17. What is the oxidation number of UinZnUO:(C2H so a. +6 correct answer. b. +2

Question

17. What is the oxidation number of UinZnUO:(C2H so a. +6 correct answer. b. +2 d. +4 18. Given the equation: 3 P 20 HNOMas 20 N Which species is oxidized and which species is reduced? a. Pu, is oxidized and H2O is reduced. lb. Pi is oxidized and HNOsa is reduced. correct answer c. Osea is oxidized and P is reduced. HN d. HN is oxidized and H20 is reduced. 19. What is the molecular formula for a compound consisting of62.78%Hg, 17.19 %Mn. and 2003 if the molar mass of the compound is 639.06? This compound may not actually exist!) b. HgMnoa correct answer d. HgMno: 20. What is the molar mass of (vo) (PO)2 d. 99993 correct answer CHEM 1110 Page 6 PRACTICE EXAM 3 2l How many grams of N are there in 23.5g(NH) C'ou? MW 124.ij a 10.6 g b. 5.31 g c, 1.33 g d. 265g

Explanation / Answer

17) Given compound ZnUO2(C2H3O2)4

Sum of the oxidation states of all the atoms in a nuetral molecule = 0

Let X be the oxidation state of U

1(+2) + 1 (X) + 2(-2) + 4 [ 2(0) + 3(+1) + 2(-2)] =0

X= +6

18) Oxidation state of P on the reactant(lef) side is 0 (it is in elemental form) but its oxidation state changed to +5 on the right hand side so it is oxidised. Where as oxidation state of Nitrogen has been reduced from +5 to +2 Hence HNO3 reduced.

19) Given molar mass of compound = 639.06

% Hg = 62.78 % = (62.78/100) 639.06 = 401.2

    atomic weight of Hg = 200.59 , no of atoms of Hg in the compound = 401.2/200.59 = 2.0001 = 2 atoms of g

So C is the only option with 2 atoms of Hg

20) Molar mass of (UO2)3(PO4)2

     No of atoms of Unranium = 3 (atomic weight = 238)

    No of atom of Oxygen = 14 (atmoic weught = 16)

   No fo atmos of phosphorus = 2( atmic weight = 31)

Molar mass = (3x238) + (16x14) + (2x31) = 1000 g/mol (option d)

21) Given molecule is (NH4)2C2O4 and mol wt = 124.1

28 g of Nitrogen is present in 124.1

so amount of nitrogen present in 23.5 g of the given compound = (23.5 x 28)/124.1 = 5.31 g

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