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IrinaVladis [17]
3 years ago
15

Which of the following molecules experience dipole-dipole forces as its strongest IMF? A) H2 B) SO2 C) NH3 D) CF4 E) BCl3

Chemistry
1 answer:
Xelga [282]3 years ago
5 0

Answer: NH_3

Explanation:

a) H_2: This  is a non polar covalent compound which are held by weak vanderwaal forces of attraction.

b) SO_2: This  is a covalent compound which is polar due to the presence of lone pair of electrons and are held by dipole-dipole forces of attraction.

c) NH_3: These are joined by a special type of dipole dipole attraction called as hydrogen bond. It forms between electronegative nitrogen atom and hydrogen atom and is the strongest interaction.

d) CF_4: This  is a covalent compound and is non polar which are held by weak vanderwaal forces of attraction.

e)  BCl_3: This  is a covalent compound and is non polar which are held by weak vander waal forces of attraction.

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Explain the difference between qualitative and quantitative properties.
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Qualitative properties are properties that are observed and can generally not be measured with a numerical result. They are contrasted to quantitative properties which have numerical characteristics.
3 0
2 years ago
Calculate the standard potential for the following galvanic cell: Fe(s) | Fe2+(aq) || Ag+(aq) | Ag(s) which has the overall bala
Nataly [62]

Answer:

E° = 1.24 V

Explanation:

Let's consider the following galvanic cell: Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s)

According to this notation, Fe is in the anode (where oxidation occurs) and Ag is in the cathode (where reduction occurs). The corresponding half-reactions are:

Anode: Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻

Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)

The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E° = E°red, cat - E°red, an

E° = 0.80 V - (-0.44 V) = 1.24 V

6 0
3 years ago
Calculate the work done when an ideal gas expands isothermally and reversibly in a piston and cylinder assembly for expansion of
pantera1 [17]

Answer:

W=5743.1077\ J

Explanation:

The expression for the work done is:

W=RT \ln \left( \dfrac{P_1}{P_2} \right)

Where,

W is the amount of work done by the gas

R is Gas constant having value = 8.314 J / K mol

T is the temperature

P₁ is the initial pressure

P₂ is the final pressure

Given that:

T = 300 K

P₁ = 10 bar

P₂ = 1 bar

Applying in the equation as:

W=8.314\times 300 \ln \left( \dfrac{10}{1} \right)

W=300\times \:8.314\ln \left(10\right)

W=2.30258\times \:2494.2

W=5743.1077\ J

5 0
3 years ago
What volume (in L) of oxygen will be required to produce 77.4 L of water vapor in the reaction below?
Inga [223]

Answer:

90.3 L

Explanation:

Given data:

Volume of water produced = 77.4 L

Volume of oxygen required = ?

Solution:

Chemical equation:

2C₂H₆ + 7O₂  →  4CO₂ + 6H₂O

It is known that,

1 mole = 22.414 L

There are 7 moles of oxygen = 7×22.414 = 156.9 L

There are 6 moles of water = 6×22.414 = 134.5 L

Now we will compare:

                               H₂O           :              O₂    

                               134.5         :              156.9

                                 77.4         :             156.9/134.5×77.4 =90.3 L

So for the production of 77.4 L water 90.3 L oxygen is required.

8 0
3 years ago
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