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3 years ago

15

According to valence bond theory, which orbitals on N and H overlap in the NH3 molecule? 2p on N overlaps with 2s on H 2s on N o

verlaps with 1s on H sp3 on N overlaps with 1s on H 2p on N overlaps with 1s on H sp3 on N overlaps with sp on H

1 answer:

Alecsey [184]3 years ago

8 0

Answer:

$sp^{3}$ on N overlaps with 1s on H.

Explanation:

Hybridization of central atom generally can be found by finding the hybridization number.Hybridization number is the sum of number of lone pairs and number of sigma bonds.

If hybridization number is 2 then the central atom is sp hybridized.

If it is 3 it is $sp^{2}$ hybridized.

If it is 4 it is $sp^{3}$ hybridized and so on.

Here in $NH_{3}$ molecule there are 3 sigma bonds and one lone pair(see the figure attached).

Hence hybridization number is 4 and hence it is s $p^{3}$ hybridized.

And in case of hydrogen atom the only electron of it resides in the 1s orbital.

Therefore $sp^{3}$ on N overlaps with 1s on H.

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How many joules of heat are required to heat 25.0 g of isopropyl alcohol from the prevailing room temperature, 21.2 oC, to its b

valentinak56 [21]

Answer:

3984.12 J joules of heat are required to heat 25.0 g of isopropyl alcohol to its boiling point.

Explanation:

$Q=m\times c\Delta T=m\times c\times (T_2-T_1)$

Where:

Q = heat absorbed or heat lost

c = specific heat of substance

m = Mass of the substance

ΔT = change in temperature of the substance

$T_1$ = Initial temperature of the substance

$T_2$ = Final temperature of the substance

We have mass of isopropyl alcohol = m = 25.0 g

Specific heat of isopropyl alcohol = c = 2.604 J/g°C

Initial temperature of the isopropyl alcohol = $T_1=21.2^oC$

Final temperature of the isopropyl alcohol = $T_2=82.4 ^oC$

Heat absorbed by the isopropyl alcohol to boil:

$Q=25.0 g\times 2.604 J/g^oC\times (82.4^oC-21.2^oC)=3984.12 J$

3984.12 J joules of heat are required to heat 25.0 g of isopropyl alcohol to its boiling point.

7 0

3 years ago

Find the volume of 6.45 moles of gas present at a temperature of 27.0oC and a pressure of 675 torr.

maw [93]

<u>Answer:</u> The volume of the gas is 178.76 L

<u>Explanation:</u>

The ideal gas equation is given as:

.......(1)

where

P = pressure of the gas = 675 torr

V = volume of gas = ?

n = number of moles of gas = 6.45 moles

R = Gas constant = 62.36 L.torr/mol.K

T = temperature of the gas = $27^oC=[27+273]K=300K$

Putting values in equation 1, we get:

$675torr\times V=6.45mol\times 62.36L.torr/mol.K\times 300K\\\\V=\frac{6.45\times 62.36\times 300}{675}=178.76 L$

Hence, the volume of the gas is 178.76 L

3 0

3 years ago

Please answer asap!

kaheart [24]

200 ml is 1/5 of a liter, so the answer is five times the number of moles present in the solution. 0.6 moles/0.2 liter = x moles/1.0 liter. Solving for x gives 0.2 x = 0.6 or x = 3.0 M

so the answer is c

3 0

3 years ago

write equations to show the chemical processes which occur when the first ionization and the second ionization energies of lithi

diamong [38]

Answer:

First ionization of lithium:

$\text{Li}\;(g)\to \text{Li}^{+} \; (g) + \text{e}^{-}$ .

Second ionization of lithium:

$\text{Li}^{+}\;(g) \to\text{Li}^{2+} \;(g) + \text{e}^{-}$ .

Explanation:

The ionization energy of an element is the energy required to remove the outermost electron from an atom or ion of the element in gaseous state. (Refer to your textbook for a more precise definition.) Some features of the equation:

Start with a gaseous atom (for the first ionization energy only) or a gaseous ion. Write the gaseous state symbol $(g)$ next to any atom or ion in the equation.
The product shall contain one gaseous ion and one electron. The charge on the ion shall be the same as the order of the ionization energy. For the second ionization energy, the ion shall carry a charge of +2.
Charge shall balance on the two sides of the equation.

First Ionization Energy of Li:

The products shall contain a gaseous ion with charge +1 $\text{Li}^{+}\;(g)$ as well as an electron $\text{e}^{-}$ .
Charge shall balance on the two sides. There's no net charge on the product side. Neither shall there be a charge on the reactant side. The only reactant shall be a lithium atom which is both gaseous and neutral: $\text{Li}\;(g)$ .

Hence the equation: $\text{Li}\;(g) \to \text{Li}^{+}\;(g) + \text{e}^{-}$ .

Second Ionization Energy of Li:

The product shall contain a gaseous ion with charge +2: $\text{Li}^{2+}\;(g)$ as well as an electron $\text{e}^{-}$ .
Charge shall balance on the two sides. What's the net charge on the product side? That shall also be the charge on the reactant side. What will be the reactant?

The equation for this process is $\text{Li}^{+} \; (g) \to \text{Li}^{2+}\;(g) + \text{e}^{-}$ .

5 0

3 years ago

A 25.00 mL sample of vinegar was titrated with 39.27 mL of 0.4293 M NaOH. Calculate the concentration of acetic acid in the vine

QveST [7]

Answer:

0.6743 M

Explanation:

HC₂H₃O₂ + NaOH → NaC₂H₃O₂ + H₂O

First we <u>calculate how many NaOH moles reacted</u>, using the <em>definition of molarity</em>:

Molarity = moles / volume

moles = Molarity * volume

0.4293 M * 39.27 mL = 16.86 mmol NaOH

<em>One NaOH moles reacts with one acetic acid mole</em>, so <u>the vinegar sample contains 16.86 mmoles of acetic acid as well</u>.

Finally we <u>calculate the concentration (molarity) of acetic acid</u>:

16.86 mmol HC₂H₃O₂ / 25.00 mL = 0.6743 M

4 0

3 years ago