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snow_lady [41]
2 years ago
15

A compound is found to contain 25.24 % sulfur and 74.76 % fluorine by weight. The molecular weight for this compound is 254.14 g

/mol. What is the molecular formula for this compound
Chemistry
1 answer:
Savatey [412]2 years ago
4 0

Answer:

S_2F_{10}

Explanation:

Hello!

In this case, since the empirical formula of a chemical compound is determined by assuming the by-mass percentages are masses, we can firstly compute the moles of sulfur and fluorine based on their atomic masses:

n_S=25.24gS*\frac{1molS}{32.01gS}=0.789molS\\\\n_F=74.76gF*\frac{1molF}{19.00gF}=3.93molF

Next, we compute the subscript of each element in the formula by dividing each moles by the fewest moles:

S=\frac{0.789}{0.789}=1\\\\F=\frac{3.93}{0.789} =5

Whose molar mass is 127.01 g/mol. Now, we can compute the ratio between the molecular and empirical formulas as follows:

ratio=\frac{254.14}{127.01} =2

Thus, the molecular formula is:

S_2F_{10}

Regards!

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If temperature is increased , the number of collision per second
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Answer: increases

Explanation:

Increase in the temperature of a reaction system will cause the molecules of the reactants to possess higher kinetic energy which they would use to travel more randomly in the system, colliding more frequently with other excited molecules and with the wall of the containing vessel.

Thus, if temperature is increased, the number of collision per second also increases.

8 0
3 years ago
Sulfur dioxide, SO 2 ( g ) , can react with oxygen to produce sulfur trioxide, SO 3 ( g ) , by the reaction 2 SO 2 ( g ) + O 2 (
aleksley [76]

<u>Answer:</u> The amount of heat produced by the reaction is -21.36 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]

For the given chemical reaction:

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(SO_3(g))})]-[(2\times \Delta H_f_{(SO_2(g))})+(1\times \Delta H_f_{(O_2(g))})]

We are given:

\Delta H_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H_f_{(SO_3(g))}=-395.7kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(2\times (-395.7))]-[(2\times (-296.8))+(1\times (0))]\\\\\Delta H_{rxn}=-197.8kJ/mol

To calculate the number of moles, we use ideal gas equation, which is:

PV=nRT

where,

P = pressure of the gas = 1.00 bar

V = Volume of the gas = 2.67 L

n = number of moles of gas = ?

R = Gas constant = 0.0831\text{ L. bar }mol^{-1}K^{-1}

T = temperature of the mixture = 25^oC=[25+273]K=298K

Putting values in above equation, we get:

1.00bar\times 2.67L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1\times 2.67}{0.0831\times 298}=0.108mol

To calculate the heat released of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,

q = amount of heat released = ?

n = number of moles = 0.108 moles

\Delta H_{rxn} = enthalpy change of the reaction = -197.8 kJ/mol

Putting values in above equation, we get:

-197.8kJ/mol=\frac{q}{0.108mol}\\\\q=(-197.8kJ/mol\times 0.108mol)=-21.36kJ

Hence, the amount of heat produced by the reaction is -21.36 kJ

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  5. sp

Refer to the sketch attached.

<h3>Explanation</h3>

The hybridization of a carbon atom depends on the number of electron domains that it has.

Each chemical bond counts as one single electron domain. This is the case for all chemical bonds: single, double, or triple. Each lone pair also counts as one electron domain. However, lone pairs are seldom seen on carbon atoms.

Each carbon atom has four valence electrons. It can form up to four chemical bonds. As a result, a carbon atom can have up to four electron domains. It has a minimum of two electron domains, with either two double bonds or one single bond and one triple bond.

  • A carbon atom with four electron domains is sp³ hybridized;
  • A carbon atom with three electron domains is sp² hybridized;
  • A carbon atom with two electron domains is sp hybridized.

Starting from the left end (H₂C=CH-) of the molecule:

  • The first carbon has three electron domains: two C-H single bonds and one C=C double bond; It is sp² hybridized.
  • The second carbon has three electron domains: one C-H single bond, one C-C single bond, and one C=C double bond; it is sp² hybridized.
  • The third carbon has three electron domains: two C-C single bonds and one C=O double bond; it is sp² hybridized.
  • The fourth carbon has two electron domains: one C-C single bond and one C≡C triple bond; it is sp hybridized.
  • The fifth carbon has two electron domains: one C-H single bond and one C≡C triple bond; it is sp hybridized.

8 0
3 years ago
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