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erastovalidia [21]
3 years ago
8

Balance the following equations:

Chemistry
1 answer:
Brilliant_brown [7]3 years ago
6 0

Answer:

1) 0 C2H4O2 + 0 O2 -> 0 CO2 + 0 H2O (balanced)

2) V2O5 + CaS -> CaO + V2S5

<em>just additional info: V2O5 </em><em>is</em><em> </em><em>divanadium</em><em> </em><em>pentaoxide</em>

LHS (Left hand side)

V: 2

O: 5

Ca: 1

S: 1 x 5 [to balance with the right hand side of the equation]

RHS (Right hand side)

V: 2

O: 1 x 5 [to balance with the left hand side of the equation]

Ca: 1

S: 5

When you balance any elements, you have to balance the whole chemical compound.

Thus,

V2O5 + <em><u>5</u></em> CaS -> <em><u>5</u> CaO</em> + V2S5

LHS CHECK:

V: 2

O: 5

Ca: 5

S: 5

RHS CHECK:

V: 2

O: 5

Ca: 5

S: 5

3) S8 + O2 -> SO2

LHS:

S: 8

O: 2

RHS:

S: 1 x 8 [to balance with LHS]

O: 2

When you balance any elements, you have to balance the whole chemical compound.

S8 + O2 -> <em><u>8</u></em><em><u> </u></em>SO2

When we add 8 to the RHS, it gives us 8S, 16 O.

In order to balance that into the RHS, I need to multiply the O2 by 8, which will give 8(O2) = 16 O particles.

Therefore, <em><u>S8</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>8</u></em><em><u> </u></em><em><u>O2</u></em><em><u> </u></em><em><u>-</u></em><em><u>></u></em><em><u> </u></em><em><u>8</u></em><em><u> </u></em><em><u>SO2</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>final</u></em><em><u> </u></em><em><u>answer</u></em><em><u> </u></em><em><u>for</u></em><em><u> </u></em><em><u>(</u></em><em><u>3</u></em><em><u>)</u></em><em><u>.</u></em>

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SO in electrochemical series who has most negative or less reduction potential value tends to be a good reducing agent than the other.

Hope it helps...

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Which law is associated with inertia
Doss [256]
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5 0
3 years ago
Could someone explain how they got this answer, explain step by step plz
gulaghasi [49]

Answer:

6.018 amu

Explanation:

Let 6–Li be isotope A.

Let 7–Li be isotope B.

Let the abundance of 6–Li be A%

Let the abundance of 7–Li be B%

The following data were obtained from the question:

Atomic mass of isotope A (6–Li) =.?

Atomic mass of isotope B (7–Li ) = 7.015 amu.

Abundance of 7–Li (B%) = 92.58%

Abundance of 6–Li (A%) = 100 – B% = 100 – 92.58 = 7.42%

Atomic mass of Lithium = 6.941amu

The atomic mass of isotope A (6–Li) can be obtained as follow:

Atomic mass = [(Mass of A x A%)/100] + [(Mass of B x B%)/100]

6.941 = [(mass of A x 7.42)/100] + [(7.015x92.58)/100]

6.941 = [(mass of A x 7.42)/100] + 6.494487

(mass of A x 7.42)/100 = 6.941 – 6.494487

(mass of A x 7.42)/100 = 0.446513

Mass of A x 7.42 = 100 x 0.446513

Mass of A x 7.42 = 44.6513

Divide both side by 7.42

Mass of A = 44.6513 / 7.42

Mass of A = 6.018 amu

Therefore, the mass of 6–Li is 6.018 amu

7 0
3 years ago
The reaction A + 2B → products was found to follow the rate law: rate = k[A] 2[B]. Predict by what factor the rate of reaction w
Alexus [3.1K]

Answer:

By a factor of 12

Explanation:

For the reaction;

A + 2B → products

The rate law is;

rate = k[A]²[B]

As you can see, the rate is proportional to the square of the concentration of  A  and the of the concentration of  B .

Let's say initially, [A] = x, [B] = y

The rate law in this case is equal to;

rate1 = k. x².y

Now you double the concentration of A and triple the concentration of B.

[A] = 2x, [B] = 3y

The new rate law is given as;

rate2 = k . (2x)². (3y)

rate2 = k . 4x² . 3y

rate2 = 12 k . x² . y

Comparing rate 2 and rate 1, the ratio is given as; rate 2/ rate 1 = 12

Therefore the rate has increased by a factor of 12.

5 0
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