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2 years ago

A student places four identical cells into four different liquids. Over time which cell will look the smallest WXY or Z

Chemistry

1 answer:

marta [7]2 years ago

3 0

Answer: W

Explanation:

The cell in liquid W will look the smallest out of the four because of Osmosis.

Osmosis is a process by which water molecules move from an area of relatively higher concentration of water to an area with a relatively lower concentration through a selectively permeable membrane(cell membrane).

With W being saltier than the cell, water molecules will move from the cell to liquid W to balance the concentrations inside and outside the cell which will lead to the cell in W being smaller.

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1) Aluminum sulphate can be made by the following reaction: 2AlCl3(aq) + 3H2SO4(aq) Al2(SO4)3(aq) + 6 HCl(aq) It is quite solubl

kolezko [41]

Answer:

88.9%

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2AlCl3(aq) + 3H2SO4(aq) —> Al2(SO4)3(aq) + 6HCl(aq)

Step 2:

Determination of the masses of AlCl3 and H2SO4 that reacted and the mass of Al2(SO4)3 produced from the balanced equation.

Molar mass of AlCl3 = 27 + (35.5x3) = 133.5g/mol

Mass of AlCl3 from the balanced equation = 2 x 133.5 = 267g

Molar mass of H2SO4 = (2x1) + 32 + (16x4) = 98g/mol

Mass of H2SO4 from the balanced equation = 3 x 98 = 294g

Molar mass of Al2(SO4)3 = (27x2) + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96] = 342g/mol

Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g

Summary:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4 to produce 342g of Al2(SO4)3.

Step 3:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4.

Therefore, 25g of AlCl3 will react with = (25 x 294)/267 = 27.53g of H2SO4.

From the calculations made above, we see that only 27.53g out 30g of H2SO4 given were needed to react completely with 25g of AlCl3.

Therefore, AlCl3 is the limiting reactant and H2SO4 is the excess.

Step 4:

Determination of the theoretical yield of Al2(SO4)3.

In this case we shall be using the limiting reactant because it will produce the maximum yield of Al2(SO4)3 since all of it is used up in the reaction.

The limiting reactant is AlCl3 and the theoretical yield of Al2(SO4)3 can be obtained as follow:

From the balanced equation above,

267g of AlCl3 reacted to produce 342g of Al2(SO4)3.

Therefore, 25g of AlCl3 will react to produce = (25 x 342) /267 = 32.02g of Al2(SO4)3.

Therefore, the theoretical yield of Al2(SO4)3 is 32.02g

Step 5:

Determination of the percentage yield of Al2(SO4)3.

This can be obtained as follow:

Actual yield of Al2(SO4)3 = 28.46g

Theoretical yield of Al2(SO4)3 = 32.02g

Percentage yield of Al2(SO4)3 =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 28.46/32.02 x 100

Percentage yield = 88.9%

Therefore, the percentage yield of Al2(SO4)3 is 88.9%

3 0

2 years ago

If the symbol X represents a central atom, Y represents outer atoms, and Z represents lone pairs on the central atom, the struct

MA_775_DIABLO [31]

Answer:

1. sp = XY₂

2. sp² = XY₂Z, XY₃

3. sp3³ = XY₄, XY₂Z₂, XY₃Z

4. sp³d = XY₅, XY₂Z₃, XY₃Z₂, XY₄Z

5. sp³d² = XY₆, XY₄Z₂, XY₅Z

Explanation:

this is quite dicey, so it should be looked into carefully.

we would classify each of the abbreviation according to their hybridization and it electron domain.

⇒ sp hybridization = XY₂

in this, we can see that the central atom X is bonded to two outer atoms Y.

this makes the no of hybrid orbitals and the no of sigma bonds both 2.

electron domain = 2.

⇒ sp² hybridization = XY₂Z, XY₃

Here we can see the central atom X bonded with three outer atoms Y in XY₂Z and in XY₃. For XY₂Z molecule, the no of sigma bonds is 2 and the no of hybrid orbitals is 3. While for XY₃ molecule, the no of sigma bonds is 3 while the no of hybrid orbital is 3.

electron domain = 3.

⇒ sp³ hybridization = XY₄, XY₂Z₂, XY₃Z

for XY₄ molecule, the central atom X is bonded with four outer atoms Y. It has 4 numbers of both the sigma and orbital atoms.

In XY₂Z₂, the central atom X is bonded to 2 outer atoms Y, and has 2 lone pairs Z. From this, the no of hybrid orbitals is 4 and the no of sigma bonds is 2, with 2 lone pairs causing the sp³ hybridization.

⇒ sp³d hybridization = XY₅, XY₂Z₃, XY₃Z₂, XY₄Z

for all the molecules listed above, the sum of both the lone pairs and the outer atoms both give a total of 5, hence have the sp³d structure, viz;

XY₂Z₃:

total electron domain = 2+3 = 5

XY₃Z₂:

total electron domain = 2+3 = 5

XY₄Z:

total electron domain = 1+4 = 5

⇒ sp³d² hybridization = XY₆, XY₄Z₂, XY₅Z

Same thing goes for the above molecule, where the sum of both the outer atoms and the lone pairs gives a total of 6 as can be seen in the example below.

XY4Z2:

total electron domain = 2+4 = 6

cheers, i hope this helps.

5 0

2 years ago

Determine the number of bonding electrons and the number of nonbonding electrons in the structure of xef2. enter the number of b

lakkis [162]

Answer : The number of bonding electrons and the number of non-bonding electrons are (4, 18).

Explanation :

The number of bonding electrons and non-bonding electrons in the structure of $XeF_2$ is determined by the Lewis-dot structure.

Lewis-dot structure : It tell us about the number of valence electrons of an atom within a molecule and it is also shows the bonding between the atoms of a molecule and the lone-pair of electrons.

In the given structure, 'Xe' is the central atom and 'F' is the terminal atom.

Xenon has 8 valence electrons and fluorine has 7 valence electrons.

Total number of valence electrons in $XeF_2$ = 8 + 2(7) = 22 electrons

From the Lewis-dot structure, we conclude that

The number of electrons used in bonding = 4

The number of electrons used in non-bonding (lone-pairs) = 22 - 4 = 18

Therefore, the number of bonding electrons and the number of non-bonding electrons are (4, 18).

The Lewis-dot structure of $XeF_2$ is shown below.

4 0

2 years ago

Read 2 more answers

If a gas has a volume of 750 mL at 25oC, what would the volume of the gas be at 55oC?

blagie [28]

Answer:

The answer to your question is V2 = 825.5 ml

Explanation:

Data

Volume 1 = 750 ml

Temperature 1 = 25°C

Volume 2= ?

Temperature 2 = 55°C

Process

Use the Charles' law to solve this problem

V1/T1 = V2/T2

-Solve for V2

V2 = V1T2 / T1

-Convert temperature to °K

T1 = 25 + 273 = 298°K

T2 = 55 + 273 = 328°K

-Substitution

V2 = (750 x 328) / 298

-Simplification

V2 = 246000 / 298

-Result

V2 = 825.5 ml

7 0

2 years ago

Consider the picture of a gas pump. Which type of gasoline has the highest percentage of octane (the main component of gasoline)

Readme [11.4K]

Answer:

premium: 91 octane rating

Explanation:

Octane number refers to the percentage or volume fraction of isooctane in a fuel.

The octane number gives a picture of how safe a fuel is for an engine. The higher the octane rating the lesser the tendency of the fuel to cause knocking of the engine.

The type of gasoline with the highest percentage of octane among the options is premium.

5 0

2 years ago

Read 2 more answers