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stiv31 [10]
3 years ago
6

PLEAASSEEE Explain, really confused

Chemistry
1 answer:
Alex17521 [72]3 years ago
4 0
We can get the mass of the lead metal by calculating the difference between the Mass of evaporating dish and lead metal and the mass of empty evaporating dish. The mass of lead metal is 1.9627 grams. The mass of sulfide is obtained by calculating the difference between Mass of evaporating dish and lead sulfide and the mass of evaporating dish and lead metal. The mass of sulfur added is 0.558 grams. We convert masses to moles. Lead is 0.0093 moles while sulfur is 0.017 moles. We divide each with the less amount between the two. Hence, lead is 1 an sulfur 1.83 or 2. The formula is PbS2 or lead (IV) sulfide. 
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6. Calculate the molarity if 35.0 grams of KBr are dissolved in 700. ml?
olga55 [171]

Answer:

To calculate molarity, divide the number of moles of solute by the volume of the solution in liters. If you don't know the number of moles of solute but you know the mass, start by finding the molar mass of the solute, which is equal to all of the molar masses of each element in the solution added together.

Explanation:

try starting with 35.0 and dived it by the volume

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3 years ago
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Describe the steps used to create Lewis dot structures to represent covalent bonds.
antiseptic1488 [7]
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2 years ago
Iron is extracted from iron oxide in the Blast Furnace: Fe 2 O 3 + 3 CO → 2 Fe + 3 CO 2
arsen [322]

a. mass of iron = 69.92 g

b. percent yield = 93%

<h3>Further eplanation </h3>

Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated

General formula:

Percent yield = (Actual yield / theoretical yield )x 100%

An actual yield is the amount of product actually produced by the reaction. A theoretical yield is the amount of product that you calculate from the reaction equation according to the product and reactant coefficients

a.

Reaction

Fe₂O₃+3CO⇒2Fe+3CO₂

MW Fe₂O₃ :  159.69 g/mol

mol Fe₂O₃

\tt \dfrac{100}{159,69}=0.626

mol Fe₂O₃ : mol Fe = 1 : 2

mol Fe :

\tt \dfrac{2}{1}\times 0.626=1.252

mass of Fe(Ar=55.845 g/mol) :

\tt 1.252\times 55.845=69.92~g

b.

actual yield = 65 g

theoretical yield = 69.92 g

percent yield :

\tt =\dfrac{65}{69.92}=0.93=93\%

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