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masha68 [24]
2 years ago
13

A 44.6 mg sample of carbon reacts with sulfur to form 124 mg of the compound. part a what is the empirical formula of the carbon

sulfide?
Chemistry
1 answer:
Inga [223]2 years ago
4 0

The empirical formula of carbon sulfide = C3S2

calculation

calculate the mass of S used = 124-44.6= 79.4 grams

find the moles of each element

that is carbon(C)= 44.6/12= 3.717 moles    

          sulfur (S) = 79.4/32= 2.48 moles

find the mole ratio by dividing each mole with the smallest mole( 2.48)

that is carbon = 3.717/2.48 = 1.5

        sulfur = 2.48/2.48= 1

multiply by 2 to remove the decimal

= C=1.5 x2=3

S= 1/2 =2

therefore the empirical formula = C3S2

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<span>then plug into the formula M = moles/liters </span>

<span>0.035 M = moles/ 0.040L </span>

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2 years ago
What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (–97.6˚C) and 1 atm
Rudiy27

Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K

Explanation :

Formula used :

\Delta S=\frac{\Delta H_{freezing}}{T_f}

where,

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As we know that:

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Now put all the given values in the above formula, we get:

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\Delta S=\frac{-3170J/mol}{175.4K}

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4 0
3 years ago
HELP ASAP ITS TIMED!!!
Tpy6a [65]

Answer:

C would be the right answer

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