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antoniya [11.8K]
3 years ago
11

Aluminum has a density of 2.70 g/mL. Calculate the mass (in grams) of a piece of aluminum having a volume of 381 mL . Express yo

ur answer to three significant figures.
Chemistry
2 answers:
laiz [17]3 years ago
4 0

Answer:

Mass ≈ 1030g (Three significant Figures)

Explanation:

We were given;

Density = 2.70 g/mL

Mass = ?

Volume = 381 mL

The relationship between all three parameters mentioned above is given by the formular below;

Density = Mass / Volume

Upon solving for Mass, we have;

Mass = Density * Volume

Substituting the respective values;

Mass = 2.7 * 381

Mass = 1028.7g

Mass ≈ 1030g (Three significant Figures)

BaLLatris [955]3 years ago
3 0
Multiply the volume by the density. 1 ml = 1 cm^3

1.11 g/cm^3 * 1 cm^3/mL * 386 mL = 428 g

All your units cancel out except for g which is what you would expect since you want units of mass.
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A chemistry student needs 60.0 g of tetrahydrofuran for an experiment. By consulting the CRC Handbook of Chemistry and Physics,
Evgen [1.6K]

Answer:

67.492cm^3

Explanation:

To get the volume of tetrahydrofuran needed, what the student need to do is to apply mathematical calculations.

We know quite well that to obtain the volume of a substance, given the mass of the substance and the density of the substance is possible.

The volume of the substance is simply the mass of the substance divided by the density of the substance.

This is thus volume = mass/density

The mass given is the question is 60g while the density given is 0.889g/cm^3

The volume = 60/0.889 = 67.492cm^3

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2 years ago
4. Identify What do scientists call a substance that forms during a chemical reaction?
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Answer:

4. Reactant

5. Left

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7 0
2 years ago
Please help me out i'm new here but i really need help for science (if you need help with any hw like ela i can help!) (50 point
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2 years ago
Which part of Earth most likely has the highest albedo value?
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2 years ago
If 15.6 grams of copper (ii) chloride react with 20.2 grams of sodium nitrate how many grams of sodium chloride can be formed? W
olasank [31]

Answer:

- 13.56 g of sodium chloride are theoretically yielded.

- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.

- 0.50 g of sodium nitrate remain when the reaction stops.

- 92.9 % is the percent yield.

Explanation:

Hello!

In this case, according to the question, it is possible to set up the following chemical reaction:

CuCl_2+2NaNO_3\rightarrow 2NaCl+Cu(NO_3)_2

Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:

m_{NaCl}^{by\ CuCl_2}=15.6gCuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2} *\frac{2molNaCl}{1molCuCl_2} *\frac{58.44gNaCl}{1molNaCl} =13.56gNaCl\\\\m_{NaCl}^{by\ NaNO_3}=20.2gNaNO_3*\frac{1molNaNO_3}{84.99gNaNO_3} *\frac{2molNaCl}{2molNaNO_3} *\frac{58.44gNaCl}{1molNaCl} =13.89gNaCl

Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:

m_{NaNO_3}^{by\ NaCl}=13.56gNaCl*\frac{1molNaCl}{58.44gNaCl}*\frac{2molNaNO_3}{2molNaCl} *\frac{84.99gNaNO_3}{1molNaNO_3}=19.72gNaNO_3

Therefore, the leftover of sodium nitrate is:

m_{NaNO_3}^{leftover}=20.2g-19.7g=0.5gNaNO_3

Finally, the percent yield is computed via:

Y=\frac{12.6g}{13.56g} *100\%\\\\Y=92.9\%

Best regards!

6 0
2 years ago
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