All categories

3 years ago

Classify which compounds will dissolve in water and which ones will not dissolve in water.

Chemistry

1 answer:

vredina [299]3 years ago

6 0

Answer:

Dissolves in water: Acetone, 1-propanol, Methanol

Do not dissolve in water: Hexane, Decane

Explanation:

From the principle that like dissolves like,polar substances will dissolve in polar solutions, while non-polar substances will dissolve in non-polar substances.

Water is a polar solution, therefore, polar substances will dissolve in it.From the given options:

Acetone is a polar molecule, therefore, it will dissolve in water.

1-propanol is a polar molecule, therefore, it will dissolve in water.

Hexane is a non-polar molecule, therefore, it will not dissolve in water.

Methanol is a polar molecule, therefore, it will dissolve in water.

Decane is a non-polar molecule, therefore, it will not dissolve in water

You might be interested in

11C. Potassium hydrogen phthalate is a solid, monoprotic acid frequently used in the laboratory to standardize strong base solut

Llana [10]

For the reactants,

The oxidation number of hydrogen = +1
The oxidation number of oxygen = -2
The oxidation number of arsenic = +5
The oxidation number of carbon = +3

For the products,

The oxidation number of hydrogen = +1
The oxidation number of oxygen = -2
The oxidation number of arsenic = +3
The oxidation number of carbon = +4

Here, arsenic (+5 to +3) and carbon (+3 to +4) are the only oxidation numbers changing.

Note that an increase in oxidation number means electrons are lost. Thus oxidation is occurring, and a decrease in oxidation number means electrons are being gained, and thus reduction is occurring.

Also, the compound that contains the element being oxidized is the reducing agent, and the compound that contains the element being reduced is the oxidizing agent.

So, the answers are:

name of the element oxidized: Carbon

name of the element reduced: Arsenic

formula of the oxidizing agent: $\text{H}_{3}\text{AsO}_{4}$

formula of the reducing agent: $\text{H}_{2}\text{C}_{2}\text{O}_{4}$

6 0

2 years ago

A chemist mixes 75.0 g of an unknown substance at 96.5°C with 1,150 g of water at 25.0°C. If the final temperature of the system

AleksAgata [21]

To do this problem it is necessary to take into account that the heat given by the unknown substance is equal to the heat absorbed by the water, but considering the correct sign:

$-m\cdot c_e\cdot \Delta T = m_w\cdot c_e_w\cdot \Delta T_w$

Clearing the specific heat of the unknown substance:

$c_e = \frac{m_w\cdot c_e_w\cdot \Delta T_w}{m\cdot \Delta T} = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot (37.1 - 25.0)^\circ C}{75\ g\cdot (37.1 - 96.5)^\circ C}$

$c_e = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot 12.1^\circ C}{75\ g\cdot (-59.4)^\circ C} = \bf 13.07\frac{J}{g\cdot ^\circ C}$

6 0

3 years ago

Which component is missing from the process of photosynthesis?

lisabon 2012 [21]

I believe the answer is 1-Water :)

3 0

3 years ago

Read 2 more answers

In the experiment, 40 mL of 3 M sodium hydroxide is used to extract the benzoic acid. In order to recover the benzoic acid from

nikdorinn [45]

Answer:

20 mL OF 6 M HYDROCHLORIC ACID WILL BE NEEDED

Explanation:

M1 V1 = M2 V2

M1 = Molarity of sodium hydroxide = 3 M

V1 = volume of sodium hydroxide = 40 mL

M2 = Molarity of hydrochloric acid = 6 M

V2 = Volume of hydrochloric acid = unknown

Rearranging the equation, we have:

V2 = M1 V1 / M2

V2 = 3 * 40 mL / 6

V2 = 120 / 6

V2 = 20 mL

To precipitate the benzoic acid by 6 M of hydrochloric acid, 20 mL volume will be needed.

6 0

3 years ago

It has been established that 8.0x10^4 tons of gold (Au) have been mined. Assume gold cost $350 per ounce. Whats the total worth

bazaltina [42]

So first lets find out how much gold there is. 10^4=10,000so that means 8.0x10,000= 8,000. now we have to find out how much money that is. All we have to do is multiply 8,000 by 350. When multiplied you get $2,800,000

7 0

3 years ago