How would you find the density of a can of soda pop?

A. Find the mass of the can of soda pop and then multiply by the number of cubic centimeters in the can
B. Find the mass of the can of soda pop and then divide by the number of cubic centimeters in the can
C. Convert a gallon into cubic centimeters and then divide by the mass of the can of soda pop
D. Convert a gallon into cubic centimeters and then subtract the mass of the can of soda pop

Answers

Answer 1

Answer:

it's A.

Explanation:

have uh good day ma :)))))))


Related Questions

5). At what temperature (K) will 0.854 moles of neon gas occupy 12.3 L at 1.95
atmospheres?

Answers

Answer:

338.38 K

Explanation:

Applying,

PV = nRT............... Equation 1

Where P = pressure, V = Volume, R = Temperature, n = number of moles, T = temperature.

Make T the subject of the equation,

T = PV/nR............. Equation 2

From the question,

Given: P = 1.95 atm, V = 12.3 L, n = 0.854 moles

Constant: R = 0.083 L.atm/K.mol

Substitute these values into equation 2

T = (1.95×12.3)/(0.854×0.083)

T = 338.38 K

why do we need to rinse the mouth before collecting the saliva​

Answers

The rinsing of mouth is to remove food particles and minimize mucous

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(11) Deduce the number of dyes in food colouring H.

(iii) Suggest why food colouring F does not move during the experiment.

(iv) Explain which two food colourings contain the dye that is likely to be the most soluble the solvent.

(b) Determine which food colouring contains a dye with R, value closest to 0.67
Show your working.

Answers

Answer:

(ii) 1 dye

(iii) Food coloring F is insoluble in the solvent

(iv) 'E' and 'H'

(b) Food colouring G

Explanation:

Paper chromatography principle is based on the rates of migration of chemicals across a sheet of paper which are different and it consists of a stationary phase such as the water in the paper and a mobile phase such as the solvent resulting in the partitioning of the components of the mixture across the paper

The solution components are positioned to start in one place from where they migrate and separate out on the chromatography paper

(ii) The number of components into which the food colouring 'H' separates into = 1

Therefore, the number of dyes in food colouring 'H' = 1 dye

(iii) Food coloring 'F' does not move because it is insoluble in the solvent, which is the mobile phase

(iv) The food colouring that contains the dye that is likely to be most soluble in the solvent are does for which the dyes travel furthest, which are;

Food coloring 'E' and 'H'

(b) Using a similar question solution found on 'tutor my self' website, we have;

The [tex]R_f[/tex] values are given as follows;

[tex]R_f = \dfrac{Distance \ moved \ by \ dye}{Distance \ moved \ by \ solvent}[/tex]

The distance moved by the solvent = 5 units

The distance moved by dyes in food colouring 'E' and 'H' = 4 units

The distance moved by dye in food colouring 'G' = 3.3 units

The distance moved by the second dye in food colouring 'E' = 2.7 units

By inspection, we get;

[tex]R_f[/tex] dye in food colouring 'G' = 3.3/5 = 0.66,

Therefore, the dye with [tex]R_f[/tex] value closest to 0.67 is the dye in food colouring 'G'.

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