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Misha Larkins [42]
2 years ago
8

Draw the correct Lewis structure of this molecule by placing atoms on the canvas and connecting them with bonds. Include all hyd

rogen atoms and lone pairs of electrons.
Chemistry
1 answer:
torisob [31]2 years ago
4 0

Answer:

hello your question is incomplete attached below is the complete question

answer : attached below

Explanation:

The Lewis structure given in the question is incorrect attached below is the correct Lewis structure.

Lewis structure of a molecule is the representation of the valence electrons in a molecule

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A process at constant T and P can be described as spontaneous if ΔG < 0 and nonspontaneous if ΔG > 0. Over what range of t
creativ13 [48]

Answer:

Incomplete question, it is lacking the data it makes reference. The missing data from Chegg is:

                              2 SO3(g)   →          2 SO2(g) + O2(g)

ΔHf° (kJ mol-1)  -395.7                        -296.8

S° (J K-1 mol-1)  256.8                         248.2              205.1

ΔH° =  kJ

S° =  J K⁻¹

Explanation:

The method to solve this problem calls for the use of the Gibbs standard free energy change:

ΔG = ΔrxnH - TΔSrxn

We know a reaction is spontaneous when ΔG is < 0, so to answer this question we need to solve for the temperature, T, at which ΔG becomes negative.

Now as mentioned in the hint, we need to determine  ΔrxnH and ΔSrxn, which are given by

ΔrxnH = ∑ ν x ΔfHº products - ∑ ν x ΔfHº reactants

where  ν  is the stoichiometric coefficient in the balanced chemical equation.

For ΔS we have likewise

ΔrxnS =  ∑ ν x ΔSº products - ∑ ν x ΔSº reactants

Thus,

ΔrxnH(kJmol⁻¹) =  2 x (-296.8) - 2 x ( -395.7 ) = 197.8 kJ

ΔrxnS ( JK⁻¹) = 2 x 248.2 + 205.1 - 2 x 256.8 = 187.9 JK⁻¹ = 0.1879 kJK⁻¹

So ΔG kJ =  197.8 - T(0.1879)

and the reaction will become spontaneous when the term  T(0.1879)  becomes greater that 197.8,

0 = 197.8 - 0.1879 T  ⇒ T = 1052 K

so the reaction is spontaneous at temperatures greater than 1052 K (780 ºC)

4 0
3 years ago
Suppose you wanted to dissolve 40.0 g NaOH in enough H2O to make 6.00 dm3 of solution
dezoksy [38]

Molarity of solution = 1.6 M

<h3>Further explanation</h3>

Given

40 g NaOH

6 L solution

Required

Steps to solve the problem of molarity

Solution

No additional information about the question.

If you want to make the solution above, then we just need to put the existing NaOH (40 g) into 6 L of water, then do the stirring (in a warm temperature above the hot plate will speed up the NaOH dissolving process)

But if you want to know the molarity of a solution, then

  • 1. we calculate the moles of NaOH

\tt mol=\dfrac{mass}{MW}

MW(molecular weight) of NaOH=

Ar Na+ Ar O + Ar H

23 + 16 + 1 = 40 g/mol

so mol NaOH :

\tt mol=\dfrac{40~g}{40~g/mol}=1~mol

  • 2. Molarity(M)

\tt M=\dfrac{n}{V}\\\\M=\dfrac{1}{6}\\\\M=0.16

5 0
3 years ago
What is the equilibrium constant, KC if the reaction is a gas phase reaction? (Ans.: Gas: KC = 0.328 dm3/mol)
valentina_108 [34]

Answer:

Equilibrium constant Kc = Qc = quotient of reactant(s) and product(s)

Kc = [C]x[D]y..../[A]m[B]n..... = 0.328dm3/mol, where [C]x[D]y is the product and [A]m[B]n is the reactant(Both in gaseous states)

Explanation:

When a mixture of reactants and products of a reaction reaches equilibrium at a given temperature, its reaction quotient always has the same value. This value is called the equilibrium constant (K) of the reaction at that temperature. As for the reaction quotient, when evaluated in terms of concentrations, it is noted as Kc.

That a reaction quotient always assumes the same value at equilibrium can be expressed as:

Qc (at equilibrium) = Kc =[C]x[D]y…/[A]m[B]n…

This equation is a mathematical statement of the law of mass action: When a reaction has attained equilibrium at a given temperature, the reaction quotient for the reaction always has the same value.

6 0
2 years ago
CaC12 + NaCO3–&gt;CaCO3 +NaC1
lapo4ka [179]
I think it would be c
7 0
3 years ago
C. How is EM created?
Arada [10]
EM is created by moving charges back and forth will produce oscillating electric and magnetic fields, and these travel at the speed of light
6 0
3 years ago
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