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r-ruslan [8.4K]
2 years ago
15

HELP ME PLEASE I ONLY HAVE 14 MINUTES LEFT

Chemistry
1 answer:
Yanka [14]2 years ago
5 0

Answer:

B po sagot dyan nasagutan ko na yan

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A student balances the following redox reaction using half-reactions.
ICE Princess25 [194]
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
Read 2 more answers
A student needs to prepare 100. mL of 0.612 M Cu(NO3)2 solution. What mass, in grams, of copper(II) nitrate should the student u
Temka [501]

Answer: 11.5 grams

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution

Molarity=\frac{n\times 1000}{V_s}

where,

Morality = 0.612 M

n= moles of solute  

V_s = volume of solution in ml = 100 ml

Now put all the given values in the formula of molarity, we get

0.612=\frac{n\times 1000}{100ml}

n=0.0612moles

Mass={\text {moles of solute }}{\times {\text {molar mass}}=0.0612moles\times 187.56g/mol=11.5g

Therefore, the mass of copper (II)nitrate required is 11.5 grams

3 0
3 years ago
Can someone please help me with my chem final!!
scoundrel [369]

Answer:

direct

Explanation:

the more dense it is the more pressure it will exert

6 0
3 years ago
Write the Henderson-Hasselbalch equation for a solution of formic acid. Calculate the quotient [HCO2]/[HCO2H] at (a) pH 3.000; (
Elena L [17]

Answer:

a. 0.182

b. 1.009

c. 1.819

Explanation:

Henderson-Hasselbach equation is:

pH = pKa + log [salt / acid]

Let's replace the formula by the given values.

a. 3 = 3.74 + log [salt / acid]

3 - 3.74 = log [salt / acid]

-0.74 = log [salt / acid]

10⁻⁰'⁷⁴ = 0.182

b. 3.744 = 3.74 + log [salt / acid]

3.744 - 3.74 = log [salt / acid]

0.004 = log [salt / acid]

10⁰'⁰⁰⁴ = 1.009

c. 4 = 3.74 + log [salt / acid]

4 - 3.74 = log [salt / acid]

0.26 = log [salt / acid]

10⁰'²⁶ = 1.819

3 0
3 years ago
Which of the following is the most common gas in the troposphere
Effectus [21]
The most common gas in the troposphere is nitrogen. 

I hope this helps!
7 0
3 years ago
Read 2 more answers
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