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RUDIKE [14]
2 years ago
12

Which element would have the largest radius; Br or Se? Explain

Chemistry
1 answer:
dusya [7]2 years ago
5 0
The atomic radium increases to the left and downwards on the periodic table. 
By that logic, Se has the largest radius.
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All physical changes can be identified at the macroscopic level. true or false
Aliun [14]

Answer:

True

Explanation:

The physical changes are reversible in most cases and these changes are not the chemical changes which means that it is only the change in its state not in their nature. Just take the example of water, on cooling it becomes solid and change in color can be seen which is white in solid form and colorless in liquid form. This is also reversible and is a physical change. This means that physical changes can be identified at macroscopic level. Hence the answer is true.

5 0
3 years ago
What is the concentration of the base (NaOH) in this titration?
dalvyx [7]

The answer is B for fact

8 0
3 years ago
Read 2 more answers
f the Ksp for HgBr2 is 2.8×10−14, and the mercury ion concentration in solution is 0.085 M, what does the bromide concentration
goldfiish [28.3K]

Answer:

0.057 M

Explanation:

Step 1: Given data

Solubility product constant (Ksp) for HgBr₂: 2.8 × 10⁻⁴

Concentration of mercury (II) ion: 0.085 M

Step 2: Write the reaction for the solution of HgBr₂

HgBr₂(s) ⇄ Hg²⁺(aq) + 2 Br⁻

Step 3: Calculate the bromide concentration needed for a precipitate to occur

The Ksp is:

Ksp = 2.8 × 10⁻⁴ = [Hg²⁺] × [Br⁻]²

[Br⁻] = √(2.8 × 10⁻⁴/0.085) = 0.057 M

7 0
3 years ago
2Ag + H2S ? Ag2S + H2 Which statement is true about this chemical equation? A) Ag (silver) is oxidized and H (hydrogen) is reduc
nadezda [96]

Answer:

It is A).

Explanation:

Silver (Ag) goes from the  pure metal to Ag+ losing 1 electron so it is oxidised.

The hydrogen ion  gains electrons and is reduced.

6 0
2 years ago
Read 2 more answers
5. What concentration of acid must be added to change the pH of 1 mM phosphate buffer from 7.4 to 7.3 (pKas of the phosphate buf
mr_godi [17]

Explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and pK_{a} is as follows.

               pH = pK_{a} + log \frac{base}{acid}

where,     pH = 7.4 and pK_{a} = 7.21

As here, we can use the pK_{a} nearest to the desired pH.

So,      7.4 = 7.21 + log \frac{base}{acid}

             0.19 = log \frac{base}{acid}

            \frac{base}{acid} = 1.55

1 mM phosphate buffer means [HPO_{4}] + [H_{2}PO_{4}] = 1 mM

Therefore, the two equations will be as follows.

           \frac{HPO_{4}}{H_{2}PO_{4}} = 1.55 ............. (1)

  [HPO_{4}] + [H_{2}PO_{4}] = 1 mM ........... (2)        

Now, putting the value of [HPO_{4}] from equation (1) into equation (2) as follows.

             1.55[H_{2}PO_{4}] + [tex][H_{2}PO_{4}] = 1 mM

                        2.55 [H_{2}PO_{4}] = 1 mM

                             [H_{2}PO_{4}] = 0.392 mM

Putting the value of [H_{2}PO_{4}] in equation (1) we get the following.

                     0.392 mM + [HPO_{4}] = 1 mM

                          [HPO_{4}] = (1 - 0.392) mM

                              [HPO_{4}] = 0.608 mM

Thus, we can conclude that concentration of the acid must be 0.608 mM.

7 0
3 years ago
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