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V125BC [204]
3 years ago
15

Chromium(VI) oxide

Chemistry
2 answers:
Vaselesa [24]3 years ago
5 0

Answer:

chromium(III) oxide

CR2O3

I hope this will help you

plz make me brainliest

barxatty [35]3 years ago
4 0

Answer:

Cro=(II)

CrO2=(IV)

CrO3=(VI)

Cr2O3=(III)

Explanation:

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Please help !!!
stiv31 [10]

Answer:

2KClO3 》》2KCl +3O2

C+ O2》》CO2

number of C moles

Required O2 moles (According to the mole ratio )

Relevant to the first equation, find the moles the KClO3, which is used to produce that amount of O2 moles

Now you can find the mass of KClO3

I mentioned the useful steps which can guide you to get the answer.

Explanation:

5 0
3 years ago
What was life like in the United States during the 1930s?
liberstina [14]

The 1930s saw natural disasters as well as manmade ones: For most of the decade, people in the Plains states suffered through the worst drought in American history, as well as hundreds of severe dust storms, or "black blizzards," that carried away the soil and made it all but impossible to plant crops.

3 0
3 years ago
10. In what part of the world might you find Mystery Island (note: you will be given full-credit for simply attempting this ques
OLga [1]
Well countries near the quitter of warmer than countries than the poles. So the mystery island would probably be near the equator near Florida or in Florida
4 0
3 years ago
Help me please i need to sleep
lara [203]

Cloud is a gas.

ice is a solid.

snow is a solid.

steam is a gas.

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6 0
3 years ago
An electrochemical cell is constructed using electrodes with the following half-cell reactions.
marshall27 [118]

Answer:

E cell = +1.95 V

Explanation:

At Anode : Oxidation reaction takes place

At Cathode : Reduction reaction takes place

The reaction with lower value of reduction potential will undergo Oxidation

Mn^{2+}(aq)+2e^{-}\rightarrow Mn(s)   E = -1.18 V

This equation undergo oxidation reaction and become:

Anode(Oxidation-Half) :

Mn(s)\rightarrow Mn^{2+}(aq)+2e^{-}      E = +1.18 V

Cathode(Reduction-Half) :

Fe^{3+}(aq)+e^{-}\rightarrow Fe^{2+}    E =+0.77 V

To balance the reaction multiply reduction-Half with 2.We get :

Fe^{3+} +2Mn(s)+\rightarrow Fe^{2+} + 2Mn^{2+}

Note that E is intensive property , do not multiply E  of oxidation-half with 2

Ecell =  0.77 -(-1.18)

E = +1.95 V

3 0
3 years ago
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