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yan [13]
3 years ago
6

__ H₂ + ___ O₂ → ____H₂O

Chemistry
2 answers:
Scilla [17]3 years ago
6 0
2H2+O2=2H2O

This is the chemical formula for water. Hope this helps.
Alex777 [14]3 years ago
4 0

Answer:

2H2+O2=2H2o

Explanation:

We have 2 hydrogens, 2 oxygens on the left-hand side.

But only 1 O on the right-hand side.

let's place a 2 before H2O (water)

Do you drink H2O every day?

But balancing the O unbalanced H

So the balanced chemical equation will look like this:

2H2O+O2=2H2O

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Reduction occurs at which electrode?
OLEGan [10]

Answer:

The electrode that removes ions from solution

Explanation:

Each electrochemical cell consists of an anode and a cathode. Oxidation occurs at the anode and reduction occurs at the cathode.

At the anode, ions move from the electrode into the solution while at the cathode ions move from the solution to the electrode.

At the cathode, metal ions accept electron(s) and become deposited on the electrode hence this electrode removes ions from solution. This is reduction.

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3 years ago
Which of the following has the most energy?
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3 years ago
Calculate ΔH o rxn for the following: CH4(g) + Cl2(g) → CCl4(l) + HCl(g)[unbalanced] ΔH o f [CH4(g)] = −74.87 kJ/mol ΔH o f [CCl
anzhelika [568]

Answer:  ΔH for the reaction is -277.4 kJ

Explanation:

The balanced chemical reaction is,

CH_4(g)+Cl_2(g)\rightarrow CCl_4(l)+HCl(g)

The expression for enthalpy change is,

\Delta H=\sum [n\times \Delta H(products)]-\sum [n\times \Delta H(reactant)]

\Delta H=[(n_{CCl_4}\times \Delta H_{CCl_4})+(n_{HCl}\times B.E_{HCl}) ]-[(n_{CH_4}\times \Delta H_{CH_4})+n_{Cl_2}\times \Delta H_{Cl_2}]

where,

n = number of moles

Now put all the given values in this expression, we get

\Delta H=[(1\times -139)+(1\times -92.31) ]-[(1\times -74.87)+(1\times 121.0]

\Delta H=-277.4kJ

Therefore, the enthalpy change for this reaction is, -277.4 kJ

4 0
3 years ago
Good Morning This is a question I need help on pleas help
elena55 [62]

Answer:

S.

Explanation:

North and South must always be opposite of each other.

8 0
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