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Ghella [55]
3 years ago
10

H2, N2, O2 molecules. . . A)must be polar. must be nonpolar. . B)can be polar or nonpolar depending on geometric . C)configurati

on. . D)indeterminate
Chemistry
2 answers:
Ivanshal [37]3 years ago
7 0
I think the correct answer from the choices listed above is the option that states that must be nonpolar. H2, N2, O2 molecules are nonpolar molecules. <span>Nonpolar molecules have "balanced" pulls: one end has the same electron affinity value as does the other end. </span>
LenKa [72]3 years ago
6 0

Answer: The correct option is A.

Explanation: The given molecules are the molecules of same element.

These molecules are considered as diatomic species.

Polar molecules are the molecules in which some polarity is present in the bond. These molecules are formed when there is some difference in the electronegativities of the elements. Example: HCl

Non-polar molecules are the molecules where no polarity is present in the bond. These molecules are formed when there is no difference in the electronegativities of the elements. Example: H_2, O_2

The given molecules are non-polar in nature.

Hence, these molecules must be non-polar. So, the correct option is A.

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the copper anode is weighed before and after the electrolysis reation. if the copper anode is not completely dry when it is weig
Dmitrij [34]

Answer:

Faraday's constant will be smaller than it is supposed to be.

Explanation:

If the copper anode was not completely dry when its mass was measured, mass of the copper must be heavier than it should have been. Hence, the calculated Faraday’s constant would be smaller than it is supposed to be since when calculating Faraday’s Constant, the charge transferred is divided by the moles of electrons.

5 0
3 years ago
Write the equilibrium-constant expression for the reactionA(s)+3B(l)↽−−⇀2C(aq)+D(aq)A(s)+3B(l)↽−−⇀2C(aq)+D(aq)in terms of [A], [
Ilia_Sergeevich [38]

Answer:

K_{c} = [\text{C}]^{2}[\text{[D]}

Explanation:

\rm A(s)+3B(l) \, \rightleftharpoons \, 2C(aq)+D(aq)

The general formula for an equilibrium constant expression is

K_{c} = \dfrac{[\text{Products}]}{[\text{Reactants}]}

Solids and liquids are not included in the equilibrium constant expression.

Thus, for this reaction,  

K_{c} = [\textbf{C}]^{\mathbf{2}}\textbf{[D]}

3 0
3 years ago
A ridged steel tank filled with 62.7L of nitrogen gas of 85.0L atm
mylen [45]

The final gas pressure : 175.53 atm

<h3>Further explanation</h3>

Maybe the complete question is like this :

A ridged steel tank filled with 62.7 l of nitrogen gas at 85.0 atm and 19 °C is heated to 330 °C while the volume remains constant. what is the final gas pressure?

The volume remains constant⇒Gay Lussac's Law  

<em>When the volume is not changed, the gas pressure in the tube is proportional to its absolute temperature  </em>

\tt \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}

P₁=85 atm

T₁=19+273=292 K

T₂=330+273=603 K

\tt P_2=\dfrac{P_1\times T_2}{T_1}\\\\P_2=\dfrac{85\times 603}{292}\\\\P_2=175.53~atm

8 0
3 years ago
You make two dilutions. You take 83.52 mL of the stock solution and dilute to 500.0 mL to make Solution #1. You then add 52.27 m
kati45 [8]

Answer:

The answer to the question is;

The concentration of the Solution #1 in terms of molarity is

0.16704X moles/litre.

Explanation:

Let the concentration of the stock solution be X moles/liter

Therefore, 83.52 ml of the stock solution contains

83.52×(X/1000) moles

Dilution of 83.52 ml of X to 500 ml gives solution 1 with a concentration of

500 ml of solution 1 contains 83.52×(X/1000) moles

Therefore 1000 ml or 1 litre contains 2×83.52×(X/1000) moles = 0.16704X moles/litre

The molarity of solution 1 is 0.16704X moles/litre.

8 0
4 years ago
Which equation represents the combined gas law?
Keith_Richards [23]
<span>((P1V1/T1)=(P2V2/T2))</span>
8 0
4 years ago
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