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Leya [2.2K]
3 years ago
10

Which of the following does not have a uniform composition throughout? A) Compound B) Homogeneous mixture C) Heterogeneous mixtu

re D) Solvent E) Element The mass of a sample is 550 milligrams. Which of the following expresses that mass in kilograms? A) 5.5 times 10^5 kg B) 5.5 times 10^-4 kg C) 5.5 times 10^-1 kg D) 5.5 times 10^8 E) 5.5 times 10^-6 kg What is the volume of a container that contains 14.3 g of a substance having a density of 0.988 g/cm^3? A) 691 cm^3 B) 14.1 cm^3 C) 14.5 cm^3 D) 141 cm^3 E) 0.0691 cm^3 How many neutrons are there in an atom of lead whose mass number is 208? A) 126 B) 82 C) 290 D) 208 E) none of them
Chemistry
1 answer:
Mumz [18]3 years ago
5 0

Answer:

Explanation:

Heterogeneous mixture = C

1 gram = 1000 milligram

1000gram= 1kg

1000x1000 milligram = 1kg

1,000,000milligram = 1kg

550milligram = 550/1,000,000 = 5.5 x 10 ^ -4

5.5 times 10^-4 kg= B

Density = Mass / Volume

Volume = mass / density = 14.3/0.988 = 14.47cm^3

Lead has 82 protons,

An atom comprises of proton and neutrons.

the number of neutrons in an atom of lead given as 208 will then

Atom = proton + neutron

208 = 82 + neutron

208 -82 = 126 neutrons

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The pH of a liquid substance is calculated through the equation,
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Substituting the given concentration of the hydronium ion to the equation above,
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A bottle of wine contains 9.81 grams of C2H5OH, dissolved in 87.5 grams of water. The final volume of the solution is 100.0 mL.
dimulka [17.4K]

Answer:

[EtOH] = 2.2M and Wt% EtOH = 10.1% (w/w)

Explanation:

1. Molarity = moles solute / Volume solution in Liters

=> moles solute = mass solute / formula weight of solute = 9.8g/46g·mol⁻¹ = 0.213mol EtOH

=> volume of solution (assuming density of final solution is 1.0g/ml) ...

volume solution =  9.81gEtOH + 87.5gH₂O = 97.31g solution x 1g/ml = 97.31ml = 0.09731 Liter solution

Concentration (Molarity) = moles/Liters = 0.213mol/0.09731L = 2.2M in EtOH

2. Weight Percent EtOH in solution (assuming density of final solution is 1.0g/ml)

From part 1 => [EtOH] = 2.2M in EtOH = 2.2moles EtOH/1.0L soln

= {(2.2mol)(46g/mol)]/1000g soln] x 100% = 10.1% (w/w) in EtOH.

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