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3 years ago

Assign oxidation states to each atom in each of the following species.?

Chemistry

1 answer:

DochEvi [55]3 years ago

3 0

1. CuCl2
Cl: 2(-1)= 2-
Cu: 2+
~ON of Cl: 1-
~ON of Cu: 2-

2.CH4
H: 4(1)= 4+
C: 4-

~ON of H: 1+
~ON of C: 4-

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Based on the position of the elements in the periodic table, what is the charge of the ions they are most likely to form?

Rainbow [258]

Explanation:

The charge on an ion denotes the amount of electrons lost or gained or even shared by an atom.

An atom will lose, gain or share an equal amount of electrons that will make it stable and achieve an octet and perfect configuration.

This is often synonymous with the component number of electrons in their outermost shell. The valence shell.

For metals on the periodic table, they are always willing to give up electrons due to their large electropositivity.

Metals will give up the number of electrons in their outermost shell to become stable.

Groups 1 and Group 2 will have charges +1 and +2 on them.

For non-metals , they will gain the exact number of electrons that will make them stable. We must note that half -filled and fully filled orbitals are equally stable.

Elements in groups 6 and 7 are electronegative and will have a charge of -2 and -1 respectively.

For those in groups 3, 4 and 5 they either gain, lose or share commensurate amount of electrons that will make them stable.

Group 8 elements are stable and have no charges.

Generally on the periodic table, metals are to the left and are always positively charged. Non - metals are to the right and are negatively charged.

learn more:

Periodic table brainly.com/question/2690837

#learnwithBrainly

4 0

3 years ago

Calculate hydrochloric acid (umol)in 200 ul of a 0.5173Msolution of acid?

strojnjashka [21]

Answer: The moles of hydrochloric acid is $1.0346\times 10^{-4}\mu mol$

Explanation:

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

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Or,

$\text{Molarity of the solution}=\frac{\text{Micro moles of solute}\times 10^6}{\text{Volume of solution (in }\mu L)}}$

We are given:

Molarity of solution = 0.5173 M

Volume of solution = $200\mu L$

Putting values in above equation, we get:

$0.5173M=\frac{\text{Micro moles of HCl}\times 10^6}{200\mu L}\\\\\text{Micro moles of HCl}=1.0346\times 10^{-4}\mu mol$

Hence, the moles of hydrochloric acid is $1.0346\times 10^{-4}\mu mol$

7 0

3 years ago

A 0.4272 g sample of an element contains 2.241 x 10 ^21 atoms . what is the symbol of the element?

grin007 [14]

Answer:

Likely $\rm In$ (indium.)

Explanation:

Number of atoms: $N = 2.241 \times 10^{21}$ .

Dividing, $N$ , the number of atoms by the Avogadro constant, $N_A \approx 6.023 \times 10^{23} \; \rm mol^{-1}$ , would give the number of moles of atoms in this sample:

$\begin{aligned} n &= \frac{N}{N_{A}} \\ &\approx \frac{2.241 \times 10^{21}}{6.023 \times 10^{23}\; \rm mol^{-1}} \approx 3.72 \times 10^{-3}\; \rm mol \end{aligned}$ .

The mass of that many atom is $m = 0.4272\; \rm g$ . Estimate the average mass of one mole of atoms in this sample:

$\begin{aligned}M &= \frac{m}{n} \\ &\approx \frac{0.4272\; \rm g}{3.72 \times 10^{-3}\; \rm mol} \approx 114.82\; \rm g \cdot mol^{-1}\end{aligned}$ .

The average mass of one mole of atoms of an element ( $114.82\; \rm g \cdot mol^{-1}$ in this example) is numerically equal to the average atomic mass of that element. Refer to a modern periodic table and look for the element with average atomic mass $114.82$ . Indium, $\rm In$ , is the closest match.

5 0

3 years ago

How many electrons can each individual orbital hold?

Hatshy [7]

Answer:

Orbitals of equal energy are each occupied by one electron before any orbital is occupied by a second electron, and all electrons in singly occupied orbitals must have the same spin state. Noble-gas configuration refers to an outer main energy level occupied, in most cases, by eight electrons.

Explanation:

I hope this helps you.

3 0

3 years ago

Read 2 more answers

A 51.0 mL aliquot of a 1.50 M solution is diluted to a total volume of 208 mL. A 104 mL portion of that solution is diluted by a

S_A_V [24]

Answer:

The final concentration is 0.226 M.

Explanation:

In this problem the dilution has occurred two times. So the symbol initial concentration, concentration after first dilution and the concentration after second dilution is given as C1, C2, and C3 respectively. Same can be done for the volume i.e, V1, V2, and V3 will be the initial volume, the volume taken after first dilution, and the volume after second dilution.

So, let's use the dilution formula for the first dilution

$C_1 \times V_1 = C_2 \times V_2$

$C_2 = \frac{C_1\times V_1}{V_2}\\\\C_2 = \frac{1.50\times 51.0}{208}\\\\C_2 = 0.368 M$

Now, for the second dilution

$C_2 \times V_2 = C_3 \times V_3\\\\C_3 = \frac {C_2 \times V_2}{V_3}\\\\C_3 = \frac {0.368 \times 104}{169}\\\\C_3 = 0.226 M$

6 0

4 years ago