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kenny6666 [7]
3 years ago
15

A 1.540 gram sample of an alloy containing only tin and zinc was reacted with excess fluorine gas to produce 2.489 grams in tota

l of a mixture of tin IV fluoride and zinc fluoride. Calculate the percent composition by mass of the two metals in the alloy.
Chemistry
1 answer:
sineoko [7]3 years ago
6 0

Answer:

Tin: 54.3%

Zinc: 45.7%

Explanation:

The molar masses of the elements are:

Tin: Sn = 117.710 g/mol

Zinc: Zn = 65.409 g/mol

Fluorine: F = 18.998 g/mol

The fluorine gas in excess, so the reaction consumes all the alloy, and all the tin is converted to SnF₄ and all the zinc is converted to ZnF₂. The molar masses of the fluorides are:

SnF₄ = 117.710 + 4*18.998 = 193.702 g/mol

ZnF₂ = 65.409 + 2*18.998 = 103.405 g/mol

If we call x the number of moles of SnF₄, and y the number of moles of ZnF₂, the total mass can be calculated knowing that the mass is the number of moles multiplied by the molar mass:

193.702x + 103.405y = 2.489

The number of moles of Sn is the same as SnF₄ (1:1), and also the number of moles of Zn is the same as ZnF₂ (1:1), so the mass of the alloy:

117.710x + 65.409y = 1.540

if we multiply it by -1.581 and sum with the other equation:

117.710x*(-1.581) + 65.409y*(-1.581) + 193.702x + 103.405y = 1.540*(-1.581) + 2.489

7.60249x = 0.05426

x = 0.0071 mol of Sn

117.710*0.0071 + 65.409y = 1.540

65.409y = 0.704259

y = 0.0108 mol of Zn

The masses are the molar mass multiplied by the number of moles:

Sn: 117.710*0.0071 = 0.836 g

Zn: 65.409*0.0108 = 0.704 g

The percent composition is the mass of the substance divided by the total mass multiplied by 100%:

Sn: (0.836/1.540)*100% = 54.3%

Zn: (0.704/1.540)*100% = 45.7%

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Find the empirical formula of the compound ribose, a simple sugar often used as a nutritional supplement. A 14.229 g sample of r
MakcuM [25]

Answer:

CH2O

Explanation:

Firstly, we need to convert the masses of the elements to percentage compositions. This can be done by placing the mass of each element over the total mass multiplied by 100% . We can start with carbon.

C = 5.692/14.229 * 100 = 40%

O = 7.582/14.229 * 100 = 53.29%

H = 0.955/14.229 * 100 = 6.71%

We then proceed to divide each percentage composition by their atomic mass of 12, 16 and 1 respectively.

C = 40/12 = 3.333

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H = 6.71/2 = 6.71

Dividing by the smaller value which is 3.33

C = 3.33/3.33 = 1

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The empirical formula of the compound ribose is CH2O

6 0
3 years ago
Read 2 more answers
When molecules have the same number of each element but those atoms are arranged differently they are known as____________.
Karo-lina-s [1.5K]

Answer: Structural Isomers

Explanation:

The compounds having similar molecular formula but different arrangement of atoms or groups in space are called isomers and the phenomenon is called as isomerism.  

Isomers are of two types: structural isomers and stereo isomers.

Structural isomers are isomers in which molecules with the same molecular formula have different bonding patterns.

Stereoisomers are isomers in which molecules have the same molecular formula and sequence of bonded atoms but differ in the three-dimensional orientations of their atoms in space.

Thus when molecules have the same number of each element but those atoms are arranged differently they are known as Structural isomers.

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3 years ago
The equilibrium constant (K p) for the interconversion of PCl 5 and PCl 3 is 0.0121:
aksik [14]

Answer: At equilibrium, the partial pressure of PCl_{3} is 0.0330 atm.

Explanation:

The partial pressure of PCl_{3} is equal to the partial pressure of Cl_{2}. Hence, let us assume that x quantity of PCl_{5} is decomposed and gives x quantity of PCl_{3} and x quantity of Cl_{2}.

Therefore, at equilibrium the species along with their partial pressures are as follows.

                         PCl_{5}(g) \rightarrow PCl_{3}(g) + Cl_{2}(g)\\

At equilibrium:  0.123-x          x              x

Now, expression for K_{p} of this reaction is as follows.

K_{p} = \frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]}\\0.0121 = \frac{x \times x}{(0.123 - x)}\\x = 0.0330

Thus, we can conclude that at equilibrium, the partial pressure of PCl_{3} is 0.0330 atm.

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