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sveticcg [70]
2 years ago
5

Which of the following describes a scenario in which the molecular geometry would have the smallest bond angle(s)? (A) four bond

ing domains (or bonding pairs) and zero nonbonding domains (or lone pairs) (B) two bonding domains (or bonding pairs) and two nonbonding domains (or lone pairs) (C) three bonding domains (or bonding pairs) and one nonbonding domain (or lone pair) (D) two bonding domains (or bonding pairs) and zero nonbonding domains (or lone pairs) (E) two bonding domains (or bonding pairs) and one nonbonding domain (or lone pair) (F) three bonding domains (bonding pairs) and zero nonbonding domains (or lone pairs)
Chemistry
1 answer:
maw [93]2 years ago
6 0

Answer: The correct answer is "B" two bonding domains(or bonding pairs) or two non bonding domains(or lone pairs)

Explanation:

Molecular geometry/structure is a three dimensional shape of a molecule. It is basically an arrangement of atoms in a molecule.It is determined by the central atom, its surrounding atoms and electron pairs.According to VSEPR theory, there are 5 basic shapes of a molecule: linear, trigonal planar, tetrahedral, trigonal bipyramidal and octahedral.

A)Four bonding domains and zero non bonding domains: shape is tetrahedral and bond angle is 109.5°

B)Two bonding domains and two non bonding domains(lone pairs): shape is bent and bond angle is 104.5°

C)Three bonding domains and one non bonding domain: shape is trigonal pyramidal and bond angle is 107°

D)Two bonding domain and zero non bonding domain: shape is linear and bond angle is 107°

E)Two bonding domain and one non bonding domain: bent shape and bond angle is 120°

F)Three bonding domains and zero nonbonding domain: shape is trigonal planar and bond angle is 120°

Hence Two bonding domains and two non bonding domains have the smallest bond angle.

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If total pressure of multiple gases is 512 mmHg, and the pressure of oxygen gas is 332 mmHg and the pressure of carbon monoxide
deff fn [24]

Answer: 67 mmHg

Explanation:

According to Dalton's Gas Law, the total pressure of a mixture of gases is the sum of the pressure of each individual gas.

i.e Ptotal = P1 + P2 + P3 + .......

In this case,

Ptotal = 512 mmHg

P(oxygen) = 332 mmHg

P(carbon mono-oxide) = 113 mmHg

Remaining pressure (P3) = ?

To get P3, apply Dalton's Gas Law formula

Ptotal = P(oxygen) + P(carbon mono-oxide) + P3

512 mmHg = 332 mmHg + 113 mmHg + P3

512 mmHg = 445 mmHg + P3

P3 = 512 mmHg - 445 mmHg

P3 = 67 mmHg

Thus, the remaining pressure is 67 mmHg

5 0
3 years ago
At room temperature, a melting solid produces a whitish liquid. A laser beam shined through the liquid is unaffected and produce
navik [9.2K]
The answer is a colloid.
8 0
3 years ago
Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 6.9 g of hexane is mi
MrRa [10]

Answer:

There, there are no leftover for C6H14 instead, an additional mass of 4g of C6H14 is needed to completely react with 38.4g of O2.

Explanation:

Step 1:

We'll begin by writing the balanced equation for the reaction. This is shown below:

2C6H14 + 19O2 —> 12CO2 + 14H2O

Step 2:

Let us calculate the masses of C6H14 and O2 that reacted from the balanced equation. This is illustrated below:

2C6H14 + 19O2 —> 12CO2 + 14H2O

Molar Mass of C6H14 = (12x6) + (14x1) = 72 + 14 = 86g/mol

Mass of C6H14 from the balanced equation = 2 x 86 = 172g

Molar Mass of O2 = 16x2 =32g/mol

Mass of O2 from the balanced equation = 19 x 32 = 608g

From the balanced equation above, 172g of C6H14 reacted with 608g of O2.

Step 3.

Now, let us determine the mass of C6H14 that will react with 38.4 g of oxygen. This is illustrated below:

From the balanced equation above, 172g of C6H14 reacted with 608g of O2.

Therefore, Xg of C6H14 will react with 38.4g of O2 i.e

Xg of C6H14 = (172 x 38.4) /608

Xg of C6H14 = 10.9g

From the calculations made above, we can see clearly that the mass of C6H14 is limited as the reaction requires 10.9g of C6H14 and only 6.9g was given. There, there are no leftover for C6H14 instead, an additional mass ( 10.9 - 6.9 = 4g) of 4g of C6H14 is needed to completely react with 38.4g of O2.

