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Serga [27]
3 years ago
7

If 11.2 g of naphthalene, C10H8, is dissolved in 107.8 g of chloroform, CHCl3, what is the molality of the solution?

Chemistry
1 answer:
Art [367]3 years ago
5 0

Answer: 0.811m

Explanation: Molarity is the moles of solute present per kg of solvent.

Molar mass of C10H8 = . 128.17 g/mol

No. of moles = given mass/molar mass

11.2g C10H8 = 11.2/128.17 g/mol C10H8

= 0.0874 mol C10H8

NOW 107.8g solvent (chloroform) contains 0.0879 molecules (naphthalene)

1000g   ''    will contain (0.0879/107.8 x 1000) mole solute

= 0.811 mole

Molarity of the solution is 0.811 m (option A)

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C) is the best description, although a scientific theory is not necessarily a "fact". They are, however, based on evidence.

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Which gas law (choices are Charles' Law, Gay-Lussac's Law, or Boyle's Law) explains each scenario:________.
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Answer:

A.  Boyle's Law

B.  Charles' Law

C. Gay-Lussac's Law

Explanation:

An air bag inflates due to the decomposition of sodium azide or NaN₃ to completely fill the bag with nitrogen gas which is an example of Boyle's law, which states that the pressure of a given mass of gas is inversely proportional to its volume, hence due to the estricted volume of the airbag, the pressure of the nitrogen gas in the bag increses protecting the occupants of a cr from injuries in a crash

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During lab, a student used a Mohr pipet to add the following solutions into a 25 mL volumetric flask. They calculated the final
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Answer:

(FeSCN⁺²) = 0.11 mM

Explanation:

Fe ( NO3)3 (aq) [0.200M] + KSCN (aq) [ 0.002M] ⇒ FeSCN+2

M (Fe(NO₃)₃  = 0.200 M

V (Fe(NO₃)₃ =  10.63 mL

n (Fe(NO₃)₃ = 0.200*10.63 = 2.126 mmol

M (KSCN) =  0.00200 M

V (KSCN) = 1.42 mL

n (KSCN) =  0.00200 * 1.42 = 0.00284 mmol

Total volume = V (Fe(NO₃)₃  + V (KSCN)

                       = 10.63 + 1.42

                       = 12.05 mL

Limiting reactant = KSCN

So,

FeSCN⁺² = 0.00284 mmol

M (FeSCN⁺²) = 0.00284/12.05

                     = 0.000236 M

Excess reactant = (Fe(NO₃)₃

n(Fe(NO₃)₃ =  2.126 mmol -  0.00284 mmol

                  =2.123 mmol

For standard 2:

n (FeSCN⁺²) = 0.000236 * 4.63

                    =0.00109

V(standard 2) = 4.63 + 5.17

                       = 9.8 mL

M (FeSCN⁺²)  = 0.00109/9.8

                      = 0.000111 M = 0.11 mM

Therefore, (FeSCN⁺²) = 0.11 mM

7 0
3 years ago
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