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timama [110]
3 years ago
13

Rank the following compounds in order of decreasing solubility in water: CH3CH2−OH Ethanol CH3CH2CH2CH2−OH Butanol CH3CH2CH2CH2C

H2CH2−OH Hexanol
Chemistry
1 answer:
Alik [6]3 years ago
6 0
Order of Solubility of given alcohols is as follow,

                     Ethanol                  Butanol                   Hexanol

Highly Soluble ---------->--------------------->------------------------>---Slightly Soluble

As we know Like dissolves Like. Water is a polar solvent and dissolves readily those compounds which are polar in nature. The solubility of all these alcohols is due o the -OH (hydroxyl) group which tends to form Hydrogen Bond interactions with the water molecules. The difference comes in the alkyl group of these compounds. The akyl group is non polar in nature and immiscible in water. As the non polar chain increases from ethanol to hexanol, so the solubility also decreases from ethanol to hexanol.
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Ksju [112]
109/8.56=12.7
50+12.7
V=62.7

Mass= Volume x Density so i divided the mass and density to get the volume. and afterwards i would just add it to the mass to get my final answer

4 0
3 years ago
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27. A solution has a pH of2. Which best describes the solution? (a point)
lianna [129]

Explanation:

solution has a pH of 2 is a strong acid.

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What is the volume of 0.640 grams of Oz gas at Standard Temperature and Pressure (STP)?
Ilia_Sergeevich [38]

Answer: The volume of 0.640 grams of O_{2} gas at Standard Temperature and Pressure (STP) is 0.449 L.

Explanation:

Given: Mass of O_{2} gas = 0.640 g

Pressure = 1.0 atm

Temperature = 273 K

As number of moles is the mass of substance divided by its molar mass.

So, moles of O_{2} (molar mass = 32.0 g/mol) is as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{0.640 g}{32.0 g/mol}\\= 0.02 mol

Now, ideal gas equation is used to calculate the volume as follows.

PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

PV = nRT\\1.0 atm \times V = 0.02 mol \times 0.0821 L atm/mol K \times 273 K\\V = 0.449 L

Thus, we can conclude that the volume of 0.640 grams of O_{2} gas at Standard Temperature and Pressure (STP) is 0.449 L.

6 0
3 years ago
mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of
anygoal [31]

<u>Answer:</u> The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

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

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

<u>For nitrogen gas:</u>

Molar mass of nitrogen gas = 28 g/mol

\text{Moles of nitrogen gas}=\frac{x}{28}mol

<u>For hydrogen gas:</u>

Molar mass of hydrogen gas = 2 g/mol

\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol

Equating the moles of the individual gases to the moles of mixture:

0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g

To calculate the mass percentage of substance in mixture we use the equation:

\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100

Mass of the mixture = 13.22 g

  • <u>For nitrogen gas:</u>

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%

  • <u>For hydrogen gas:</u>

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

5 0
3 years ago
The rate constant for this second‑order reaction is
o-na [289]

Answer:

daddadaddddddddddddddddddddddddddddda

Explanation:

dddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd

3 0
3 years ago
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