All categories

2 years ago

Explain hybridization and what effects does hybridization have on bonds

1 answer:

5 0

Hybridization is the idea that atomic orbitals fuse to form newly hybridized orbitals, which in turn, influence molecular geometry and bonding properties.

You might be interested in

Which of the following bonds is the most reactive? A. C—C B. N—H C. O—O D. C—H

USPshnik [31]

I think the answer is b

6 0

2 years ago

A 75 g piece of gold (Au) at 1000 K is dropped into 200 g of H2O at 300K in an insulatedcontainer at 1 bar. Calculate the temper

Arturiano [62]

Answer:

the final temperature is T final = 308 K

Explanation:

since all heat released by gold is absorbed by water

Q gold + Q water = Q surroundings =0 (insulated)

Assuming first that no evaporation of water occurs , and denoting g as gold and w as water , then

Q gold = m g*cp g* ( T final - T initial g)

Q gold = m w*cp w* ( T final - T initial w)

where

m= mass

cp = specific heat capacity

T final = final temperature

T initial g and T initial w = initial temperature of gold and water respectively

thus

Q gold + Q water = 0

m g*cp g* ( T final - T initial g) + m w*cp w* ( T final - T initial w) =0

m g*cp g* T final + m w*cp w* T final = m g*cp g* T initial g+ m w*cp w* T initial w

T final = (m g*cp g* T initial g+ m w*cp w* T initial w)/(m g*cp g+ m w*cp w)

replacing values and assuming cp w = 1 cal/gK = 4.816 J/gK and cp g = 0.129 J/gK (from tables), then

T final = (75 g*0.129 J/gK* 1000 K + 200 g * 4.816 J/gK * 300 K )/(75 g*0.129 J/gK*+ 200 g * 4.816 J/gK ) = 308 K

T final = 308 K

since T boiling water = 373 K and T final = 308 K , we confirm that water does not evaporate

therefore the final temperature is T final = 308 K

3 0

2 years ago

Zinc sulfate is a 2-ion electrolyte,

Gekata [30.6K]

Answer: The value of i is 1.4 and 40% dissociation of 100 particles of zinc sulfate will yield 60 undissociated particles.

Explanation:

The equation used to calculate the Vant' Hoff factor in dissociation follows:

$\alpha =\frac{i-1}{n-1}$

where,

$\alpha$ = degree of dissociation = 40% = 0.40

i = Vant' Hoff factor

n = number of ions dissociated = 2

Putting values in above equation, we get:

$0.40=\frac{i-1}{2-1}\\\\0.40=i-1\\\\i=1.4$

The equation used to calculate the degee of dissociation follows:

$\alpha =\frac{\text{Number of particles dissociated}}{\text{Total number of particles taken}}$

Total number of particles taken = 100

Degree of dissociation = 40% = 0.40

Putting values in above equation, we get:

$0.40=\frac{\text{Number of particles dissociated}}{100}\\\\\text{Number of particles dissociated}=(0.40\times 100)=40$

This means that 40 particles are dissociated and 60 particles remain undissociated in the solution.

Hence, 40% dissociation of 100 particles of zinc sulfate will yield 60 undissociated particles.

7 0

2 years ago

Help ASAP!

ipn [44]

I believe the answer is D

5 0

3 years ago

A 50.0 mL sample of an aqueous H2SO4 solution is titrated with a 0.375 M NaOH solution. The equivalence point is reached with 62

vesna_86 [32]

Answer: The concentration of $H_2SO_4$ is 0.234 M

Explanation:

According to the neutralization law,

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1$ = basicity $H_2SO_4$ = 2

$M_1$ = molarity of $H_2SO_4$ solution = ?

$V_1$ = volume of $H_2SO_4$ solution = 50.0 ml

$n_2$ = acidity of $NaOH$ = 1

$M_1$ = molarity of $NaOH$ solution = 0.375 M

$V_1$ = volume of $NaOH$ solution = 62.5 ml

Putting in the values we get:

$2\times M_1\times 50.0=1\times 0.375\times 62.5$

$M_1=0.234M$

Therefore concentration of $H_2SO_4$ is 0.234 M

6 0

2 years ago