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fgiga [73]
3 years ago
14

18) Calculate the molarity

Chemistry
1 answer:
Sloan [31]3 years ago
4 0

Answer:

B. 0.92 M

Explanation:

Molarity of a solution = number of moles (n) ÷ volume (V)

According to the information provided in this question;

mass of NaCl = 42g

Volume of water = 780mL

Using mole = mass/molar mass

Molar mass of NaCl = 23 + 35.5 = 58.5g/mol

mole = 42/58.5

mole (n) = 0.72mol

Volume (V) = 780 mL = 780/1000 = 0.780 L

Hence, molarity = n/V

Molarity = 0.72/0.780

Molarity = 0.923 M

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In a laboratory experiment, pure lead metal reacted with excess sulfur to produce a lead sulfide compound. The following data wa
Ugo [173]
To get the empirical formula, we need to know the ratio of the moles of both lead and sulfur.

First, lets get the mass of each:
mass of lead = mass of evaporating dish and lead - mass of empty dish
                     = <span>26.927  - 25.000  = 1.927 gm of lead (Pb)
mass of sulfur = mass of dish and lead sulfide - mass of dish and lead
                       = </span><span>27.485  - 26.927  = 0.558 gm of sulfur (S)

The next step is to get the number of moles in 1.927 gm of Pb and in 0.558 gm of S:
From the periodic table:
molar mass of lead = </span>207.2 gm<span>
molar mass of sulfur = 32 gm

number of moles = mass / molar mass
number of moles of Pb = 1.927 / 207.2 = </span>0.0093 moles<span>
number of moles of S = 0.558 / 32 = </span><span>0.0174 moles

The third step is to get the ratio between the moles by dividing the number of moles of each by the smaller of the two:
1 mole of Pb (</span>0.0093 / 0.0093) reacts with 0.0174 / 0.0093 = 1.99 (approximately 2 moles) of sulfur.

Final step is to write the empirical formula based on the ratio:
Lead sulfide compound is written as : PbS2
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Gekata [30.6K]

Answer:

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Explanation:

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