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Stella [2.4K]
4 years ago
12

When an antifreeze like salt is used, the freezing point of a solution is lowered. The amount by which it is lowered is called t

he _____.
*freezing point depression
*boiling point elevation
*cooling point declension
*freezing point elevation
Chemistry
2 answers:
Ivan4 years ago
5 0
Freezing point depression.
Lena [83]4 years ago
5 0

Answer : The correct option is, freezing point depression

Explanation :

Depression in freezing point : The depression in freezing point is depends on the molarity of the solution that is the number of moles of solute dissolved in 1000 gram of the solvent. It does not depend on the mature of the solute.

When an antifreeze like salt is used, the freezing point of a solution is lowered. The amount by which it is lowered is called the freezing point depression.

Hence, the correct option is, freezing point depression

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The activation energy of a chemical reaction is the energy that
Svetlanka [38]

Explanation:

The activation energy of a chemical reaction is the amount of energy that must be added to go from the energy level of the reactants to the energy level of transition state.

3 0
3 years ago
What is the molarity of a solution containing 72.9 grams of HCl in enough water to make
Karolina [17]

Answer:

3.9mole/liter

Explanation:

Rtsjsimkymkdmdjmkkmkdnjsdm

6 0
3 years ago
PLEASE PLEASE HELP ME!! I WILL BRAINLIST YOU!! A 28.0 g sample of N2 is in a rigid 4.50 L container at 32 °C. Calculate the pres
chubhunter [2.5K]

Answer:

P=4184.36 torr

Explanation:

For this problem we can use the idea gas law,

PV=nRT

where P is pressure, V is volume, n is moles of substance, R is the constant, and T is temperature, We will need to manipulate it and a couple values so that we can get our answer in torr. To do this, let's rearrange the equation to solve for P.

P=\frac{nRT}{V}

Next, let's convert T (32) to kelvin. To do this, add 273 to the value.

32+273=305

Next, let's convert 28g of N2 to moles of N2. To do this, divide 28 by the molar mass of N2 (28.02)

\frac{28.0}{28.01} \\=.99 moles of N2

Next, we need to find out what value of R we need to use. Because you want your answer in torr, we will need to use the value of

R=62.36\frac{torr}{mol*K}

Now that we have all of our values, let's plug them into the equation (I'll be excluding units for simplicity but they cancel out to leave units of torr in the answer)

P=\frac{(.99)(62.36)(305)}{4.50}\\P=4184.36

P=4184.36 torr

<em>Very briefly</em>, I don't know what your periodic table looks like, but mine has nitrogen's molar mass as 14.01. If you have a different mass of nitrogen, this pretty drastically impacts the value. For example, if your nitrogen's molar mass was rounded to 14.0 (making N2 28g) then it would be one mole instead of .99 moles, raising the answer to 4228.70 torr. If you have any different values just plug them in in their proper spots. A similar concept applies to the Kelvin conversion.

6 0
3 years ago
How much (in mL) 12.1 M solution of hydrochloric acid would be required to make 500 mL of a 2.5 M solution
victus00 [196]

Answer:

Volume of the concentrated solution, which is needed is 103.30 mL

Explanation:

Let's apply the formula for dilutions to solve the problem

Conc. Molarity . Conc. volume = Dil. Molarity . Dil volume

12.1 M . Conc. volume = 2.5 M . 500 mL

Conc. volume = (2.5 M . 500 mL) / 12.1M

Conc. volume = 103.30 mL

6 0
3 years ago
When 16.4 g of steam at 100°C condenses how much heat is released?
OLEGan [10]

Answer:

37064 J

Explanation:

Data Given:

mass of Steam (m) = 16.4 g

heat released (Q) = ?

Solution:

This question is related to the latent heat of condensation.

Latent heat of condensation is the amount of heat released when water vapors condenses to liquid.

Formula used

            Q = m x Lc . . . . . (1)

where

Lc = specific latent heat of condensation

Latent heat of  vaporization of water is exactly equal to heat of condensation with - charge

So, Latent heat of  vaporization of water have a constant value

Latent heat of  vaporization of water = 2260 J/g

So

Latent heat of condensation of water will be = - 2260 J/g

Put values in eq. 1

          Q = (16.4 g) x (- 2260 J/g)

          Q = - 37064 J

So,  37064 J of heat will be released negative sign indicate release of energy

4 0
3 years ago
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