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8090 [49]
3 years ago
6

The normal boiling point corresponds to the temperature at which both the liquid and gas are under the standard atmospheric pres

sure of 1 atm.
TRUE

FALSE
Chemistry
1 answer:
ANTONII [103]3 years ago
3 0
TRUE. (Lorenzo Romano Amadeo Carlo Avogadro) Ideal Gas Law that defined as one in which all collisions between atoms or molecules are perfectly elastic in which there are no intermolecular attractive forces.  In such a gas, all the internal energy is the form of kinetic energy and any change in internal energy is accompanied by a change in temperature. That characterized by three state variables: absolute pressure (P) = 1 atm, volume (V) = 22.4 L and absolute temperature (T) = 273 K.
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15 g of anhydrous calcium chloride is dissolved in 185 mL of water. What is molarity of the prepared solution? 81.1 M O0.73 M 0.
geniusboy [140]

Answer:

When you prepared the solution, you will find that the molarity is 0.73M

Explanation:

First of all you should get by the periodic table, molar mass in the anhydrouds calcium.

CaCl2 · 0H20 = 110.98 g/m

Now we have to find out how many mols are 15 g.

So 15g / 110.98 g/m = 0.135 moles

This moles are in 185 ml of water. Molarity as you should know are moles of solute in 1 L of solution (either 1000 ml)

185 ml ______ 0.135 moles

1000 ml _____ x    x = (1000*0.135) /185 = 0.730M

3 0
3 years ago
Ammonia (nh3) is widely used as a fertilizer and in many household cleaners. how much ammonia is produced when 6.44 mol of hydro
attashe74 [19]

The answer would be the <span>Chemical formula for ammonia by hydrogen and nitrogen gas: 1 N2 + 3 H2 --> 2 NH3 t<span>hen use stoichiometry. You know you have enough nitrogen gas to react, so you can just straight convert mol H2 to mol NH3, where 3 mol H2 = 2 mol NH3. 
<span>3.44 mol H2 * (2 mol NH3 / 3 mol H2) = 2.29 mol NH3</span></span></span>

8 0
3 years ago
PLEASE HELP!!!!! The table below shows the volume of two samples, X and Y, when placed in three containers of different volumes.
Dahasolnce [82]

Answer:

B m8

Explanation:

7 0
3 years ago
At 298 K, the osmotic pressure of a glucose solution (C6H12O6 (aq)) is 12.1 atm. Calculate the freezing point of the solution. T
Anarel [89]

<u>Answer:</u> The freezing point of solution is -0.974°C

<u>Explanation:</u>

  • To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

where,

\pi = osmotic pressure of the solution = 12.1 atm

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of the solution = 298 K

Putting values in above equation, we get:

12.1atm=1\times M\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\M=\frac{12.1}{1\times 0.0821\times 298}=0.495M

This means that 0.495 moles of glucose is present in 1 L or 1000 mL of solution

  • To calculate the mass of solution, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of solution = 1.034 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

1.034g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.034g/mL\times 1000mL)=1034g

  • To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of glucose = 0.495 moles

Molar mass of glucose = 180.16 g/mol

Putting values in above equation, we get:

0.495mol=\frac{\text{Mass of glucose}}{180.16g/mol}\\\\\text{Mass of glucose}=(0.495mol\times 180.16g/mol)=89.18g

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 = 1 (For non-electrolytes)

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

m_{solute} = Given mass of solute (glucose) = 89.18 g

M_{solute} = Molar mass of solute (glucose) = 180.16  g/mol

W_{solvent} = Mass of solvent (water) = [1034 - 89.18] g = 944.82 g

Putting values in above equation, we get:

0-\text{Freezing point of solution}=1\times 1.86^oC/m\times \frac{89.18\times 1000}{180.16g/mol\times 944.82}\\\\\text{Freezing point of solution}=-0.974^oC

Hence, the freezing point of solution is -0.974°C

8 0
3 years ago
Which polymer structure should be used if a strong and heat - resistant adhesive is required?
alexira [117]

Cross-linked polymers are strong and heat - resistant hence they can be used for this purpose.

A polymer is obtained by the combination of small molecules called monomers. A polymer consists of a regular repeating unit of small molecules called monomers.

A cross-linked polymer is a polymer in which covalent bonds are used to join polymer chains together. Cross-linked polymers are strong and heat - resistant hence they can be used for this purpose.

Learn more: brainly.com/question/17638582

3 0
2 years ago
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