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Lostsunrise [7]
3 years ago
5

A compound is 2.00% H by mass, 32.7% S by mass, and 65.3% O by mass. What is its empirical formula? The second step is to calcul

ate the number of moles of each element present in the sample. To do this, you must divide the mass of each element by its molar mass. The masses of the elements present are: mass H = g mass S = g mass O = g
Chemistry
2 answers:
Free_Kalibri [48]3 years ago
8 0
Lets take 100 g of this compound,
so it is going to be 2.00 g H, 32.7 g S and 65.3 g O.

2.00 g H *1 mol H/1.01 g H ≈ 1.98 mol H
32.7 g S *1 mol S/ 32.1 g S ≈ 1.02 mol S
65.3 g O * 1 mol O/16.0 g O ≈ 4.08 mol O

1.98 mol H : 1.02 mol S : 4.08 mol O = 2 mol H : 1 mol S : 4 mol O

Empirical formula
H2SO4
lbvjy [14]3 years ago
3 0

the masses of the elements present are:

mass H = 2.00 g

mass S = 32.7 g

mass O = 65.3 g

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Answer:

D. It predicts whether or not a reaction will be spontaneous.

Explanation:

<em>What does Gibbs free energy (ΔG) predict? .</em>

A. It predicts what the rate of the reaction will be. NO. ΔG is a thermodynamical parameter and it is not related to the kinetics of the reaction.

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C. It predicts if entropy will increase or decrease. NO. ΔG depends on the entropy but not the other way around.

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5 0
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saul85 [17]

<u>Answer:</u> The \Deltas H^o_{formation} for the reaction is -1406.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical reaction for the formation reaction of AlCl_3 is:

2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3 (s)    \Delta H^o_{formation}=?

The intermediate balanced chemical reaction are:

(1) HCl(g)\rightarrow HCl(aq.)    \Delta H_1=-74.8kJ    ( ×  6)

(2) H_2(g)+Cl_2(g)\rightarrow 2HCl(g)    \Delta H_2=-185kJ     ( ×  3)

(3) AlCl_3(aq.)\rightarrow AlCl_3(s)    \Delta H_3=+323kJ     ( ×  2)

(4) 2Al(s)+6HCl(aq.)\rightarrow 2AlCl_3(aq.)+3H_2(g)    \Delta H_4=-1049kJ

The expression for enthalpy of formation of AlCl_3 is,

\Delta H^o_{formation}=[6\times \Delta H_1]+[3\times \Delta H_2]+[2\times \Delta H_3]+[1\times \Delta H_4]

Putting values in above equation, we get:

\Delta H^o_{formation}=[(-74.8\times 6)+(-185\times 3)+(323\times 2)+(-1049\times 1)]=-1406.8kJ

Hence, the \Deltas H^o_{formation} for the reaction is -1406.8 kJ.

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fenix001 [56]

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Explanation:

The properties of Rubidium are more similar to those of cesium compared to strontium.

Elements in the same group on the periodic table have similar chemical properties.

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  • Other elements in this group are lithium, sodium, potassium and francium
  • Strontium belongs to the second group on the periodic table.
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Learn more:

Sodium brainly.com/question/6324347

Periodic table brainly.com/question/1704778

#learnwithBrainly

#learnwithBrainly

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