Which of the following statements describes the elements in family 16 of the periodic table

tester [92]

Because there is a permanent dipole between the Br and Cl atom. The Cl atom has a higher electronegativity, so it will tend to pull the "shared" electrons formed in the bond to its pole, creating a permanent dipole and partial electron deficiency on the bromine atom.

5 0

3 years ago

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A buffer is a solution that is a mixture of either a weak acid and its conjugate base or a weak base and its conjugate acid. Whe

babymother [125]

Answer:

b. The weak base of an alkaline buffer will accept hydrogen protons when a strong acid is added to the solution

d.The conjugate acid of an alkaline buffer will donate hydrogen protons when a strong base is added to the solution.

Explanation:

A buffer is a solution that resist pH change, it shows minimal change upon addition of small amount of strong acid or strong base. An alkaline buffer will have a basic pH, above 7. It is made by mixing a weak base and its salt with a strong acid. An example of an alkaline buffer is carbonate-bicarbonate buffer that is prepared using varying amount of anhydrous sodium carbonate and volume of solution of sodium bicarbonate to get pH range between 9.2 to 10.7

Within the buffer,the salt is completely ionized while the weak base is partly ionized. on addition of acid, the released protons will be removed by the bicarbonate ion to form sodium carbonate; on addition of base, the hydroxide ion released by the base will be removed by the hydrogen ions to form water and the pH remains relatively the same

6 0

2 years ago

5. A 5.00 g sample of an unknown substance was heated from 25.2 C to 55.1 degrees * C , and it required 133 to do so. Identify t

Leviafan [203]

Answer:

Aluminum

Explanation:

Given

$T_1 = 25.2^oC$

$T_2 = 55.1^oC$

$m = 5.00g$

$\triangle Q= 133J$

See attachment for chart

Required

Identify the unknown substance

To do this, we simply calculate the specific heat capacity from the given parameters using:

$c = \frac{\triangle Q}{m\triangle T}$

This gives:

$c = \frac{\triangle Q}{m(T_2 - T_1)}$

So, we have:

$c = \frac{133J}{5.00g * (55.1C - 25.2C)}$

$c = \frac{133J}{5.00g * 29.9C}$

$c = \frac{133J}{149.5gC}$

$c = 0.89\ J/gC$

From the attached chart, we have:

$Al(s) = 0.89\ J/gC$ --- The specific heat capacity of Aluminum

Hence, the unknown substance is Aluminum

7 0

3 years ago

Which of the following terms best describes any solution?

baherus [9]

I believe D hopefully this helps

8 0

3 years ago

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2.A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a s

Aleks04 [339]

Answer:

In order to prepare 10 mL, 5 μM; 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.

In order to prepare 10 mL, 10 μM; 4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

In order to prepare 10 mL, 15 μM; 6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

In order to prepare 10 mL, 20 μM; 8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

Explanation:

Using the dilution equation:

no of moles before dilution = no of moles after dilution.

Molarity x volume (initial)= Molarity x volume (final).

In order to prepare 10 mL, 5 μM from 25 μM solution,

Final molarity = 5 μM, final volume = 10 mL, initial molarity = 25 μM, initial volume = ?

25 x initial volume = 5 x 10

Initial volume = 50/25

= 2 mL

2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

In order to prepare 10 mL, 10 μM from 25 μM stock,

25 x initial volume = 10 x 10

Initial volume = 100/25 = 4 mL

4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

In order to prepare 10 mL, 15 μM from 25 μM stock,

25 x initial volume = 15 x 10

initial volume = 150/25 = 6 mL

6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

In order to prepare 10 mL, 20 μM from 25 μM stock,

25 x initial volume = 20 x 10

initial volume = 200/25 = 8 mL

8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

6 0

2 years ago