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FinnZ [79.3K]
3 years ago
12

The electron pair geometry of a molecule is tetrahedral. What is its bond angle if it shows one lone pair of electrons and three

bonding pairs?
A) Less than 109.5°
B) More than 109.5°
C) Equal to 109.5°
D) Zero

Chemistry
2 answers:
den301095 [7]3 years ago
3 0
I think is Gonna be C
murzikaleks [220]3 years ago
3 0

Answer:  A) Less than 109.5°

Explanation: The electron pair geometry of a molecule is tetrahedral , which means the number of electrons is 4 that means the hybridization will be sp^3. and the bond angle will be 109.5°.

But as one lone pair of electrons and three bonding pairs are present, the fourth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the bond angle will decrease from 109.5° and molecular geometry will be trigonal pyramidal.

As in ammonia molecule, the bond angle reduces to 107° due to the presence of lone pair of electron in tetrahedral geometry.

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What group is the element barium(Ba) In?<br>A. 1<br>B. 6<br>C. 56<br>D. 4​
LenaWriter [7]
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7 0
2 years ago
4. Can 200 ml of fluid be transferred to a 1-quart container? Explain the process that you used to arrive at your answer.
sukhopar [10]

Answer:

  • <u><em>Yes, 200 ml of fluid can be transferred to a 1-quart container.</em></u>

Explanation:

You must compare the two volumes, 200 ml and 1 quart. If 200 ml is less than or equal to 1 quart, then 200 ml of fluid can be transferred to a 1-quart container, else it is not possible.

To compare, the two volumes must be on the same system of units.

Quarts is a measure of volume equivalent to 1/4 of gallon.

One gallon is approximately 3.785 liters.

3.785 liter = 3.785 liter × 1,000 ml/liter

Then, to convert 1 quart to ml use the unit cancellation method:

  • (1/4)gallon × 3.785 liter/gallon × 1,000ml / liter = 946.25 ml

Thus, you get that a 1-quart container has volume of 946.25 ml, which allows that 200ml of fluid be transferred to it.

7 0
3 years ago
One gram of liquid benzene is burned in a bomb calorimeter. The temperature before ignition was 20.826 C, and the temperature af
Licemer1 [7]

Answer:

fH = - 3,255.7 kJ/mol

Explanation:

Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.

Qsystem + Qcombustion = 0

Qsystem = heat capacity*ΔT

10000*(25.000 - 20.826) + Qc = 0

Qcombustion = - 41,740 J = - 41.74 kJ

So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:

n = mass/molar mass = 1/ 78

n = 0.01282 mol

fH = -41.74/0.01282

fH = - 3,255.7 kJ/mol

4 0
4 years ago
What would happen to the volume of the container if the pressure is increased by a factor of 2​
telo118 [61]

Answer:

volume of the container will decreases if pressure increases.

Explanation:

According to Boyle's law:

Pressure is inversely proportional to volume which means if pressure of a gas increases the volume of the gas will decreases as gas molecules will collide and come closer forcefully so volume will decreases. And its formula for determining volume and pressure is:

<em>PV=nRT</em>

where "R" is a ideal gas constant

"T" is temperature and

"n" is number of particles given in moles while "V" is volume and "P" is pressure.

8 0
3 years ago
If the temperature on 244 mL of a gas is changed to 488 mL and 6 atm, at constant
Fittoniya [83]

Answer:

<h2>12 atm</h2>

Explanation:

To find the initial pressure we use the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the initial pressure

P_1 =  \frac{P_2V_2}{V_1}  \\

From the question we have

P_1 =  \frac{488 \times 6}{244}  =  \frac{2928}{244}  \\

We have the final answer as

<h3>12 atm</h3>

Hope this helps you

6 0
3 years ago
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