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taurus [48]
3 years ago
9

To balance a chemical equation, one: a. changes the subscripts of the formulas. b. changes the coefficients of the formulas. c.

eliminates some of the products or reactants. d. changes the chemical symbols of the elements.
Chemistry
1 answer:
AfilCa [17]3 years ago
4 0
<span>B. Changes the coefficients of the formulas.</span>
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What is typically displayed on the y-axis of a solubility curve?
Evgen [1.6K]
X axis would be time and y axis would probably be grams of solute dissolved
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How many electrons are in a atom of zicronium
MrRissso [65]

Answer:

There are 40 electrons in one atom of Zirconium.

Explanation:

Note: The word is not zicronium, it is Zirconium.

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3 years ago
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Consider the redox reaction below.
vovangra [49]

Answer:

Zn(s) → Zn⁺²(aq) + 2e⁻

Explanation:

Let us consider the complete redox reaction:

Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

This is a redox reaction because, both oxidation and reduction is simultaneously taking place.

  • Oxidation (loss of electrons or increase in the oxidation state of entity)
  • Reduction (gain of electrons or decrease in the oxidation state of the entity)
  • An element undergoes oxidation or reduction in order to achieve a stable configuration. It can be an octet configuration. An octet configuration is that of outer shell configuration of noble gas.

Here Zn(s) is undergoing oxidation from OS 0 to +2

And H in HCl (aq) is undergoing reduction from OS +1 to 0.

Therefore, for this reaction;

Oxidation Half equation is:

Zn(s) → Zn⁺²(aq) + 2e⁻

Reduction Half equation is:

2H⁺ + 2e⁻ → H₂(g)

4 0
3 years ago
The greater the [H*], the stronger the acid.<br> True<br> False
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Trueeeeeeeeeeeeeeeee!!!!!
4 0
3 years ago
in an experiment 3.425g of lead oxide was reduced to form 3.105g of lead the empirical formula of the lead oxide is?​
Romashka-Z-Leto [24]

Answer:

Pb3O4

Explanation:

According to this question, 3.425g of lead oxide was reduced to form 3.105g of lead in an experiment. Since lead oxide contains both lead (Pb) and oxygen (O) element,

Mass of lead oxide = 3.425g

Mass of lead = 3.105g

Mass of oxygen = (3.425g - 3.105g) = 0.320g

Next, we convert each mass value to mole by dividing by respective molar mass

Pb = 3.105g ÷ 207.2 = 0.0149mol

O = 0.320g ÷ 16 = 0.02mol

Next, we divide each mole value by the smallest (0.0149)

Pb = 0.0149mol ÷ 0.0149mol = 1

O = 0.02mol ÷ 0.0149mol = 1.342

Multiply each ratio value by 3 to get:

Pb = 1 × 3 = 3

O = 1.342 × 3 = 4.026

The whole number ratio, approximately, of Pb and O is 3:4, hence, their empirical formula is Pb3O4.

4 0
3 years ago
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