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murzikaleks [220]
3 years ago
15

A 60g sample of tetraethyl lead, a gasoline additive, is found to contain 38.43g lead, 17.83g carbon and 3.74g hydrogen. Determi

ne the compounds empirical formula.
Chemistry
1 answer:
anastassius [24]3 years ago
6 0

Answer:

Empirical formula (which matches the molecular formula) is = PbC₈H₂₀

Explanation:

Our sample: 60 g of tetraethyl lead

In order to determine the compound empirical formula we need the centesimal composition:

(Mass of element / Total mass) . 100 =

(38.43 g lead / 60g ) . 100 = 64.05%

(17.83 g C / 60g) . 100 = 29.72%

(3.74 g H / 60g) . 100 = 6.23 %

These % are the mass of the elements in 100 g of compound. Let's find out the moles of them:

64.05 g / 207.2 g/mol = 0.309 moles

29.72 g / 12 g/mol = 2.48 moles

6.23 g/ 1 g/mol = 6.23 moles

Next, we divide the moles, by the lowest value of them (0.309)

0.309 / 0.309 = 1 mol Pb

2.48 / 0.309 = 8 mol C

6.23 / 0.309 = 20 mol H

There, we have our formula PbC₈H₂₀

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how many moles of sodium hydroxide (NaOH) are required to completely neutralize 2 mol of nitric acid (HNO3)
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2 moles of sodium hydroxide will be needed.

<h3><u>Explanation</u>:</h3>

Sodium hydroxide is a compound which is a base and nitric acid is the acid. The formula of the nitric acid is HNO3 and that of sodium hydroxide is NaOH.

The reaction between them are

NaOH +HNO3 =NaNO3 +H2O.

So here we can see that 1 mole of sodium hydroxide reacts with 1 mole of nitric acid to produce 1 mole of sodium nitrate and 1 mole of water.

So for 2 moles of nitric acid, 2 moles of sodium hydroxide will be required.

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3 years ago
Identify the reactants 4 A g + 2 H 2 S + O 2​
Arturiano [62]
2Ag2S + 2H2O—>4Ag+2H2S+O2

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8 0
2 years ago
In lab, a dialysis bag was filled with a 10% sucrose solution and placed in an unlabeled beaker filled with clear liquid. The di
DaniilM [7]

Answer:

Of lower concentration or less concentrated

Explanation:

Osmosis is the movement of solvent from a region of lower concentration of solute to a region of higher concentration of solute through a semipermeable membrane in order to equalize the concentration of the solutions on both sides.

Since the membrane of the bag is semipermeable, then the fact that the bag in the beaker decreased in size, lost volume, and became flaccid indicates that the solution in the bag is of lower solute concentration than the solution in the beaker hence the movement of water molecules into the beaker by osmosis.

5 0
2 years ago
A 150.0 mL sample of an aqueous solution at 25°C contains 15.2 mg of an unknown nonelectrolyte compound. If the solution has an
inysia [295]

<u>Answer:</u> The molar mass of the unknown compound is 223.2 g/mol

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

where,

\pi = osmotic pressure of the solution = 8.44 torr

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = ?

R = Gas constant = 62.3637\text{ L torr }mol^{-1}K^{-1}

T = temperature of the solution = 25^oC=[273+25]=298K

Putting values in above equation, we get:

8.44torr=1\times M\times 62.3637\text{ L. torr }mol^{-1}K^{-1}\times 298K\\\\M=\frac{8.44}{1\times 62.3637\times 298}=4.54\times 10^{-4}M

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Molarity of solution = 4.54\times 10^{-4}M

Given mass of unknown compound = 15.2 mg = 0.0152 g   (Conversion factor:  1 g = 1000 mg)

Volume of solution = 150.0 mL

Putting values in above equation, we get:

4.54\times 10^{-4}M=\frac{0.0152\times 1000}{\text{Molar mass of unknown compound}\times 150.0}\\\\\text{Molar mass of unknown compound}=\frac{0.0152\times 1000}{150.0\times 4.54\times 10^{-4}}=223.2g/mol

Hence, the molar mass of the unknown compound is 223.2 g/mol

5 0
3 years ago
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