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Vinil7 [7]
4 years ago
6

The atomic mass of 12C is 12.000000. Multiply the atomic mass of 12C by its abundance. Report the number to 5 significant digits

.
Chemistry
2 answers:
Anna35 [415]4 years ago
5 0

Answer:

1187.2

Explanation:

Given data:

Atomic mass of C₁₂ = 12.000000.

Multiplication of its abundance and mass = ?

Solution:

The natural abundance of carbon-12 is 98.93%.

98.93 × 12. 000000 = 1187.16

Answer in five significant figures:

1187.2

aleksley [76]4 years ago
3 0

Answer : The average mass of carbon is, 11.872 amu

Explanation : Given,

Atomic mass of C-12 = 12.000000 amu

Natural abundance of C-12 = 98.93 % = 0.9893

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Average mass of carbon = Atomic mass of C-12 × Natural abundance of C-12

Average mass of carbon = 12.000000 × 0.9893

Average mass of carbon = 11.872 amu

Thus, the average mass of carbon is, 11.872 amu

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Camphor, a white solid with a pleasant odor, is extracted from the roots, branches, and trunk of the camphor tree. Assume you di
katrin [286]

<u>Answer:</u> The molarity of solution is 0.799 M , molality of solution is 1.02 m, mole fraction of camphor is 0.045 and mass percent of camphor in solution is 13.43 %

<u>Explanation:</u>

  • <u>Calculating the molarity of solution:</u>

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

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Given mass of camphor = 70.0 g

Molar mass of camphor = 152.2 g/mol

Volume of solution = 575 mL

Putting values in above equation, we get:

\text{Molarity of camphor}=\frac{70\times 1000}{152.2\times 575}\\\\\text{Molarity of camphor}=0.799M

  • <u>Calculating the molarity of solution:</u>

To calculate the mass of ethanol, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of ethanol = 0.785 g/mL

Volume of ethanol = 575 mL

Putting values in above equation, we get:

0.785g/mL=\frac{\text{Mass of ethanol}}{575mL}\\\\\text{Mass of ethanol}=(0.785g/mL\times 575mL)=451.38g

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

\text{Molality of solution}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

m_{solute} = Given mass of solute (camphor) = 70 g

M_{solute} = Molar mass of solute (camphor) = 152.2  g/mol

W_{solvent} = Mass of solvent (ethanol) = 451.38 g

Putting values in above equation, we get:

\text{Molality of camphor}=\frac{70\times 1000}{152.2\times 451.38}\\\\\text{Molality of camphor}=1.02m

  • <u>Calculating the mole fraction of camphor:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

<u>For camphor:</u>

Given mass of camphor = 70 g

Molar mass of camphor = 152.2 g/mol

Putting values in equation 1, we get:

\text{Moles of camphor}=\frac{70g}{152.2g/mol}=0.459mol

<u>For ethanol:</u>

Given mass of ethanol = 451.38 g

Molar mass of ethanol = 46 g/mol

Putting values in equation 1, we get:

\text{Moles of ethanol}=\frac{451.38g}{46g/mol}=9.813mol

Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

Moles of camphor = 0.459 moles

Total moles = [0.459 + 9.813] = 10.272 moles

Putting values in above equation, we get:

\chi_{(camphor)}=\frac{0.459}{10.272}=0.045\

  • <u>Calculating the mass percent of camphor:</u>

To calculate the mass percentage of camphor in solution, we use the equation:

\text{Mass percent of camphor}=\frac{\text{Mass of camphor}}{\text{Mass of solution}}\times 100

Mass of camphor = 70 g

Mass of solution = [70 + 451.38] = 521.38 g

Putting values in above equation, we get:

\text{Mass percent of camphor}=\frac{70g}{521.38g}\times 100=13.43\%

Hence, the molarity of solution is 0.799 M , molality of solution is 1.02 m, mole fraction of camphor is 0.045 and mass percent of camphor in solution is 13.43 %

3 0
3 years ago
Rolls of foil are 300mm wide and 2.020mm thick. (The density of foil is 2.7 g/cm^3). What maximum length of foil can be made fro
horrorfan [7]

Answer:  8556 mm, or 855.6 cm (8560 mm to 3 sig figs)

Explanation:  Convert mm to cm by dividing by 10 (1cm/10mm)

Find the area of the foil face in cm^2 (30cm*0.2020cm) = 0.606 cm^2

Calculate the volume occupied by 1.40 kg of foil in cm^3.  1.40kg = 1400g

1.400g/(2.7 g/cm^3) = 518.5 cm^3 for 1.40 kg Au

   Volume = Area (of the face) * Length  

We want Length:

Length = Volume/Area

L = (518.5 cm^3/0.606 cm^2)

L = 855.6 cm (8556 mm)  Round to 3 sig figs (856 cm and 8560 mm)

5 0
3 years ago
What explanation can you give for why the sodium-potassium pump does not run out of ions to move in or out of the cell
rewona [7]

The sodium-potassium pump does not run out of ions since ion exchange is essential for the action potential to take place and to maintain homeostasis.

The cell has variable concentrations of different substances compared to the environment that surrounds it, with significant differences with sodium and potassium.

  • The main function of the sodium-potassium pump is to maintain homeostasis of the intracellular medium, controlling the concentrations of these two ions.

  • In order to carry out the adequate exchange of sodium and potassium ions in the extra and intracellular medium, the cells need an active transport process that is carried out thanks to the sodium potassium pump.

  • This process is needed for the maintenance and functioning of cells, and it is essential for the action potential to be executed, necessary for the transmission of electrical impulses from neuron to neuron.

Therefore, we can conclude that the sodium potassium pump produces an exchange of potassium ions for sodium ions which keeps the cellular system functioning properly.

Learn more here: brainly.com/question/24336764

6 0
3 years ago
Most plants and all animals depend on _____ to meet their need for nitrogen.
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They depend on nitrogen-fixing bacteria, which convert atmospheric nitrogen into a usable form.
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Es algo de ciencias
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