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2 years ago

How many bonds would be needed when bonding 2 atoms of nitrogen to achieve a full octet for both atoms?

Chemistry

1 answer:

andriy [413]2 years ago

8 0

Answer:

3 bonds are needed.

Explanation:

The electrons that are involved in chemical bonding are those in the outer shell of the highest energy level of the atom. The electron configuration of nitrogen (N) is 1s²2s²2p³. That means thy at each nitrogen atom has 5 valence electrons: 2 electrons in the 2s orbital and 3 electrons in the 2p orbital. To fullfil the octet, each nitrogen atom needs 3 electrons. So, they can share each other 3 electrons to form 3 simple bonds. Therefore, the nitrogen molecule (N₂) has 3 bonds involving 6 bonding electrons or a triple bond.

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20 mL of an approximately 10% aqueous solution of ethylamine, CH3CH2NH2, is titrated with 0.3000 M aqueous HCl. Which indicator

Eduardwww [97]

Answer:

Therefore, The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6

Bromocresol green, color change from pH = 4.0 to 5.6

Explanation:

The equation for the reaction is :

$C_2H_5NH_2_(_a_q_) + H^+_(_a_q_) --- C_2H_5NH_{3(aq)}^+$

concentration of $C_2H_5NH_{2(aq)$ = 10%

10 g of $C_2H_5NH_{2(aq)$ in 100 ml solution

molar mass = 45.08 g/mol

number of moles = 10 / 45.08

= 0.222 mol

Molarity of $C_2H_5NH_2(aq) = 0.222 \times \frac{1000}{100}mL$

= 2.22 M

number of moles of $C_2H_5NH_{2(aq)$ in 20 mL can be determined as:

$= 20 mL \times 2.22 M= 44*10^{-3} mole$

Concentration of $C_2H_5NH_2(aq) = \frac{44*10^{-3}*1000}{20}$

= 2.22 M

Similarly, The pKa Value of $C_2H_5NH_{2(aq)$ is given as 10.75

pKb value will be: 14 - pKa

= 14 - 10.75

= 3.25

the pH value at equivalence point is,

$pH= \frac{1}{2}pKa - \frac{1}{2}pKb-\frac{1}{2}log[C]$

$pH = \frac{14}{2}-\frac{3.25}{2}-\frac{1}{2}log [2.22]$

$pH = 5.21$

Therefore, The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6

8 0

2 years ago

How many mols are in 100.g of Fe?

PolarNik [594]

Answer:

1.79 moles

Explanation:

1.79 moles

3 0

2 years ago

Read 2 more answers

How many atoms of iodine are in 12.75g of CaI2? Hint. How many Iodines are there in one CaI2 particle?

xenn [34]

Answer:

$5.225x10^{22}atoms\ I$

Explanation:

Hello!

In this case, since 12.75 g of calcium iodide has the following number of moles (molar mass = 293.89 g/mol):

$n_{CaI_2}=12.75gCaI_2*\frac{1molCaI_2}{293.89gCaI_2}=0.0434molCaI_2$

In such a way, since 1 mole of calcium iodide contains 2 moles of atoms of iodine, and one mole of atoms of iodine contains 6.022x10²³ atoms (Avogadro's number), we compute the resulting atoms as shown below:

$atoms\ I=0.0434molCaI_2*\frac{2molI}{1molCaI_2} *\frac{6.022x10^{23}atoms\ I}{1molI} \\\\atoms\ I = 5.225x10^{22}atoms\ I$