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3 years ago

Compared to the normal freezing point and boiling point of water, a 1-molar solution of sugar in water will have a

Chemistry

1 answer:

marysya [2.9K]3 years ago

6 0

Answer :

Boiling point of the sugar solution will be higher than that of water's boling point.
Freezing point of the sugar solution will be lower than that of water's freezing point.

Explanation:

Boiling point of a liquid is defined as temperature at which vapor pressure of liquid becomes equal to the atmospheric pressure.

Boiling point of solution is always higher than that of the pure solvent

Vapor pressure increases with increase in temperature which means sugar solution will be heated more to make vapor pressure equal to atmospheric pressure.

Freezing point is defined as temperature at which solid and liquid phase are at equilibrium or temperature at which vapor pressure of liquid becomes equal to the vapor pressure in its solid phase.

Freezing point of solution is always lower than that of the pure solvent.

Lower the temperature, lower will be the vapor pressure which sugar solution solution will get freeze at lower temperature than that of the water.

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.................. are microorganism used to improve soil fertility. plz ams it correct

slavikrds [6]

Answer:

Some bacteria like rhizobium and blue green algae are able to fix nitrogen gas from the atmosphere to enrich the soil with nitrogen compounds and increase its fertility. The nitrogen-fixing bacteria and blue green algae are called biological nitrogen fixers.

5 0

3 years ago

Read 2 more answers

An alloy with an average grain diameter of 35 μm has a yield strength of 163 Mpa, and when it undergoes strain hardening, the gr

Lera25 [3.4K]

Answer:

$\sigma_y\ =210.2\ MPa$

Explanation:

Given that

d= 35 μm ,yield strength = 163 MPa

d= 17 μm ,yield strength = 192 MPa

As we know that relationship between diameter and yield strength

$\sigma_y=\sigma_o+\dfrac{K}{\sqrt d}$

$\sigma_y\ =Yield\ strength$

d = diameter

K =Constant

$\sigma_o\ =material\ constant$

So now by putting the values

d= 35 μm ,yield strength = 163 MPa

$163=\sigma_o+\dfrac{K}{\sqrt 35}$ ------------1

d= 17 μm ,yield strength = 192 MPa

$192=\sigma_o+\dfrac{K}{\sqrt 17}$ ------------2

From equation 1 and 2

$192-163=\dfrac{K}{\sqrt 17}-\dfrac{K}{\sqrt 35}$

K=394.53

By putting the values of K in equation 1

$163=\sigma_o+\dfrac{394.53}{\sqrt 35}$

$\sigma_o\ =96.31\ MPa$

$\sigma_y=96.31+\dfrac{394.53}{\sqrt d}$

Now when d= 12 μm

$\sigma_y=96.31+\dfrac{394.53}{\sqrt 12}$

$\sigma_y\ =210.2\ MPa$

4 0

3 years ago

Both HNO3(aq) and CH3COOH(aq) can be classified as *

lord [1]

Answer:

acids

Explanation:

HNO3 is a strong acid (Nitric Acid)

CH3COOH is a weak acid (Acetic Acid

6 0

3 years ago

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Consider the reaction between iron (III) oxide, Fe2O3 and carbon monoxide, CO.

Lana71 [14]

Fe2O3 + 3CO --> 2Fe + 3CO2

m(Fe2O3)=213 g
m(CO)=140 g
_______________

n(Fe2O3)=?
m(Fe)=?
n(Fe2O3)=?
n(CO)=?
n(CO2)=?

n(Fe2O3)=m(Fe2O3) / M(Fe2O3)
n(Fe2O3)= 213 g / 159,7 gmol-1 = 1,33 mol

n(CO)= m(CO) / M(CO) n(CO)= 140 g / 28,01 gmol-1 = 4,99 mol

3 0

3 years ago

The pH of a .0727 M aqueous sodium cyanide, KCN, solution at 25 degrees C

dybincka [34]

Kb = [HA} [OH-] / [A-] where [A-] represents the concentration of CN- (.068M)

Kb = Kw / Ka = 1 x10-14 / 4.9 x 10-10 = 2 x 10-5

Since this is a salt solution which could be considered to have formed from the neutralization of a strong base (NaOH) and a weak acid (HCN), the Na+ will have no effect on the pH of the solution while the CN- ion will undergo hydrolysis:

CN- + H2O --> HCN + OH-

Based on this equation, the quantities of HCN and OH- produced must be the same and therefore [HCN]=[OH-]. We will set this equal to x.

Plugging into the original equation yields:

2 x 10-5 = x2 / .068 M

Solving for x yields 1.2 x 10-3 whidh is equal to the [OH-]

The pOH then is equal to -log (1.2x10-3) = 2.9

The pH of the solution would be 14 - 2.9 = 11.1

8 0

3 years ago