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PilotLPTM [1.2K]
3 years ago
8

The combustion of 2.95 grams of a compound that contains only C, H and S yields 5.48

Chemistry
1 answer:
lesya692 [45]3 years ago
4 0

The empirical formula of the compound is C₃H₃S.

Explanation:

molecular weight of CO₂ = 44 g/mole

molecular weight of H₂O = 18 g/mole

Knowing the molecular weights of the compounds we devise the following reasoning:

if in           44 g of CO₂ there are 12 g of C

then in  5.48 g of CO₂ there are X g of C

X = (5.48 × 12) / 44 = 1.49 g C

if in           18 g of H₂O there are 2 g of H

then in    1.13 g of H₂O there are Y g of H

Y= (1.13 × 2) / 18 = 0.126 g H

mass of compound = mass of C + mass of H + mass of S

mass of S = mass of compound - mass of C - mass of H

mass of S = 2.95 - 1.49 - 0.126 = 1.33 g S

Now to find the empirical formula of the compound we use the fallowing algorithm:

we devise each mass by the molar weight of the element:

for C   1.49 / 12 = 0.124

for H   0.126 / 1 = 0.126

for S   1.33 / 32 = 0.0416

now we divide the result by the lowest number which is 0.0416

for C    0.124 / 0.0416 ≈ 3

for H    0.126 / 0.0416 ≈ 3

for S    0.0416 / 0.0416 = 1

The empirical formula of the compound is C₃H₃S.

Learn more about:

empirical formula

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What is the molariity of a 50.0 mL aqueous solution containing 10.0 grams of table salt, NaCl?
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Answer:

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Explanation:

You should know or have the equation to solve for Molarity which is;

M = n/v           (M: Molarity) (n: moles of solute) (v: Liters of solute)

You can start off differently but I would start by converting the mL to L. This is your "v" value.

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Now, you have to convert grams to moles in order to solve for molarity (M).

1.) On the periodic table find the molecular weights of Na and Cl.

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2.) Add them together to have their combined molecular weights.

22.99 + 35.45= 58.44 g/mol

3.) Now, you're going to use the "picket fence method" or whichever your teacher taught you to convert from grams to moles. This will be your "n" value. (I cannot show it on here without it looking weird, so my sincere apologies.)

10.0 g/ 58.44 g = 0.17111 mol

4.)You are now going to plug in your answers into the equation for Molarity.

M= 0.17111 mol / 0.05 L = 3.4222

5.) I am sure your professor might be a stickler so for sig figs sake when you multiply or divide use the smallest amount of sig figs you see which is 1. Round 3.4222 to 3 mol/L

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The equilibrium constant, Kp, for the following reaction is 9.52Ã10-2 at 350 K: CH4(g) + CCl4(g) 2CH2Cl2(g). Calculate the equil
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Answer:

A) p CH₄       = 0.732 atm

B) p CCl₄      = 0.732 atm

c)  p CH₂Cl₂  = 0.22 atm

Explanation:

we have the equilibrium constant for this problem, along the initial pressures for the reactants, and  need to find the partial pressures at equilibrium. So lets setup the equilibrium:

CH₄ (g) + CCl₄ (g)   ⇔  2 CH₂Cl₂ (g)  Kp =  9.50 x 10⁻²

where Kp is given by :

Kp = p CH₂Cl₂ ² / p CH₄ x p CCl₄

where p are the partial pressures

                               p CH₄  atm          p CCl₄  atm              p CH₂Cl₂  atm

initial                             0.844              0.844                           0

change                           - x                     - x                           +2x

equilibrium               0.844 - x            0.844 - x                      2 x

Kp = 9.52 x 10⁻² = ( 2x )²/  (( 0.844 - x ) x ( 0.844 - x ))

9.52 x 10⁻² =  (2x)² / ( 0.844 - x )²

Taking square root to both sides of the equation:

√9.52 x 10⁻²  = 2x / (0.844 - x )

0.309 = 2x / (0.844 - x)

0.260 - 0.309 x = 2x

0.260 = 2.309 x   ⇒ x = 0.112

So the partial pressures are at equilibrium are:

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p CH₂Cl₂  = 2 x (0.112 atm) = 0.224 atm

You can check your answer is correct by plugging this values and comparing  with the given Kp.

6 0
3 years ago
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