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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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Write the charge and full ground-state electron configuration of the monatomic ion most likely to be formed by each:
Sladkaya [172]

Answer:

Part A:

Charge is P^{3-}

Configuration is 1s^2 2s^22p^63s^23p^6

Part B:

Charge is Mg^{2+}

Configuration is 1s^2 2s^22p^6

Part C:

Charge is Se^{2-}

Configuration is 1s^2 2s^22p^63s^23p^64s^23d^{10}4p^6

Explanation:

Monatomic ions:

These ions consist of only one atom. If they have more than one atom then they are poly atomic ions.

Examples of Mono Atomic ions: Na^+, Cl^-, Ca^2^+

Part A:

For P:

Phosphorous (P) has 15 electrons so it require 3 more electrons to stabilize itself.

Charge is P^{3-}

Full ground-state electron configuration of the mono atomic ion:

1s^2 2s^22p^63s^23p^6

Part B:

For Mg:

Magnesium (Mg) has 12 electrons so it requires 2 electrons to lose to achieve stable configuration.

Charge is Mg^{2+}

Full ground-state electron configuration of the mono atomic ion:

1s^2 2s^22p^6

Part C:

For Se:

Selenium (Se) has 34 electrons and requires two electrons to be stable.

Charge is Se^{2-}

Full ground-state electron configuration of the mono atomic ion:

1s^2 2s^22p^63s^23p^64s^23d^{10}4p^6

8 0
3 years ago
The following reaction produces ethanoic acid (CHACOOH) from methanol (CH3OH) and carbon
tigry1 [53]

Answer:

\boxed{\text{300 g}}

Explanation:

We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

M_r:         32                          60

           CH₃OH + CO ⟶ CH₃COOH

m/g:        160

(a) Moles of CH₃OH

\text{Moles of CH$_{3}$OH} = \text{160 g CH$_{3}$OH }\times \dfrac{\text{1 mol CH$_{3}$OH }}{\text{32 g CH$_{3}$OH}}= \text{5.00 mol CH$_{3}$OH}

(b) Moles of CH₃COOH

\text{Moles of CH$_{3}$COOH} = \text{5.00 mol CH$_{3}$OH } \times \dfrac{\text{1 mol CH$_{3}$COOH}}{\text{1 mol CH$_{3}$OH }} = \text{5.00 mol CH$_{3}$COOH}

(c) Mass of CH₃COOH

\text{Mass of CH$_{3}$COOH} =\text{5.00 mol CH$_{3}$COOH} \times \dfrac{\text{60 g CH$_{3}$COOH}}{\text{1 mol CH$_{3}$COOH}} = \textbf{300 g CH$_{3}$COOH}\\\\\text{The maximum mass of ethanoic acid that can be produced is $\boxed{\textbf{300 g}}$}

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3 years ago
HELP Which of the following fractions can be used in the conversion of 32 m3 to the unit mm3?
poizon [28]
We know that each millimeter contains 10⁻³ meters. Writing this as a ratio:
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We require a conversion from m³ to mm³, so we must take the cube of the ratio we have made:
1 mm³ = (10⁻³)³ m³

Therefore, the conversion used will be:
(1 mm / 10⁻³ m)³

When we multiply by this conversion, we will get:
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Answer:

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Explanation:

Step 1

To find a, take the number and move a decimal place to the right one position.

Original Number: 3,010,000

New Number: 3.010000

Step 2

Now, to find b, count how many places to the right of the decimal.

3 0
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Why is salt an ionic bond? Why is salt an ionic bond?
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