, 3. Describe how to prepare 25.0 mL of 0 solution. (8 points) Given the following balanced chemical equation: How many moles of Na 4. NazCOs(aq)+2H(a2NaCI(aq)+ CO2(g)+ H200) CI are produced by the complete reaction of 5.00 ml of 0.125 A 5. A3L sample of methane gas is heated from 20°C to 120°C at constant pressure. W 6. 0.0250 g of AI(OH)s(s) is dissolved in enough water to produce 50.0 mL of solution. and excess HCl(aq)? (7 points) final volume of the gas? (5 points) a) What is the molar concentration of Al(OH)3(aq) in the solution? (4 points) b) What is the molar concentration of OH (aq) ions in the solution? (4 points) 7. Calcium (Ca) is more reactive than Silver (Ag). a) Write the products of the following reaction and balance the equation. (5 points) Ca(s) + AgNO3(aq)- b) Write the molecular equation from part a in the total ionic and netionic forms, and ic spectator ions. (8 points) occupies a volume of 2.50 liters at STP. What will be the pressure (in atm 8. An ideal gas volume 15.0 liters of the same sample of gas at-10 °C? (8 points) 9. (3 points) Balance the following equation. H2SO,+ 10. For the following reaction: 2A(s) +3H,SO,(aq) -+ Al(Son(a)+3H ??Na2SO4 (g a) Give the oxidation number for each (and eve b) Identify which atom is oxidized and which is reduced. (2 points) c) Identify the oxidizing agent and the reducing agent. (4 points) ry) atom. (8 points) 11. MULTIPLE CHOICE points) In the following chemical equation, which of the follo a. HCHO2 and CHO2 b. H2O and H0 c. 4 points)on, which of the following are B ????? + H20-3H30' + CH02. ronsted acids? o' c. HCHO2 and H30 d. H20 and CHO CHO2 and H30 d. H2o Please Turn Over >>>

Explanation / Answer

3. We need 25 mL of 0.0130 M NaBr. Now ,

0.0130 M solution means

1000 ml solution has NaBr = 0.0130 mol. So,

25 ml solution has NaBr = (0.0130 × 25)/1000 = 0.000325 mol

This means we need 0.000325 mol of NaBr.

Now the given solution is 2.50 M NaBr

2.50 M solution means

2.50 mole of NaBr is present in solution = 1000 ml

0.000325 mol will be present in solution

=(1000 × 0.000325)/2.50 =0.13 ml

So,

To prepare 25 ml of solution take 0.13 ml of given solution and add water hoti the mark of 25 ml

4. Volume of Na2CO3 solution used = 5 ml

Concentration of solution = 0.125 M. 0.125 M solution means

1000 ml solution has Na2CO3 = 0.125 mol

5 ml solution has Na2CO3 =( 0.125 × 5)/1000 = 0.000625 mol

From chemical reaction it is clear that

1 mol of Na2CO3 give NaCl = 2 mol

.000625 mol Na2CO3 give NaCl=2× .000625 mol = .00125mol

Hence, 0.00125 mol of NaCl is produced.

5. Since pressure is kept constant, so Charles law can be used to calculate final volume. According to Charles law: at constant pressure

V1/T1 (initial condition)= V2/T2 (final condition)

V1 = 7.3 L, V2 =?,T1 = 20? = (273+20)K = 293 K,T2 = 393K

So, V2 =( V1×T2)/T1 = 7.3 L × 393 K/293K = 9.8 L

6.a) Molar concentration of Al(OH)3 = moles of it/volume in L

Moles of Al(OH)3 = given mass / molar mass

Molar mass of Al(OH)3 = 78 g/mol

So, moles of Al(OH)3 = 0.0250 g/(78 g/mol) = 0.00032 mol

Volume given = 50 ml =0.050L

So concentration = 0.00032 mol/0.050 L = 0.0064 mol/L

So concentration of Al(OH)3 = 0.0064 M

b) 1 unit of Al(OH)3 has 3 units of OH-

So, concentration of OH- ions = 3 × concentration of Al(OH)3

So, concentration of OH- = 3× 0.0064 M = 0.0192 M

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