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3 years ago

Any five uses of solution.....

2 answers:

7 0

Household cleaners like bleach are solutions, they help us to make our houses clean. Beverages such as fizzy drinks, mineral water and tea are solutions. 3.. In hydroponics, the roots of plants are immersed in solutions containing dissolved mineral salts

laiz [17]3 years ago

3 0

Answer:

The majority of chemical processes are reactions that occur in solution. Important industrial processes often utilize solution chemistry. "Life" is the sum of a series of complex processes occurring in solution. Air, tap water, tincture of iodine, beverages, and household ammonia are common examples of solutions.Turpentine as a solvent are used in the production of paints, inks and dyes. ↔Water as a solvent is used in the making of food, textiles, soaps and detergents. ↔Alloys are solid solutions that are used in the manufacture of cars, aerospace and other vehicles.Household cleaners like bleach are solutions, they help us to make our houses clean. Beverages such as fizzy drinks, mineral water and tea are solutions. In hydroponics, the roots of plants are immersed in solutions containing dissolved mineral salts.

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15. If you dilute a 6 M solution of HCl from 5 mL to 50mL, what is the concentration of this new solution? (M1V1 = M2V2)

Andreyy89

Answer:

B) 0.6M

Explanation:

I apologize in advance if it is not correct :l

The (M1V1= M2V2) is given for you to plug in the correct numbers so let's jot this down.

(M1*V1= M2*V2)

so they give us 6M which would be our (M1), from this we can also conclude that 5mL is also V1; ( if you notice the M1's and V1's are always found next to eachother). This leads us to our 50mL, this would be our V2 because the volume went from 5mL to 50mL. Now lets put this in order based on what we know.

M1= 6M (M1*V1= M2*V2)

V1= 5mL

M2= ?

V2= 50mL

now we plug in what we know into the equation to find the unknown (M2)

(6M*5mL= M2*50mL)

now we could do the long math, but I don't think your on brainly to do the hard way. so lets keep it simple!

We are going to put the 50mL under the (6M*5mL) for division.

$\frac{(6M*5mL)}{(50mL)}$ This is honestly MUCH easier, than manually answering. you just put that in the calculator and it'll give you B) 0.6M

honestly though I might not know what I'm doing cuz im currently doing my test and decided to answer this question ;)

Good Luck!

6 0

3 years ago

Help pls :) I am stuck on this chemistry question about percentage yields!

Charra [1.4K]

You just switch them around

4 0

3 years ago

Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres

yulyashka [42]

<u>Answer:</u> The freezing point of solution is -0.454°C

<u>Explanation:</u>

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 2

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (KCl) = 5.0 g

$M_{solute}$ = Molar mass of solute (KCl) = 74.55 g/mol

$W_{solvent}$ = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC$