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Sveta_85 [38]
3 years ago
10

Two aqueous solutions are prepared: 1.00 m Na2CO3 and 1.00 m LiCl. Which of the following statements is true?

Chemistry
1 answer:
svlad2 [7]3 years ago
7 0

Answer:

The Na_{2}CO_{3} solution has a higher osmotic pressure and higher boiling point than LiCl solution.

Explanation:

As concentrations of two aqueous solutions are same therefore we can write:

                                \Delta P\propto i , \Delta T_{b}\propto i and \pi \propto i

where \Delta P, \Delta T_{b} and \pi are lowering of vapor pressure, elevation in boiling point and osmotic pressure of solution respectively. i is van't hoff factor.

i = total number of ions generated from dissolution of one molecule of a substance (for strong electrolyte).

Here both Na_{2}CO_{3} and LiCl are strong electrolytes.

So, i(Na_{2}CO_{3})=3 and i(LiCl)=2

Hence, lowering of vapor pressure, elevation in boiling point and osmotic pressure will be higher for Na_{2}CO_{3} solution.

Therefore the Na_{2}CO_{3} solution has a higher osmotic pressure and higher boiling point than LiCl solution.

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A standard room temperature and pressure, most of the elements on the periodic table are
Maslowich

Answer:

Solids

Explanation:

At standard room temperature and pressure, most of the elements are solids.

Just a few of the elements are liquids and gases at this temperature.

The periodic table is made up of metals, metalloids and non-metals. Most of these substances are actually solid.

Some non-metals are gaseous at standard room temperature.

8 0
2 years ago
What is the molarity of 5.39 grams of CaCl2 dissolved in 609 mL of water? (molar mass = 110.98 g/mol)
Ivenika [448]
Use google it works well
3 0
3 years ago
Thermal energy would naturally flow between objects at which temperatures?
marissa [1.9K]

Answer:

i think the answer is letter C. From 35°c to 45°c

Explanation:

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4 0
3 years ago
A chemist titrates 160.0mL of a 0.3403M aniline C6H5NH2 solution with 0.0501M HNO3 solution at 25°C . Calculate the pH at equiva
maxonik [38]

Answer : The pH of the solution is, 5.24

Explanation :

First we have to calculate the volume of HNO_3

Formula used :

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the initial molarity and volume of C_6H_5NH_2.

M_2\text{ and }V_2 are the final molarity and volume of HNO_3.

We are given:

M_1=0.3403M\\V_1=160.0mL\\M_2=0.0501M\\V_2=?

Putting values in above equation, we get:

0.3403M\times 160.0mL=0.0501M\times V_2\\\\V_2=1086.79mL

Now we have to calculate the total volume of solution.

Total volume of solution = Volume of C_6H_5NH_2 +  Volume of HNO_3

Total volume of solution = 160.0 mL + 1086.79 mL

Total volume of solution = 1246.79 mL

Now we have to calculate the Concentration of salt.

\text{Concentration of salt}=\frac{0.3403M}{1246.79mL}\times 160.0mL=0.0437M

Now we have to calculate the pH of the solution.

At equivalence point,

pOH=\frac{1}{2}[pK_w+pK_b+\log C]

pOH=\frac{1}{2}[14+4.87+\log (0.0437)]

pOH=8.76

pH+pOH=14\\\\pH=14-pOH\\\\pH=14-8.76\\\\pH=5.24

Thus, the pH of the solution is, 5.24

4 0
3 years ago
A chemist prepares a solution of mercury(I) chloride Hg2Cl2 by measuring out
Genrish500 [490]

Answer:

1.26 × 10^-8 M

Explanation:

We are given;

Number of moles of mercury (i) chloride as 0.000126 μmol

Volume is 100 mL

We are required to calculate the concentration of the solution.

We need to know that;

Concentration is also known as molarity is given by;

Molarity = Number of moles ÷ Volume

Number of moles = 1.26 × 10^-10 Moles

Volume = 0.01 L

Therefore;

Concentration = 1.26 × 10^-10 Moles ÷ 0.01 L

                       = 1.26 × 10^-8 M

Thus, the molarity of the solution is 1.26 × 10^-8 M

6 0
3 years ago
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