When the vapor pressure of a liquid equals the pressure on its surface the liquid will be

The balanced chemical equation between solid chromium and solid iodine to form solid chromium(III) iodide is: 2 Cr(s) + 3 I₂(s) ⇒ 2 CrI₃(s)

Let's consider the unbalanced equation that occurs when ​solid chromium reacts with solid iodine to form solid chromium(III) iodide.​ This is a synthesis or combination reaction.

Cr(s) + I₂(s) ⇒ CrI₃(s)

We will balance it using the trial and error method.

First, we will balance I atoms by multiplying I₂ by 3 and CrI₃ by 2.

Cr(s) + 3 I₂(s) ⇒ 2 CrI₃(s)

Finally, we get the balanced equation by multiplying Cr by 2.

2 Cr(s) + 3 I₂(s) ⇒ 2 CrI₃(s)

The balanced chemical equation between solid chromium and solid iodine to form solid chromium(III) iodide is: 2 Cr(s) + 3 I₂(s) ⇒ 2 CrI₃(s)

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Answer:

2 Cr(s) + 3 I2(s) → 2 CrI3(s)

Explanation:

Reactants:

Solid Chromium

• Chromium is a transition metal with an elemental symbol → Cr

• It can exist as a stable monatomic solid → Cr(s)

Solid Iodine

• Iodine is a non-metal halide of Group 7A → I

• It exists as a diatomic solid → I2(s)

Products:

Chromium(III) iodide

• Chromium, Cr → roman numeral III → +3 charge → Cr3+

• Iodide, I → from group 7A → -1 charge → I-

• When ions combine, their charges or oxidation numbers crisscross to become subscripts

• Cr will have a subscript of 1 from I- → Cr

• I will have a subscript of 3 from Cr3+ → I3

• Subscript 1 is no longer indicated → CrI3(s)

Unbalanced Chemical Reaction: Cr(s) + I2(s) → CrI3(s)

Balancing the chemical reaction:

1. Balance I:

Cr(s) + I2(s) → CrI3(s)

• There are 2 I in the reactant and 3 I in the product side

• Balance I by adding the coefficient 3 in I2 and 2 in CrI2

• There would be a total of 6 I in the reactant and product sides

Cr(s) + 3 I2(s) → 2 CrI3(s)

2. Balance Cr:

Cr(s) + 3 I2(s) → 2 CrI3(s)

• There is 1 Cr in the reactant and 2 Cr in the product side

• Balance Cr by adding the coefficient 2 in Cr(s)

• There would be a total of 2 Cr in the reactant and product sides

2 Cr(s) + 3 I2(s) → 2 CrI3(s)

The limiting reagent is O₂.

Firstly, write the balanced chemical equation:

C₂H₄ + O₂ → CO₂ + H₂O

Then, we need to write the mole ratio of the reactants

The theoretical mole ratio is given by the coefficients of the balanced chemical equation: 3 moles of O₂ react with 1 mol of C₂H₄

Now, let's write the mole ratio of the substances available

 ≈  

We concluded that,

Since, the ratio of the O₂ to C₂H₄ available is less than the theoretical ratio of O₂ needed to react with C₂H₄, the former is not enough: it will be consumed before the other reagent and it is the limiting reactant.

Hence, the limiting reagent is O₂.

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Answer:

• The limiting reagent is O₂.

Explanation:

1. Write the balanced molecular equation:

2. Write the mole ratio of the reactants

The theoretical mole ratio is given by the coefficients of the balanced molecula equation: 3 moles of O₂ react with 1 mol of C₂H₄

3. Write the mole ratio of the substances available

4. Conclusion

Since the ratio of the O₂ to C₂H₄ available is less than the theoretical ratio of O₂ needed to react with C₂H₄, the former is not enough: it will be consumed before the other reagent and it is the limiting reactant.

Hence, the limiting reagent is O₂.

Answer:

The correct option is C

Oxygen

Carbon

Hydrogen

Nitrogen  

Sulfur

Phosphorus

Explanation:

Oxygen is considered to be the most abundant element to be found in the living cells. Usually, a living cell is made up of 65% oxygen.

Carbon is the element which forms the basis for life for all organisms which can be seen on the planet Earth.

Hydrogen combines with oxygen to form water which makes up 65 percent of the human body.

Nitrogen is an active constituent of molecules such as proteins which are of utter importance to an organism.

Phosphorus is present in the form of phospholipid layer which is an active constituent of the cell membranes.

Answer:

The correct option is C

Explanation: