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balandron [24]
2 years ago
10

What is the formula of the compound between calcium and sulfur that has the percent composition 55.6?

Physics
1 answer:
Yanka [14]2 years ago
8 0

CaS is the empirical formula of the compound between calcium and sulphur that has the percent composition 55.6.

When percentages are given, we take a total mass of 100 grams.

Therefore the mass of each element is equal to the percentage given.

Mass of Ca = 55.6 g (given) of

S Mass = 44.4 g (100 - 55.6 = 44.4)

Step 1: Convert the given mass to moles.

moles Ca = given mass Ca / molar mass Ca

moles = 55.6 / 40 = 1.39 moles

mol S = specific mass S / molar mass S

mol = 44.4 / 32 = 1.39 mol

Step 2: Divide the molar ratio of each molar value by the smallest number of moles calculated.

For Ca = 1.39 / 1.39 = 1

For S = 1.39 / 1.39 = 1

The ratio of Ca : S = 1:1

Hence the empirical formula of the given compound will be CaS.

Learn more about empirical formula here : brainly.com/question/1496676

#SPJ4

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Complete Question

The complete question is shown on the first and second uploaded image

Answer:

a

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at

the surface is \frac{(n/V)_{15\ m}}{(n/V)_{surface}} = 2.5

b

The number of moles of gas that must be released is  n= 0.3538\ mols

Explanation:

We are told from the question that the pressure at the surface is 1 atm and for each depth of 10m below the surface the pressure increase by 1 atm

 This means that the pressure at the depth of the surface would be

                P_d = [\frac{15m}{10m} ] (1 atm) + 1 atm

                      = 2.5 atm

The ideal gas equation is mathematically represented as

                PV = nRT

Where P is pressure at the surface

           V is the volume

            R is the gas constant  = 8.314 J/mol. K

making n the subject we have

        n = \frac{PV}{RT}

 Considering at the surface of the water the number of moles at the surface would be

               n_s = \frac{P_sV}{RT}

Substituting 1 atm = 101325 N/m^2 for P_s ,6L = 6*10^{-3}m^3 for volume , 8.314 J/mol. K for R , (37° +273) K for T into the equation

              n_s = \frac{(1atm)(6*10^{-3} m^3)}{(8.314J/mol \cdot K)(37 +273)K}

                   = 0.2359 mol  

To obtain the number of moles at the depth of the water we use

                n_d  = \frac{P_d V}{RT}

Where P_d \ and \ n_d \ are pressure and no of moles at the depth of the water

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              n_d = \frac{(2.5)(101325 N/m^2)(6*10^{-3}m^3)}{(8.314 J/mol \cdot K)(37 + 273)K}

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Now to obtain the number of moles released we have

             n =  n_d - n_s

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     The molar concentration at the surface  of water is

                [\frac{n}{V} ]_{surface} = \frac{0.2359mol}{6*10^-3m^3}

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    The molar concentration at the depth  of water is

           [\frac{n}{V} ]_{15m} = \frac{0.5897}{6*10^{-3}}

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         \frac{(n/V)_{15\ m}}{(n/V)_{surface}} = \frac{98.28}{39.31} =2.5

                   

                     

                     

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