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Orlov [11]
3 years ago
15

Calculate the molecular (formula) mass of each compound: (a) iron(ll) acetate tetrahydrate; (b) sulfur tetrachloride; (c) potass

ium permanganate.
Chemistry
1 answer:
Nutka1998 [239]3 years ago
5 0

Answer:

a) Iron(ll) acetate tetrahydrate: 245,68 g/mol

b) Sulfur tetrachloride: 173,87 g/mol

c) Potassium ermanganate 158,034 g/mol

Explanation:

To solve this kind of exercises you must look for the number of atoms in each molecule first, then look on the periodic table the atom weight and the multiply the atom weight times the quantity of each atom. For instance:

The molecule of Iron(II) acetate itetrahydrate is (CH3COO)2Fe•4H2O, it means that you have:

2 atoms of carbon times the atomic weight of C (12.00g/mol)= 24g

14 atoms of Hidrogen times the atomic weight of H (1,00g/mol)= 14g

6 atoms of Oxigen times the atomic weight of O (16,0g/mol)= 96

2 atoms of Iron times the atomic weight of Fe (55,84g/mol)= 111,68g

At last, you only have to add the results: 24+14+96+11,68= 245,68g/mol. This example was for the first molecule.

See you,

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In a certain industrial process involving a heterogeneous catalyst, the volume of the catalyst (in the shape of a sphere) is 10.
trasher [3.6K]

Answer:

The second run will be faster - true, the increased surface area of catalyst will increase the rate of reaction

The second run will have the same rate as the first - possible, in case there is a factor other than catalyst limiting the reaction

The second run has twice the surface area - yes, 44 sqcm to 22 sqcm

Explanation:

A catalyst is a material which speeds up a reaction without being consumed in the process. A heterogeneous catalyst is one which is of a different phase than the reactants. The effectiveness of a catalyst is dependent on the available surface area. The first step for this question is to determine the total available surface area of catalyst in both processes.

Step 1: Determine radius of large sphere

V_{L}=\frac{4}{3}*pi*{r_{L}}^{3}

10=\frac{4}{3}*pi*{r_{L}}^{3}

{r_{L}}^{3}=\frac{7.5}{pi}

r_{L}=1.337cm

Step 2: Determine surface area of large sphere

A_{L}=4*pi*{r_{L}}^{2}

A_{L}=4*pi*1.337^{2}

A_{L}=22.463 cm^{2}

Step 3: Determine radius of small sphere

V_{s}=\frac{4}{3}*pi*{r_{s}}^{3}

1.25=\frac{4}{3}*pi*{r_{s}}^{3}

{r_{s}}^{3}=\frac{0.9375}{pi}

r_{s}=0.668cm

Step 4: Determine surface area of small sphere

A_{s}=4*pi*{r_{s}}^{2}

A_{s}=4*pi*0.668^{2}

A_{s}=5.607 cm^{2}

Step 5: Determine total surface area of 8 small spheres

A_{S}=8*A_{s}

A_{S}=8*5.607

A_{S}=44.856 cm^{2}

A_{L}=22.463 cm^{2} - Surface area of 1 large sphere

A_{S}=44.856 cm^{2} - Surface area of 8 small spheres

Options:

  1. The second run will be faster - true, the increased surface area of catalyst will increase the rate of reaction
  2. The second run will be slower - false, the increased surface area of catalyst will increase the rate of reaction
  3. The second run will have the same rate as the first - possible, in case there is a factor other than catalyst limiting the reaction
  4. The second run has twice the surface area - yes, 44 sqcm to 22 sqcm
  5. The second run has eight times the surface area - no, 44 sqcm to 22 sqcm
  6. The second run has 10 times the surface area - no, 44 sqcm to 22 sqcm
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