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IceJOKER [234]
3 years ago
5

An atom has three full orbitals in its second energy level.

Chemistry
2 answers:
ICE Princess25 [194]3 years ago
7 0
Orbitals am only hold two electrons each, so 3 orbitals can hold 6 electrons
DIA [1.3K]3 years ago
7 0

Answer:

6 hope this helps

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lassify these bonds as ionic, polar covalent, or nonpolar covalent. Ionic Polar covalent Nonpolar covalent
jarptica [38.1K]

The question is incomplete, complete question is:

Classify these bonds as ionic, polar covalent, or nonpolar covalent.

Na-F

P-Cl

Cl-Cl

Answer:

Na-F  : The bond present is ionic bond.

PCl  : The bond present is polar covalent bond.

ClCl  : The bond present is non polar covalent bond.

Explanation:

1) Ionic bonds are defined as the bonds which are formed by the complete transfer of electrons from cation (positively charged ions) to anion (negatively charged ions). For Example: NaCl, BaSO_4 etc.

Ionic compound :This compound is formed when difference in electronegativity between the atoms is very large.

2) Covalent bond is defined as the bond which is formed by the sharing of electrons between the atoms. For Example: HCl, CH_4 etc.

Polar covalent compound: This compound is formed when difference in electronegativity between the atoms is present. When atoms of different elements combine, it results in the formation of polar covalent bond. For Example: CO_2,NO_2 etc..

Non-polar covalent compound: This compound is formed when there is no difference in electronegativity between the atoms. When atoms of the same element combine, it results in the formation of non-polar covalent bond. For Example: N_2,O_2 etc.

Na-F  : Two different atoms with high electronegativity difference

Ionic compound because sodium is metal and fluorine is non metal and the difference between their electronegativity is very large.The bond present is ionic bond.

PCl  : Two different atoms with low electronegativity difference

Polar covalent  compound because phosphorus and fluorine both are non metal and the difference between their electronegativity is not very large.The bond present is polar covalent bond.

ClCl  : Two same atoms with nil electronegativity difference.

Non Polar covalent compound because both are chlorine atom (non metal) and the difference between their electronegativity is zero.The bond present is non polar covalent bond.

5 0
3 years ago
. A large quantity of very dilute aqueous HCl solution is neutralized by the addition of the stoichiometric amount of a 10-mol-%
ad-work [718]

Answer:

Make the question more clear for me

Explanation:

5 0
3 years ago
Arbon dioxide is dissolved in blood (ph 7.5) to form a mixture of carbonic acid and bicarbonate. Part a neglecting free co2, wha
Aleks04 [339]

Answer : The fraction of carbonic acid present in the blood is 5.95%

Explanation :

The mixture consists of carbonic acid ( H₂CO₃) and bicarbonate ion ( HCO₃⁻). This represents a mixture of weak acid and its conjugate which is a buffer.

The pH of a buffer is calculated using Henderson equation which is given below.

pH = pKa + log \frac{[Base]}{[Acid]}

We have been given,

pH = 7.5

pKa of carbonic acid = 6.3

Let us plug in the values in Henderson equation to find the ratio Base/Acid.

7.5 = 6.3 + log \frac{[base]}{[acid]}

1.2 = log \frac{[base]}{[acid]}

\frac{[Base]}{[Acid]} = 10^{1.2}

\frac{[Base]}{[Acid]} = 15.8

[Base] = 15.8 \times [Acid]

The total of mole fraction of acid and base is 1. Therefore we have,

[Acid] + [Base] = 1

But Base = 15.8 x [Acid]. Let us plug in this value in above equation.

[Acid] + 15.8 \times [Acid] = 1

16.8 [Acid] = 1

[Acid] = \frac{1}{16.8}

[Acid] = 0.0595

[Acid] = 0.0595 x 100 = 5.95 %

The fraction of carbonic acid present in the blood is 5.95%

4 0
3 years ago
Read 2 more answers
What is the SI base unit for length?
sp2606 [1]

The SI base unit for length is meter.

In order to make smaller measurements, you can use the centi-, milli-, micro-, etc. prefixes.

When you want to reference larger measurements, you can use the kilo-, mega-, giga- and prefixes such as those.

5 0
3 years ago
If 50 ml of 0.235 M NaCl solution is diluted to 200.0 ml what is the concentration of the diluted solution
Helen [10]

This is a straightforward dilution calculation that can be done using the equation

M_1V_1=M_2V_2

where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.

Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:

M_2=\frac{M_1V_1}{V_2}.

Substituting in our values, we get

\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].

So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.

5 0
2 years ago
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