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3 years ago

4 Which pair of ions is isoelectronic?

Chemistry

1 answer:

Softa [21]3 years ago

7 0

Answer:

B. Na+ and O2-

Explanation:

Na+ plus has 10 electrons and O2- also has 10 electrons

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A heterogeneous material may be

IgorLugansk [536]

Answer:

4) a mixture.....................

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3 years ago

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The new boiling point is 101.02°C. Suppose that you add another 2.0 mol of sucrose to this solution. What do you predict the new

solniwko [45]

Mad Lad, thanks for the answer :)

7 0

3 years ago

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GarryVolchara [31]

Answer:

The value of the Golden Igloo is $227.4 million.

Explanation:

First, we need to find the inner and the outer volume of the half-spherical shell:

$V_{i} = \frac{1}{2}*\frac{4}{3}\pi r_{i}^{3}$

$V_{o} = \frac{1}{2}*\frac{4}{3}\pi r_{o}^{3}$

The total volume is given by:

$V_{T} = V_{o} - V_{i}$

Where:

$V_{i}$ : is the inner volume

$r_{i}$ : is the inner radius = 1.25/2 = 0.625 m

$V_{o}$ : is the outer volume

$r_{o}$ : is the outer radius = 1.45/2 = 0.725 m

Then, the total volume of the Igloo is:

$V_{T} = \frac{2}{3}\pi r_{o}^{3} - \frac{2}{3}\pi r_{i}^{3} = \frac{2}{3}\pi [(0.725 m)^{3} - (0.625 m)^{3}] = 0.29 m^{3}$

Now, by using the density we can find the mass of the Igloo:

$m = 19.3 \frac{g}{cm^{3}}*0.29 m^{3}*\frac{(100 cm)^{3}}{1 m^{3}} = 5.60 \cdot 10^{6} g$

Finally, the value (V) of the antiquity is:

$V = \frac{\$ 1263}{oz}*5.60 \cdot 10^{6} g*\frac{1 oz}{31.1034768 g} = \$ 227.4 \cdot 10^{6}$

Therefore, the value of the Golden Igloo is $227.4 million.

I hope it helps you!

8 0

2 years ago

What is ionic bond,,,?

andreev551 [17]

Answer:

An ionic bond is the bonding between a non-metal and a metal, that occurs when charged atoms (ions) attract.

Explanation:

Here I put the function of Iconic Bond.

Ionic bonds form so that the outermost energy level of atoms are filled. Ion. an atom or group of atoms that bring out a positive or negative electric charge as a result of having lost or gained one or more electrons.

Therefore, I hope this helps!

7 0

2 years ago

Read 2 more answers

A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

sammy [17]

Answer: The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

Explanation:

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(S_8)$ = 0.800 g

$M_{solute}$ = Molar mass of solute $(S-8)$ = 256.52 g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m$

Calculation for freezing point of solution:

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

$\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

or,

$\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm$

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC$

Hence, the freezing point of solution is 16.5°C

Calculation for boiling point of solution:

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

or,

$\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm$

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC$

Hence, the boiling point of solution is 118.2°C

5 0

3 years ago