What is the formula of the conjugate acids of these bases HC0_3^- CN^- H_20 C_2H

Question

What is the formula of the conjugate acids of these bases HC0_3^- CN^- H_20 C_2H_30_2^- Make a table and fill it in. The table is found on p. 615 # 19. How many grams of ZnS0_4 will be made from 41.0 grams each of Zn, K_2Cr_20_7 and H_2S0_4 4 Zn + K_2Cr_20_7 + 7 H_2S0_4 right arrow 4 ZnS0_4 + 2 CrS0_4 + K_2S0_4 + 7 H_20 Which is the more abundant isotope, Cl- 35 or Cl- 37 Why What is the percent O in numeric acid Formula for the acid is C_4H_40_4. Calculate the [H_30+ ] for a solution with a pOH of 11.0 A compound is found to have the following percent composition: 38.7 % K, 13.8 % N, and 47.5% 0. Find the empirical formula of this compound. A sample contains 43.7 % phosphorous and 56.3 % oxygen. Its molar mass = 384 units. The formula of the compound is P_40_5. Element # 33 is diamagnetic. In Rutherford's famous gold foil experiment the electron was discovered . The pH of a solution having a [ OH- ] of 1.0 times 10^-6 is 8. The molarity of 10.9 g of KCI dissolved into 150 mL of water would be 0.0727 Multiple choice: (#37-45) When iron rusts its oxidation number goes from 0 to +3. The iron has : been reduced been oxidized been precipitated been dissolved The oxidation number of Cr in Cr_20_7^-2 is - 2 +4 +6 +7 The complete combustion of a 0.500 g sample of a pure hydrocarbon yielded 0.973 g C0_2 and 0. 323 g of water. The empirical formula of this hydrocarbon is a C_3H_5 C_2H_3 CsH8 NONE OF THESE. How many mol of KCI are in 0.556 L of a 2.3 M KCI (aq) 0.55 1.1

Explanation / Answer

25)

a) Carbonic Acid

b) Hydrogen Cyanide

c) Hydroxyl ion

d) Acetic Acid.

27)

4 Zn + K2Cr2O7 + 7 H2SO4 4 ZnSO4 + 2 CrSO4 + K2SO4 + 7 H2O

(41 g Zn) / (65.4094 g Zn/mol) x (7/4) = 1.0969 mol H2O produced
(41 g K2Cr2O7) / (294.1850 g/mol) x (7/1) = 0.9755 mol H2O produced
(41 g H2SO4) / (98.0791 g /mol) x (7/7) = 0.4180 mol H2O produced
(*H2O was chosen arbitrarily here)

Since H2SO4 produces the least amount of H2O, H2SO4 is the limiting reagent.

(41 g H2SO4) / (98.0791 g /mol) x (4/7) x (65.4094 g Zn/mol) = 15.6246 g Zn reacted

41 g Zn - 15.6246 g Zn = 25.37 g Zn unreacted

28)

Among Cl-35 and Cl-37, Cl-35 isotope is more abundant because the atomic weight of chlorine is 35.453 which is more closer to 35. Hence Cl-35 is abundant in nature.

29)

Molecular Mass of Fumaric Acid (C4H4O4) = 116.07 g/mol

Mass percentage of oxygen = 16 x 4 / 116.07 x 100 = 55.13 %

30)

pOH = 11

pH = 14 - pOH = 3

pH = -log [H3O+]

[H3O+] = 10-pH

[H3O+] = 1 x 10-3 M

31) Assume we have 100 g of sample. Hence we have 38.7 g of K, 13.8 g of N and 47.5 g of O.

Find number of moles:

K: 38.7 / 39.09 = 0.99 mole ~ 1

N: 13.8 / 14 = 0.98 mole ~ 1

O: 47.5 / 16 = 2.96 mole ~ 3

Hence, the emperical formula would be KNO3.

32) False

33) True

34) False

35) True

36) True

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