Empirical formula for 6. 63.52% Fe, 36.48% S

A student balances the following redox reaction using half-reactions.

Answer: 6.

Explanation:

1) Aluminum

$Al^0-3e^----\ \textgreater \ Al^{3+}$

So each atom of aluminum lost 3 electrons to pass from 0 oxidation state to 3+ oxidation state.

2) Manganesium

$Mn^{2+}+2e^{-}---\ \textgreater \ Mn$

So, each ion of Mn(2+) gained 2 electrons pass from 2+ oxidation state to 0.

3) Balance

Multiply aluminum half-reaction (oxidation) by 2 and multiply manganesium half-raction (reduction) by 3:

$2Al^{0}-6e^{-}---\ \textgreater \ 2Al^{3+}

3Mn^{2+}+6e^{-}---\ \textgreater \ 3Mn^{0}$

4) Net equation

Add the two half-equations:

$2Al^{0}+3Mn^{2+}----\ \textgreater \ 2Al^{3+}+3Mn^{0}$

As you see the left side has 2 Al, 3Mn, and 3*2 positive charges.

The right side has 2 Al, 3 Mn, and 2*3 positive charges.

So, the equation is balanced.

5) Count the number of electrons involved.

As you see 2 atoms of aluminum lost 6 electrons (3 each).

That is the answer to the question. 6 electrons will be lost.

5 0

3 years ago

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Describe happens when you move any atom toward the positive nuclear charge.

bekas [8.4K]

Answer:

Across a period, effective nuclear charge increases as electron shielding remains constant. A higher effective nuclear charge causes greater attractions to the electrons, pulling the electron cloud closer to the nucleus which results in a smaller atomic radius. ... This results in a larger atomic radius.

Explanation:

Now lets d8

5 0

3 years ago

Calcium + zinc nitrate ------->

Snezhnost [94]

Answer:

c

Explanation:

there will be displacement reaction taking place

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3 years ago

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A graduated cylinder contains 20.8 mL of water. What is the new water level, in milliliters, after 35.2 g of silver metal is sub

stepladder [879]

<u>Answer:</u> The new water level of the cylinder is 24.16 mL

<u>Explanation:</u>

To calculate the volume of water displaced by silver, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of silver = 10.49 g/mL

Mass of silver = 35.2 g

Putting values in above equation, we get:

$10.49g/mL=\frac{35.2g}{\text{Volume of silver}}\\\\\text{Volume of silver}=\frac{35.2g}{10.49g/mL}=3.36mL$

We are given:

Volume of graduated cylinder = 20.8 mL

New water level of the cylinder = Volume of graduated cylinder + Volume of water displaced by silver

New water level of the cylinder = (20.8 + 3.36) mL = 24.16 mL

Hence, the new water level of the cylinder is 24.16 mL

5 0

3 years ago

A 100.0 mL solution containing 0.864 g of maleic acid (MW=116.072 g/mol) is titrated with 0.276 M KOH. Calculate the pH of the s

Lilit [14]

Answer:

pH = 1.32

Explanation:

H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺

This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:

So first calculate the moles reacted and produced:

n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M

54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH

it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.

moles H₂M left = 0.074 - 0.015 = 0.059

moles HM⁻ produced = 0.015

Using the Henderson - Hasselbach equation to solve for pH:

ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325

Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.

For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.

3 0

3 years ago