5 0
2 years ago
Explain why, when the imidazole ring of histidine is protonated, the double-bonded nitrogen is the nitrogen that accepts the pro
eduard
The important thing to note is the reason why electron react is due to the instability of the electrons. All elements wants to aim the electron configuration of the noble gases. This is the most stable form in which each of the orbitals are sufficiently filled. When it comes to bonding, the order of reactivity is: alkynes > alkenes > alkanes. Alkynes are compounds with triple bonds, alkenes with double bonds and alkanes with single bonds. The single bonds are called saturated hydrocarbons. This is because they have reached stability, so it is quite difficult to react this with reducing or oxidizing agents. Alkynes and alkenes are unsaturated hydrocarbons. They readily react with reducing and oxidizing agents so as to become saturated, as well. The underlying principle for this is that single bonds contain sigma bonds which is the head-on overlapping of electrons. These is the strongest type of covalent bond. Double and triple bonds contain pi bonds which is the side overlapping of electrons orbitals. Hence, these electrons would be easily separated making it more reactive especially during protonation.
4 0
2 years ago
A student weighs an empty flask and stopper and finds the mass to be 55.844 g. She then adds about 5 mL of an unknown liquid and
Oduvanchick [21]

Answer :

(a) The pressure of the vapor in the flask in atm is, 0.989 atm

(b) The temperature of the vapor in the flask in Kelvin is, 372.7 K

    The volume of the flask in liters is, 0.2481 L

(c) The mass of vapor present in the flask was, 0.257 g

(d) The number of moles of vapor present are 0.00802 mole.

(e) The mass of one mole of vapor is 32.0 g/mole

Explanation : Given,

Mass of empty flask and stopper = 55.844 g

Volume of liquid = 5 mL

Temperature = 99.7^oC

Mass of flask and condensed vapor = 56.101 g

Volume of flask = 248.1 mL

Barometric pressure in the laboratory = 752 mmHg

(a) First we have to determine the pressure of the vapor in the flask in atm.

Pressure of the vapor in the flask = Barometric pressure in the laboratory = 752 mmHg

Conversion used :

1atm=760mmHg

or,

1mmHg=\frac{1}{760}atm

As, 1mmHg=\frac{1}{760}atm

So, 752mmHg=\frac{752mmHg}{1mmHg}\times \frac{1}{760}atm=0.989atm

Thus, the pressure of the vapor in the flask in atm is, 0.989 atm

(b) Now we have to determine the temperature of the vapor in the flask in Kelvin.

Conversion used :

K=273+^oC

As, K=273+^oC

So, K=273+99.7=372.7

Thus, the temperature of the vapor in the flask in Kelvin is, 372.7 K

Now we have to determine the volume of the flask in liters.

Conversion used :

1 L = 1000 mL

or,

1 mL = 0.001 L

As, 1 mL = 0.001 L

So, 248.1 mL = 248.1 × 0.001 L = 0.2481 L

Thus, the volume of the flask in liters is, 0.2481 L

(c) Now we have to determine the mass of vapor that was present in the flask.

Mass of flask and condensed vapor = 56.101 g

Mass of empty flask and stopper = 55.844 g

Mass of vapor in flask = Mass of flask and condensed vapor - Mass of empty flask and stopper

Mass of vapor in flask = 56.101 g - 55.844 g

Mass of vapor in flask = 0.257 g

Thus, the mass of vapor present in the flask was, 0.257 g

(d) Now we have to determine the number of moles of vapor present.

Using ideal gas equation:

PV = nRT

where,

P = Pressure of vapor = 0.989 atm

V = Volume of vapor  = 0.2481 L

n = number of moles of vapor = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of vapor = 372.7 K

Putting values in above equation, we get:

(0.989atm)\times 0.2481L=n\times (0.0821L.atm/mol.K)\times 372.7K\\\\n=0.00802mole

Thus, the number of moles of vapor present are 0.00802 mole.

(e) Now we have to determine the mass of one mole of vapor.

\text{Mass of one mole of vapor}=\frac{\text{Mass of vapor}}{\text{Moles of vapor}}

\text{Mass of one mole of vapor}=\frac{0.257g}{0.00802mole}=32.0g/mole

Thus, the mass of one mole of vapor is 32.0 g/mole

8 0
2 years ago
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