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Ierofanga [76]
3 years ago
15

Ethanol is a common laboratory solvent and has a density of 0.789 g/ml. What is the mass, in grams, of 125 ml of ethanol?

Chemistry
1 answer:
olga55 [171]3 years ago
3 0

The density of a material is its unique property by which a unknown material can be identified and also the impurity (if present) in a material can be concluded. Mathematically the density can be expressed as-\frac{mass}{volume}. Thus from the mathematical expression we can say that the density is the mass per unit volume of a material. Here the density of ethanol is given 0.789 g/mL. Thus the weight of the 1 mL ethanol is 0.789g. Thus the weight of the 125 mL of ethanol will be (125×0.789) = 98.625 g.

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Be sure to answer all parts.
Liula [17]

A. The patch's area in square kilometers (km²) is 1.61×10⁻⁹ km²

B. The cost of the patch to the nearest cent is 734 cents

<h3>A. How to convert 16.1 cm² to square kilometers (km²)</h3>

We can convert 16.1 cm² to km² as illustrated below:

Conversion scale

1 cm² = 1×10⁻¹⁰ km²

Therefore,

16.1 cm² = 16.1 × 1×10⁻¹⁰

16.1 cm² = 1.61×10⁻⁹ km²

Thus, 16.1 cm² is equivalent to 1.61×10⁻⁹ km²

<h3>B. How to determine the cost in cent</h3>

We'll begin by converting 16.1 cm² to in². This can be obtained as illustrated below:

1 cm² = 0.155 in²

Therefore,

16.1 cm² = 16.1 × 0.155

16.1 cm² = 2.4955 in²

Finally, we shall the determine the cost in centas fo r llow:

  • Cost per in² = $2.94 = 294 cent
  • Cost of 2.4955 in² =?

1 in² = 294 cent

Therefore,

2.4955 in² = 2.4955 × 294

2.4955 in² = 734 cents

Thus, the cost of the patch is 734 cents

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6 0
1 year ago
How would you prepare 1.00 L of a 0.50 M solution of each of the following?
Lunna [17]

Answer:

Take approx 41.7 mL of 12-M HCl in a 1.00-L flask and fill the rest of the volume with distilled water.

Explanation:

Hello,

In this case, for the dilution process from concentrated  12-M hydrochloric acid to 1.00 L of the diluted 0.50M hydrochloric acid, the volume of concentrated HCl you must take is computed by considering that the moles remain constant for all dilution processes as shown below:

n_1=n_2

Which can also be written in terms of concentrations and volumes:

M_1V_1=M_2V_2

Thus, solving for the initial volume or aliquot that must be taken from the 12-M HCl, we obtain:

V_1=\frac{M_2V_2}{M_1} \\\\V_1=\frac{1.00L*0.5M}{12M}\\ \\V_1=0.0417L=41.7mL

It means that you must take approx 41.7 mL of 12-M HCl in a 1.00-L flask and fill the rest of the volume with distilled water for such preparation.

Best regards.

6 0
3 years ago
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