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Anit [1.1K]
3 years ago
9

Mass , volume, and density are all properties of what

Chemistry
1 answer:
Vladimir [108]3 years ago
4 0

Answer: Physical properties

Explanation: Can be measured without changing a substance's chemical identity.

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What’s the answer!?!!
EastWind [94]

Answer: generic material and protein coat. Have a great day

Explanation:

3 0
4 years ago
DDT was banned from use as a pesticide in the United States because
Dmitry_Shevchenko [17]
Too much money and dangerous 
5 0
3 years ago
How many milliliters of water are needed to prepare a 3.5M solution of NaOH if you have .5mol of the solute
Mamont248 [21]

Answer:

1335.12 mL of H2O

Explanation:

To calculate the mililiters of water that the solution needs, it is necessary to know that the volume of the solution is equal to the volume of the solute (NaOH) plus the volume of the solvent (H2O).

From the molarity formula we can first calculate the volume of the solution:

M=\frac{solute moles}{solution volume}

Solutionvolume=\frac{solute moles}{M} =\frac{5mol}{3.5\frac{mol}{L} } =1.429L

The volume of the solution as we said previously is:

Solution volume = solute volume + solvent volume

To determine the volume of the solute we first obtain the grams of NaOH through the molecular weight formula:

MW=\frac{mass}{mol}

Mass=MW*mol=39.997\frac{g}{mol} *5mol=199.985g

Now with the density of NaOH the milliliters of solute can be determined:

d=\frac{mass}{volume}

Volume=\frac{mass}{d} =\frac{199.985g}{2.13\frac{g}{mL} } =93.88mL of NaOH

Having the volume of the solution and the volume of the solute, the volume of the solvent H2O can be calculated:

Solvent volume = solution volume - solute volume

Solvent volume = 1429 mL -  93.88 mL = 1335.12 mL of H2O

7 0
4 years ago
You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30
dimaraw [331]

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula:\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

∴

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{1}{3})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = 0.012345

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

3 0
3 years ago
How many acetyl coa molecules may be obtained from oxidation of an 18-carbon fatty acid?
saveliy_v [14]
The answer would be 9
7 0
3 years ago
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