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kvv77 [185]
3 years ago
13

Select the statement(s) that describe the characteristics of transition metals. a. Transition metals have partially filled d sub

shells. b. Transition metals give rise to cations that have completely filled d subshells. c. Group 2B elements are not transition metals because they neither have nor readily acquire partially filled d orbitals.
Chemistry
1 answer:
yan [13]3 years ago
4 0

Answer: The correct option is A,

--> a.) Transition metals have partially filled d subshells.

Explanation:

Transition elements are all metals of economic importance. They are found in the d- lock of the periodic table between group 2 and 3. They occupy three rows, with ten elements in each row. The term 'transition metals' refers only to an element which has PARTIALLY filled d orbitals. Typical example of transition metals include iron (Fe).

They have partially filled 3d orbitals which are responsible for the special properties of the metals. These include:

--> Physical properties: the transition metals have high boiling and melting points. They are hard, dense and lustrous. They are also good conductors of heat and electricity.

--> Chemical reactivity: In the s- block and p-block, the chemical properties of the elements in the same period vary, often quite markedly, from left to right. This does not happen with the transition metals because electrons are added progressively to the inner d-orbitals.

--> Variable oxidation states: they have variable oxidation states because 3d electrons are available for bond formation.

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Excess silver(I) nitrate was added to a 8.500 g mixture containing some amount of barium chloride, and 7.123 g of silver chlorid
dsp73

Answer:

60.88%

Explanation:

The balanced equation of the reaction is given as;

2 AgNO3 (aq) + BaCl2 (aq) → 2 AgCl (s) + Ba(NO3)2 (aq)

Since AgNO3 is in excess, the limiting reactant is BaCl2. From the reaction;

1 mol of BaCl2 produces 2 mol of AgCl

Converting to masses;

Mass = Number of mol * Molar mass

BaCl2;

Mass = 1 * 208.23 g/mol = 208.23 g

AgCl;

Mass = 2 * 143.32 g/mol = 286.64 g

208.23 g BaCl2 produces 286.64 g of AgCl

x g BaCl2 produces 7.123 g of AgCl

Solving for x;

x = 7.123 * 208.23 / 286.64 = 5.1745 g

Mass percent = Mass / Total mass of Mixture  * 100

Mass Percent = 5.1745 / 8.500 = 0.6088 * 100 = 60.88%

8 0
3 years ago
How many moles of copper are needed to react with sulfur to produce 2.9 moles of copper (I) sulfide
dalvyx [7]

Answer:

5.8 moles of copper are needed to react with sulfur to produce 2.9 moles of copper (I) sulfide.

Explanation:

2Cu+S\rightarrow Cu_2S

Moles of copper(I) sufide = 2.9 mol

According to reaction, 1 mole of copper(I) sulfide is obtained from 2 moles of copper, then 2.9 moles of copper(I) sulfide will be obtained from :

\frac{2}{1}\times 2.9 mol=5.8 mol copper

5.8 moles of copper are needed to react with sulfur to produce 2.9 moles of copper (I) sulfide.

4 0
3 years ago
An acetic acid buffer solution is required to have a pH of 5.27. You have a solution that contains 0.010 mol of Acetic acid. Wha
dlinn [17]

Answer:

Molarity of sodium acetate you will need to add is 0.0324M

Explanation:

<em>Assuming volume of the buffer is 1L.</em>

<em />

The pH of a buffer can be determined using Henderson-Hasselbalch equation:

pH = pKa + log [A⁻] / [HA]

<em>Where pKa is pKa of the weak acid,  [A⁻] molar concentration of conjugate base and [HA] molar concentration of weak acid</em>

<em />

Replacing for the acetic buffer (pKa = 4.76):

pH = 4.76 + log [Sodium Acetate] / [Acetic Acid]

As you have 0.010 moles of acetic acid in 1L:

[Acetic Acid] = 0.010mol / 1L = 0.010M

And you require a pH of 5.27:

5.27 = 4.76 + log [Sodium Acetate] / [0.010M]

0.51 = log [Sodium Acetate] / [0.010M]

10^0.51 = [Sodium Acetate] / [0.010M]

3.236 =  [Sodium Acetate] / [0.010M]

3.236 [0.010M] = [Sodium Acetate]

0.0324M = [Sodium Acetate]

<h3>Molarity of sodium acetate you will need to add is 0.0324M</h3>

<em />

7 0
3 years ago
If 0.97 moles of BaCl2 are dissolved in enough water to make a 1.2 liter soulution, what is the resulting molarity
Akimi4 [234]
         If    1.2 L of solution contains 0.97 mol
then let     1 L of solution contain  x   mol
     
       ⇒ (1.2 L) x  = (0.97 mol) (1 L)

                      x  = (0.97 mol · L)  ÷  (1.2 L)

                      x  = 0.8083 mol

Thus the molarity of the Barium Chloride solution is 0.808 mol / L   OR  0.808 mol/dm³. 

5 0
3 years ago
Volume of 100ml. it’s density is 0.15g/ml. what is its mass?
11Alexandr11 [23.1K]
V=100mL\\&#10;d=0,15\frac{g}{mL}\\\\&#10;m=dV=0,15\frac{g}{mL}*100mL=15g
3 0
3 years ago
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