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3 years ago

Which is a practical application of boiling-point elevation?

1 answer:

7 0

Boiling-point is the point of a pure liquid matter starts to evaporate and change into gaseous phase. It is where the set of conditions such as the pressure and temperature enough to do so. Boiling-point elevation, on the other hand, is the phenomenon of which the boiling point of a pure liquid matter is elevated because of the dissolved substances. A great example would be the boiling point of a distilled water (pure water) which is lesser than the boiling point of a sea water because of the dissolved salts. A pure water boils at 100°C at atmospheric pressure while a salt water boils at higher temperature than 100°C at the same pressure. Thus, the answer is D.

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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

PLZ HURRY

Oxana [17]

Answer:

Explanation:

3 0

3 years ago

Read 2 more answers

Nitrogen dioxide is produced by combustion in an automobile engine. For the following reaction, 0.377 moles of nitrogen monoxide

Contact [7]

Answer:

The amount of NO₂ that can be produced 8.533 g

Explanation:

According to question

2 NO(g) + O₂(g) → 2 NO₂(g)

Given

Moles of nitrogen monoxide = 0.377

Moles of oxygen = 0.278

$'For NO'=\frac{Mole}{Stoichiometry}=\frac{0.377}{2} =0.1855\\'For O_{2} '=\frac{0.278}{1}= 0.278\\$

Since 'NO' is the limiting reagent according to this ratio.

According to equation

2 moles NO reacts to form 2 moles NO₂

So, 0.1855 moles NO give = 0.1855 moles of NO₂

Mass of 1 mole NO₂ = 46 g/mole

Mass of 0.1855 moles = 46 x 0.1855 = 8.533 g

5 0

3 years ago

What physical property makes metal pots good for cooking?

jeka94

Metal pots are good for cooking because they have heat conductivity.

4 0

3 years ago

Read 2 more answers

When a 12.8 g sample of KCL dissolves in 75.0 g of water in a calorimeter the temp. drops from 31 Celsius to 21.6 Celsius. Calcu

Delicious77 [7]

Answer:

Step 1: Calculate qsur (the surrounding is

usually the water)

qsur = ? J

m = 75.0 g water

c = 4.184 J/g

ΔT = (Tfinal- Tinitial)= (21.6 – 31.0) = -9.4 oC

qsur = m · c · (ΔT)

qsur = (75.0g) (4.184 J/g

oC) (-9.4 oC)

qsur = - 2949.72 J

First, using the information we know that we

must solve for qsur, which is the water. We know

the mass for water, 75.0g, the specific heat of

the water, 4.184 j/g

c, and the change in

temperature, 21.6-31.0 = -9.4 oC. Plugging it

into the equation, we solve for qsur.

Step 2: Calculate qsys qsys = - (qsur)

qsys = - (- 2949.72 J)

qsys = + 2949.72

In this case, the qsur is negative, which means

that the water lost energy. Where did it go? It

went to the system. Thus, the energy of the

system is negative, opposite, the energy of the

surrounding.

Step 3: Calculate moles of the substance

that is the system

Given: 12.8 g KCl

Mol system = (g system given)

(molar mass of system)

Mol system = (12.8 g KCl)

(39.10g + 35.45g)

Mol system = 12.8 g KCl

74.55 g

Mol system = 0.172

Here, we solve for the mol in the system by

using the molar mass of the material in the

system.

Step 4: Calculate ΔH ΔH = q sys .

Mol system

ΔH= + 2949.72 J

0.172 mol

ΔH= +17179.81 J/mol or +1.72 x 104

J/mol

i hope this helps

7 0

3 years ago