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uranmaximum [27]
4 years ago
13

Experimental data have shown that the rate law for the reaction 2 hgcl2(aq) + c2o4 2 -(aq) → 2 cl-(aq) + 2 co2 (g) + hg2cl2 (s)

is: rate = k[hgcl2][c2o4 2 -]2 how will the rate of reaction change if the concentration of c2o4 2 - is tripled and the concentration of hgcl2 is doubled?
Chemistry
1 answer:
NNADVOKAT [17]4 years ago
6 0
When the rate = K [ HgCl2] [C2O4 2-]^2
if the concentration of [ C2O4 2- ] is tripled so we will substitute by [3 (C2O4 2-)] instead of [ C2O4 2- ] 
and if the concentration of [HgCl2] is doubled so we will substitute by [2( HgCl2)] instead of [HgCl2] 
So the new rate will be = K [2(Hgcl2)] [3(C2O4 2-)]^2
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The empirical formula is the simplest formula of a chemical compound.

To find the empirical formula, we take the following steps;

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  • Divide the quotient of each by the lowest value obtained instep 1 above
  • Write the result of step 2 above as the subscript following each atom.

1) O - 88.10/16,      H - 11.190/1

  O - 5.5,               H - 11.19

  O - 5.5/5.5,        H - 11.19/5.5

  O -  1,                 H - 2

Empirical formula = OH2

2) C - 41.368/12  H - 8.101/1,   N - 32.162/14,   O - 18.369/16

   C - 3,               H - 8,           N - 2,                  O - 1

   C - 3/1,            H - 8/1          N - 2/1                 O - 1/1

    C - 3,             H - 8,           N - 2,                   O - 1

Empirical formula = C3H8N2O

To obtain the molecular formula where n = number of atoms of each element;

Molecular weight = 174.204 g/mol

[ 3(12) + 8(1) + 2(14) + 16]n = 174

n= 174/88

n = 2 (to the nearest whole number)

Hence, we have;

[C3H8N2O]2

The molecular formula is C6H16N4O2

3)  C - 19.999/12,  H - 6.713/1,   N - 46.646/14,   O - 26.641/16

    C - 2,                H - 7,            N -  3,                 O - 2

    C - 2/2,            H - 7/2,         N -   3/2,             O - 2/2

    C - 1,                H - 4,            N -  2,                  O - 1

Empirical formula - CH4N2O

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