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ExtremeBDS [4]
3 years ago
14

Which orbitals from a pi bond?

Chemistry
1 answer:
QveST [7]3 years ago
8 0

Answer:

B. The s orbital and two p orbitals

Explanation:

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Hydrogen peroxide (h​2​o​2​) decomposes according to the equation:h​2​o​2​(​l​) ​⇆​ h​2​o(​l​) + ½ o​2​(​g​)calculate k​p​ for t
goldenfox [79]
Hello! Let me try to answer this :)

Thanks and please correct if there are any mistakes ^ ^

6 0
3 years ago
Which of the following is an acid-base neutralization reaction? (1 point)
GuDViN [60]
It would be NaOH + HCl → <span>NaCl + H2O
</span>
NaOH is sodium hydroxide, which is a strong base. HCl  is hydrochloric acid, which is a strong acid. 

You have a strong base and a strong acid on the left side, however, at the result side, you end up with NaCl + H2O. Sodium chloride is simply table salt and H2O is just water, thus it has been neutralized. 


6 0
3 years ago
Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) 2 Ce41(aq) 1 3 I2(aq) S 2 Ce31(aq) 1 I32
Law Incorporation [45]

<u>Answer:</u>

<u>For a:</u> The standard Gibbs free energy of the reaction is -347.4 kJ

<u>For b:</u> The standard Gibbs free energy of the reaction is 746.91 kJ

<u>Explanation:</u>

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}           ............(1)

  • <u>For a:</u>

The given chemical equation follows:

2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)

<u>Oxidation half reaction:</u>   Ce^{4+}(aq.)\rightarrow Ce^{3+}(aq.)+e^-       ( × 2)

<u>Reduction half reaction:</u>   3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)

We are given:

n=2\\E^o_{cell}=+1.08V\\F=96500

Putting values in equation 1, we get:

\Delta G^o=-2\times 96500\times (+1.80)=-347,400J=-347.4kJ

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

  • <u>For b:</u>

The given chemical equation follows:

6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)

<u>Oxidation half reaction:</u>   Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-       ( × 6)

<u>Reduction half reaction:</u>   2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)

We are given:

n=6\\E^o_{cell}=-1.29V\\F=96500

Putting values in equation 1, we get:

\Delta G^o=-6\times 96500\times (-1.29)=746,910J=746.91kJ

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

7 0
3 years ago
A pure substance is found to contain 53.7% fluorine and 46.3% xenon by mass. What is the empirical formula of this substance?
evablogger [386]

Answer: XF8

Explanation:

Empirical Formular shows the simplest ratio of elements in a compound.

 Xe = 46.3%             F  = 53.7%

Divide the percentage composition of each element by the atomic  mass.

Xe = 46.3/ 131.3                      F= 53.7/ 19

      = 0.353( approx)               =  2.826 (approx)

Divide through with the smallest of the answers gotten in previous step.

 Xe = 0.353 / 0.353                F = 2.826/ 0.353

       =  1                                       = 8.0

Empirical formular = XF8

3 0
3 years ago
Which element has the largest density at stp
MArishka [77]
I believe it is Sodium. I could be wrong though.
8 0
3 years ago
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