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Rama09 [41]
3 years ago
10

The formula for the compound that contains Cu2+ actions and oxygen anions is

Chemistry
1 answer:
azamat3 years ago
7 0

Answer:

. CuO

Explanation:

Since oxygen also has -2 the valency will cancel out the +2 on Cu leaving CuO

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9yuigyu need points sorry

Explanation:

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2 years ago
The net ionic equation for the reaction between aqueous nitric acid and aqueous sodium hydroxide is _______.
ValentinkaMS [17]
HNO₃(aq) + NaOH(aq) → NaNO₃(aq) + H₂O(l)

H⁺ + NO₃⁻ + Na⁺ + OH⁻ → Na⁺ + NO₃⁻ + H₂O

H⁺ + OH⁻ → H₂O   (the net ionic equation)
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3 years ago
Be sure to answer all parts. Compounds a and b are isomers having molecular formula c5h12. Heating a with cl2 gives a single pro
Elden [556K]

Neo-pentane represents the Compound A while compound B is n-pentane.

After careful consideration we can say that compounds A and B are alkanes and also isomers of pentane. In chemistry, Isomers are defined as compounds having same empirical molecular formula but different structural formulas due to varying arrangement of atoms.

Now, as per the question statement, compound A gives a single monochlorination product upon heating with the molecule of chlorine i.e. Cl2 showing that the molecule is extremely symmetric. This molecule must be neo-pentane. Refer to image 1.

Similarly, Compound B forms 3 constitutional isomers after undergoing monochlorination. This compound must be n-pentane since three are 3 different types of carbon atoms in the structure. Refer to image 2.

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3 0
1 year ago
Read 2 more answers
Using the two models, compare the processes of nuclear fusion and nuclear fission. What do the two processes have in common?
EleoNora [17]

Answer:it’s A and B

Explanation:

7 0
3 years ago
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For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia
Fittoniya [83]

Answer : The value of E^o_{(Cu^{2+}/Cu)} is, 0.34 V

Explanation :

Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction:  Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction:  Cu^{2+}+2e^-\rightarrow Cu

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu

To calculate the E^o_{(Cu^{2+}/Cu)} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=E^o_{(Cu^{2+}/Cu)}-E^o_{(Zn^{2+}/Zn)}

Putting values in above equation, we get:

1.10V=E^o_{(Cu^{2+}/Cu)}-(-0.76V)

E^o_{(Cu^{2+}/Cu)}=0.34V

Hence, the value of E^o_{(Cu^{2+}/Cu)} is, 0.34 V

8 0
3 years ago
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