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olya-2409 [2.1K]
3 years ago
6

The melting point of the metal potassium is 63.5c, while that of titanium is 1660c. what explanation can be given for this great

difference in melting temperature?
Chemistry
1 answer:
kiruha [24]3 years ago
3 0

The melting point of potassium = 63.5^{0}C

Melting point of titanium = 1660^{0}C

Titanium has a stronger metallic bonding compared to potassium. Titanium being a transition metal has greater number of valence electrons (4 valence electrons) contributing to the valence electron sea compared to potassium which has only one valence electron. The atomic size of Titanium much lower than that of potassium, so the bonding between Titanium atoms is stronger than that of potassium. Hence, the melting point of Titanium is much higher than that of potassium.

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Starting with 250 mL of a 0.250 M solution of HBr; a) Calculate the initial pH of the solution.b) Calculate the pH after adding
maxonik [38]

Answer:

a.

pH =  0.602

b.

pH = 1.5

c.

pH = 7

d.

pH = 12.1

Explanation:

a ) To calculate the pH, use the following equation:

pH = -log [H+]

Hbr is a strong acid, so the [H+] concentration can be calculated as follow:

[HBr] = 0.250 M

As acid Hbr:

Hbr = H+ + Br-

As strong acid HBr dissociates at all, so

[HBr] = [H+] = 0.250 M

So the pH:

<u>pH = -log [0.250 M] = 0.602</u>

<u></u>

<u>b) Calculate the pH after adding 250 mL of 0.125M NaOH</u>

<u></u>

<u>I</u>n this point, the reactions starts:

<u></u>

HBr + NaOH = H2O + NaBr

- First, we gonna find the mol of each reactant:

HBr:

mol = [M] × L

mol = 0.250 M × 0.250 L

mol = 0.0625 mol HBr

NaOH:

mol = [M] × L

mol = 0.125 M × 0.250L

mol = 0.03125 mol NaOH

In base on the reaction, it’s needed 1 mol of NaOH to neutralize 1 mole of HBr, so to neutralize 0.0625 moles of HBr its needed 0.0625 mol of NaOH:

0.0625 mol HBr – 0.03125 mol HBr = 0.03125 mol HBr

These are the moles free in the solution, and we going to use them to calculate the pH:

pH = -log [ H]

pH = - log [ 0.03125 ] = 1.5

<u>c) Calculate the pH after adding 500 mL of 0.125M NaOH</u>

NaOH:

mol = [M] × L

mol = 0.125 M × 0.500L

mol = 0.0625 mol NaOH

In base on the reaction, it’s needed 1 mol of NaOH to neutralize 1 mole of HBr, so to neutralize 0.0625 moles of HBr its needed 0.0625 mol of NaOH:

0.0625 mol HBr – 0.0625 mol HBr = 0 HBr

pH = 7

<u>d) Calculate the pH after adding 600 mL of 0.125M NaOH.</u>

<u>NaOH:</u>

mol = [M] × L

mol = 0.125 M × 0.600L

mol = 0.075 mol NaOH

In base on the reaction, it’s needed 1 mol of NaOH to neutralize 1 mole of HBr, so to neutralize 0.0625 moles of HBr its needed 0.0625 mol of NaOH:

0.075 mol NaOH – 0.0625 mol NaOH = 00125 NaOH

These are the moles free in the solution, and we going to use them to calculate the pH:

pOH = -log [ OH]

pOH = - log [ 0.0125 ] = 1.90

pH + pOH = 14

pH = 14- pOH = 14 – 1.90 = 12.1

4 0
2 years ago
Which of these is a compound?<br> A)air<br> B)sugar C12H22O11<br> C)the alloy steel<br> D)nitrogen N
7nadin3 [17]
Nitrogen N I think would be the answer
6 0
3 years ago
An atom has 13 protons and 14 neutrons. What is its mass number?
Hitman42 [59]
Protons have a mass of 1
Neutrons have a mass of 1

So 13*1 + 14*1 = Mass number 27
3 0
2 years ago
Read 2 more answers
Organize the following species in terms
ser-zykov [4K]

Explanation:

  1. F Cl Br
  • ionization energy decreases we move down the group
  • so Br<Cl<F

2. Na K Li

  • ionization energy decreases we move down the group
  • so K<Na<Li

3. C N O F

  • ionization energy increases as we move across the peroid
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plz mark as brainliest if it helps

6 0
3 years ago
A particular reaction, A- products, has a rate that slows down as the reaction proceeds. The half-life of the reaction is found
Thepotemich [5.8K]

Explanation:

Half life of zero order and second order depends on the initial concentration. But as the given reaction slows down as the reaction proceeds, therefore, it must be second order reaction. This is because rate of reaction does not depend upon the initial concentration of the reactant.

a. As it is a second order reaction, therefore, doubling reactant concentration, will increase the rate of reaction 4 times. Therefore, the statement  a is wrong.

b. Expression for second order reaction is as follows:

\frac{1}{[A]} =\frac{1}{[A]_0} +kt

the above equation can be written in the form of Y = mx + C

so, the plot between 1/[A] and t is linear. So the statement b is true.

c.

Expression for half life is as follows:

t_{1/2}=\frac{1}{k[A]_0}

As half-life is inversely proportional to initial concentration, therefore, increase in concentration will decrease the half life. Therefore statement c is wrong.

d.

Plot between A and t is exponential, therefore there is no constant slope. Therefore, the statement d is wrong

8 0
3 years ago
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