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alexandr402 [8]
3 years ago
13

What volume of sample, in mL, would be needed to make 2 mL of a 5-fold dilution? (3 significant figures needed)

Chemistry
1 answer:
natulia [17]3 years ago
5 0

Answer:

0.400 mL

Explanation:

Hello, the dilution factor (in folds) is given by:

folds=\frac{total_{volume}}{sample_{volume}}

Thus, the sample volume with three significant figures is given by:

sample_{volume}=\frac{2mL}{5"folds"}=0.400 mL

Best regards.

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Which of the following is not equal to 120 centimeters?
tia_tia [17]
<h3>Answer:</h3>

0.012 dekameters (dkm)

<h3>Explanation:</h3>

<u>We are given;</u>

  • 120 centimeters( cm).

Required to identify the measurements that is not equivalent to 120 cm.

  • Centimeters are units that are used to measure length together with other units such as kilometers(km), meters (m), millimeters (mm), dekameters (dkm), etc.
  • These units can be inter-converted to one another using suitable conversion factors.
  • To do this, we are going to have a table showing the suitable conversion factor from one unit to another.

Kilometer (km)

10

Decimeter (Dm)

10

Hectometer (Hm)\

10

Meter (m)

10

Dekameter (dkm)

10

Centimeter (cm)

10

Millimeter (mm)

Therefore;

To convert cm to km

Conversion factor is 10^5 cm/km

Thus;

120 cm = 120 cm ÷ 10^5 cm/km

            = 0.0012 km

To convert cm to dkm

Conversion factor is 10 cm/dkm

Therefore,

120 cm = 120 cm ÷ 10 cm/dkm

            = 12 dkm

To convert cm to m

The suitable conversion factor is 10^2 cm/m

Thus,

120 cm = 120 cm ÷ 10^2 cm/m

           = 1.2 m

To convert cm to mm

Suitable conversion factor is 10 mm/cm

Therefore;

120 cm = 120 cm × 10 mm/cm

            = 1200 mm

Therefore, the measurement that is not equal to 120 cm is 0.012 dkm

4 0
3 years ago
What is the total number of atoms in 0.20 mol of propanone, CH3COCH3?
Nutka1998 [239]

Answer:

1.2×10²³ atoms.

Explanation:

Data obtained from the question include:

Mole of propanone = 0.20 mole

Number of atoms of propanone =.?

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.022×10²³ atoms.

This implies that 1 mole of propanone also contains 6.022×10²³ atoms.

Thus, we can obtain the number of atoms in 0.20 mole of propanone as illustrated below:

1 mole of propanone contains 6.022×10²³ atoms.

Therefore, 0.20 mole of propanone will contain = 0.2 × 6.022×10²³ = 1.2×10²³ atoms.

Thus, 0.20 mole of propanone contain

1.2×10²³ atoms.

6 0
3 years ago
Rank each of the following gases in order of increasing urms assuming equivalent amounts and all gases are at the same temperatu
anzhelika [568]
the answer is option c
5 0
3 years ago
What element was named for the scientist who discovered the nucleus of the atom using gold foil
Artist 52 [7]
<span>rutherfordium element # 104</span>
4 0
3 years ago
has a standard free‑energy change of − 3.59 kJ / mol at 25 °C. What are the concentrations of A , B , and C at equilibrium if, a
Mamont248 [21]

Answer: The concentrations of A , B , and C at equilibrium are 0.1583 M, 0.2583 M, and 0.1417 M.

Explanation:

The reaction equation is as follows.

               A + B \rightarrow C

Initial :     0.3   0.4          0

Change:  -x       -x           x

Equilbm: (0.3 - x)  (0.4 - x)  x  

We know that, relation between standard free energy and equilibrium constant is as follows.

      \Delta G = -RT ln K

Putting the given values into the above formula as follows.

      \Delta G = -RT ln K

      -3.59 kJ/mol = -8.314 \times 10^{-3} kJ/mol K ln (\frac{x}{(0.3 - x)(0.4 - x)})

                x = 0.1417

Hence, at equilibrium

  •  [A] = 0.3 - 0.1417

       = 0.1583 M

  •  [B] = 0.4 - 0.1417

       = 0.2583 M

  •  [C] = 0.1417 M
5 0
3 years ago
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