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3 years ago

What is the density, in g/mL, of a cube of lead (Pb) that weighs 0.371 kg and has a volume of 2.00 in.3

Chemistry

1 answer:

myrzilka [38]3 years ago

7 0

Answer:

11.32 g/mL

Explanation:

Given that:-

Mass = 0.371 kg = 371 g ( 1 kg = 1000 g)

Volume = 2.00 in³

The conversion of in³ to mL is as shown below:-

1 in³ = 16.3871 mL

So, Volume = $2.00\times 16.3871\ mL$ = 32.7741 mL

The expression for the calculation of density is shown below as:-

$\rho=\frac{m}{V}$

Applying the values as:-

$\rho=\frac{371\ g}{32.7741\ mL}=11.32\ g/mL$

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2 years ago

Where is almost all the mass of an atom

VashaNatasha [74]

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2 years ago

I need to make up an experiment !! Please help

katen-ka-za [31]

Question: Baking a Cake Without Flour.

Hypothesis: I think that when I remove the flour from the standard cake recipe, I'll end up with a flat but tasty cake.

Procedure: I baked two cakes during my experiment. For my control, I baked a cake following a normal recipe. I used the Double Fudge Cake recipe on page 292 of the Betty Crocker Cookbook. For my experimental cake, I followed the same recipe but left out the flour. I first obtained a 2-quart mixing bowl.

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What did I learn?/Conclusion: Since the experimental cake burned, my results did not support my hypothesis. I think that the cake burned because it had less mass, but cooked for the same amount of time. I propose that the baking time be shortened in subsequent trials.

I hope this helped :))

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2 years ago

Draw the structure of an alkane or cycloalkane that has more than three but fewer than ten carbon atoms, and only primary hydrog

goldfiish [28.3K]

Answer:

The structures are shown in the figure.

Explanation:

The primary hydrogens are those which are attached to primary carbon.

Primary carbons are the carbons which are attached to only one carbon.

Primary carbons is bonded to three hydrogens.

In order to draw such structure we will draw structures which will have carbon with three hydrogens or no hydrogens (quaternary)

The structures are shown in the figure with clear marking.

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2 years ago

A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

sammy [17]

Answer: The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

Explanation:

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(S_8)$ = 0.800 g

$M_{solute}$ = Molar mass of solute $(S-8)$ = 256.52 g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m$

Calculation for freezing point of solution:

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

$\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

or,

$\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm$

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC$

Hence, the freezing point of solution is 16.5°C

Calculation for boiling point of solution:

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

or,

$\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm$

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC$

Hence, the boiling point of solution is 118.2°C

5 0

3 years ago