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asambeis [7]
3 years ago
8

Copper oxide, CuO, reacts with hydrochloric acid, HCI, to produce copper chloride, CuCL2 and water

Chemistry
1 answer:
spayn [35]3 years ago
6 0

Explanation:

El óxido de cobre (II), también llamado antiguamente óxido cúprico ({\displaystyle {\ce {CuO}}}{\displaystyle {\ce {CuO}}}), es el óxido de cobre con mayor número de oxidación. Como mineral se conoce como tenorita.

{\displaystyle {\ce {2Cu + O2 = 2CuO}}}{\displaystyle {\ce {2Cu + O2 = 2CuO}}}

Aquí, se forma junto con algo de óxido de cobre (I) como un producto lateral, por lo que es mejor prepararlo por calentamiento de nitrato de cobre (II), hidróxido de cobre (II) o carbonato de cobre (II):

{\displaystyle {\ce {2 Cu(NO3)2 = 2 CuO + 4 NO2+ O2}}}{\displaystyle {\ce {2 Cu(NO3)2 = 2 CuO + 4 NO2+ O2}}}

{\displaystyle {\ce {Cu(OH)2 (s) = CuO (s) + H2O (l)}}}{\displaystyle {\ce {Cu(OH)2 (s) = CuO (s) + H2O (l)}}}

{\displaystyle {\ce {CuCO3 = CuO + CO2}}}{\displaystyle {\ce {CuCO3 = CuO + CO2}}}

El óxido de cobre (II) es un óxido básico, así se disuelve en ácidos minerales tales como el ácido clorhídrico, el ácido sulfúrico o el ácido nítrico para dar las correspondientes sales de cobre (II):

{\displaystyle {\ce {CuO + 2 HNO3 = Cu(NO3)2 + H2O}}}{\displaystyle {\ce {CuO + 2 HNO3 = Cu(NO3)2 + H2O}}}

{\displaystyle {\ce {CuO + 2 HCl =CuCl2 + H2O}}}{\displaystyle {\ce {CuO + 2 HCl =CuCl2 + H2O}}}

{\displaystyle {\ce {CuO + H2SO4 = CuSO4 + H2O}}}{\displaystyle {\ce {CuO + H2SO4 = CuSO4 + H2O}}}

Reacciona con álcali concentrado para formar las correspondientes sales cuprato.

{\displaystyle {\ce {3 XOH + CuO + H2O = X3[Cu(OH)6]}}}{\displaystyle {\ce {3 XOH + CuO + H2O = X3[Cu(OH)6]}}}

Puede reducirse a cobre metálico usando hidrógeno o monóxido de carbono:

{\displaystyle {\ce {CuO + H2 = Cu + H2O}}}{\displaystyle {\ce {CuO + H2 = Cu + H2O}}}

{\displaystyle {\ce {CuO + CO = Cu + CO2}}}{\displaystyle {\ce {CuO + CO = Cu + CO2}}}

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The volume of a pond being studied for the effects of acid rain is 35 kiloliters (kL). There are 1,000 liters (L) in 1 kL and 1
Lerok [7]

Answer:

35,000,000,000 mL

Explanation:

You first multiply 35 times 1000.

35,000 L

Now you multiply 35,000 times 10^6

35,000,000,000 mL

4 0
3 years ago
Write the chemical formula for sodium hypobromite.
irga5000 [103]

Answer:

             NaOBr  (or)  Na⁺ ⁻OBr

Explanation:

The Oxo-Acids of Bromine are as follow,

                           Hypobromous Acid  =  HOBr

                           Bromous Acid  =  HOBrO

                           Bromic Acid  =  HBrO₃

                           Perbromic Acid  =  HBrO₄

When these acids are converted to their conjugate bases their names are as follow,

                           Hypobromite  =  ⁻OBr

                           Bromite  =  ⁻OBrO

                           Bromate  =  ⁻OBrO₂

                           Perbromate  =  ⁻OBrO₃

According to rules, the positive part of ionic compound is named first and the negative part is named second. So, Sodium Hypobromite has a chemical formula of Na⁺ ⁻OBr or NaOBr.

6 0
3 years ago
What is the electron configuration for helium (He)?
givi [52]
Hello!

The electron configuration for helium is 1s2
3 0
3 years ago
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If the Dry bulb reads 25 degrees Celsius and the wet bulb reads 22
Nitella [24]

Answer:

Relative humidity is low .

Explanation:

The wet bulb reads low temperature because due to low humidity of atmosphere , evaporation of water takes place from the wet bulb which makes the bulb cool and therefore it reads lower temperature . In the process of evaporation , heat equal to latent heat of vaporization is taken from the bulb and it loses temperature.

6 0
3 years ago
An electrochemical cell has the following standard cell notation.
icang [17]

The cell notation is:

Mg(s)|Mg^{+2}(aq)||Ag^{+}(aq)|Ag(s)

here in cell notation the left side represent the anodic half cell where right side represents the cathodic half cell

in anodic half cell : oxidation takes place [loss of electrons]

in cathodic half cell: reduction takes place [gain of electrons]

1) this is a galvanic cell

2) the standard potential of cell will be obtained by subtracting the standard reduction potential of anode from cathode

E^{0}_{Mg}=-2.38V

E^{0}_{Ag}=+0.80V

Therefore

E^{0}_{cell}=0.80-(-2.38)=+3.18V

3) as the value of emf is positive the reaction will be spontaneous as the free energy change of reaction will be negative

ΔG^{0}=-nFE^{0}

As reaction is spontaneous and there will be conversion of chemical energy to electrical energy it is a galvanic cell.

7 0
3 years ago
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