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fgiga [73]
1 year ago
9

silicon, tin, and lead are all in the same group. the density of tin is about 7.28 g/cm3 and the density of lead is about 11.34

g/cm3. based on this group trend, what do you estimate the density of silicon to be?
Chemistry
1 answer:
Sophie [7]1 year ago
7 0

The atom's radius and mass increase from top to bottom. The density of elements from top to bottom also increases in a periodic table.

Density of Sn= Mass of Sn/Volume of Sn

                     =24.3968/7.28

                     =3.49 g/ml*ml/cm³

                     =3.49 g/cm³

It is accurate to assume that the density of Sn will fall between that of Si and lead .  the mass number and radius of the atoms increase from group to group as we move from top to bottom in a periodic table.. The density rises because the increase in mass number is greater than the increase in radius. Simply multiplying by the molar mass of the gas  allows one to derive the formula for mass density from the formula for number density. Do not forget that the mass is equal to the moles (n) times the molar mass (M).  Smaller masses can be measured in grams instead of kilograms (kg), which is the standard SI unit for mass.

To learn more about density please visit:

brainly.com/question/406690

#SPJ4

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Answer: The correct option is (c). The total pressure doubles.

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p_{O_2}=Tp_1\times X_{O_2}  (X_{O_2}=\frac{4}{4}) ( according to Dalton's law of partial pressure)

p_{O_2}=Tp_1\times 1=Tp_1....(1)

Tp_1= Total pressure when only oxygen gas was present.

Final total pressure when 4 moles of helium gas were added:

X'_{O_2}=\frac{4}{8}=\farc{1}{2},X_{He}=\frac{4}{8}=\frac{1}{2}

partial pressure of oxygen in the mixture :

Since, the number of moles of oxygen remains the same, the partial pressure of oxygen will also remain the same in the mixture.

p_{O_2}=Tp_2\times X'_{O_2}=Tp_2\times \frac{1}{2}

Tp_2= Total pressure of the mixture.

from (1)

Tp_1=Tp_2\times X'_{O_2}=Tp_2\times \frac{1}{2}

On rearranging, we get:

Tp_2=2\times Tp_1

The new total pressure will be twice of initial total pressure.

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Now add all the coefficients/moles.

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<u>Answer:</u> C) be hypertonic to Tank B.

<u>Explanation: </u>

<u> The ability of an extracellular solution to move water in or out of a cell by osmosis</u> is known as its tonicity. Additionally, the tonicity of a solution is related to its osmolarity, which is the <u>total concentration of all the solutes in the solution. </u>

Three terms (hypothonic, isotonic and hypertonic) are used <u>to compare the osmolarity of a solution with respect to the osmolarity of the liquid that is found after the membrane</u>.  When we use these terms, we only take into account solutes that can not cross the membrane, which in this case are minerals.

  1. If the liquid in tank A has a lower osmolarity (<u>lower concentration of solute</u>) than the liquid in tank B, the liquid in tank A would be hypotonic with respect to the latter.
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In the case of the problem, option A is impossible because the minerals can not cross the membrane, since it is permeable to water only. There is no way that the concentration of minerals decreases in tank A, so <u>the solution in this tank can not be hypotonic with respect to the one in Tank B. </u>

Equally, both solutions can not be isotonic and neither we can say that the solution in tank A has more minerals that the one in tank B because the liquid present in tank B is purified water that should not have minerals. Therefore, <u>options B and D are also not correct.</u>

Finally, the correct option is C, since in the purification procedure the water is extracted from the solution in tank A to obtain a greater quantity of purified water in tank B. In this way, the solution in Tank A would be hypertonic to Tank B.

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