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2 years ago

Which ion can be both an oxidizing agent and reducing agent?

Chemistry

1 answer:

a_sh-v [17]2 years ago

4 0

Answer: a) Sn 2+

Explanation:A reducing agent is a substance that can be oxidized when it loses electrons while An oxidizing agent is a substance that is reduced when it gains electrons.

Considering Cu 2+, Fe 3+, Al 3+, Here, Cu 2+ can be reduce to its Copper metal, same for Al 3+ to Aluminium metal and Fe 3+ to Fe 2+ ion and will not be oxidized further since they are in their most oxidized states. The three ions are acting as only oxidizing agents.

while Sn 2+ can be oxidized to become Sn 4+ and also be reduced to become Sn metal acting as a reducing agent and an oxidizing agent.

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What is the correct chemical equation for the reaction between methane and oxygen to produce carbon dioxide and water?

kodGreya [7K]

Answer:

The answer to your question is: CH₄ + 3/2 O₂ ⇒ CO₂ + 2 H₂O

Explanation:

Methane = CH₄

Oxygen = O

Carbon dioxide = CO₂

Water = H₂O

CH₄ + 3/2 O₂ ⇒ CO₂ + 2 H₂O

This is the balanced equation

3 0

2 years ago

A solution has a poh of 7. 1 at 10∘c. what is the ph of the solution given that kw=2. 93×10−15 at this temperature? remember to

Gnom [1K]

A solution has a pOH of 7. 1 at 10∘c. Then the pH of the solution given that kw=2. 93×10−15 at this temperature is 7.4 .

It is given that,

pOH of solution = 7.1

Kw =2.93×10^(-15)

Firstly, we will calculate the value of pKw

The expression which we used to calculate the pKw is,

pKw=-log [Kw]

Now by putting the value of Kw in this expression,

pKw =−log{2.93×10^(-15)}

pKw =15log(2.93)

pKw=14.5

Now we have to calculate the pH of the solution.

As we know that,

pH+pOH=pKw

Now put all the given values in this formula,

pH+7.1=14.5

pH=7.4

Therefore, we find the value of pH of the solution is, 7.4.

learn more about pH value:

brainly.com/question/12942138

#SPJ4

7 0

10 months ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0

3 years ago

According to rounding rules for addition the sum of 27.1, 34.538, and 37.68 is

Katena32 [7]

The answer should be...99.318!

5 0

2 years ago

When there is unequal sharing between two atoms of the electrons in a bond, which type of bond is it?

dexar [7]

Polar covalent, it is what happens when there is unequal bond between two atoms, where one has a slightly negative charge and one has a slightly positive charge.

3 0

3 years ago

Read 2 more answers