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Pavlova-9 [17]
3 years ago
12

Will give branliest!

Chemistry
1 answer:
Fiesta28 [93]3 years ago
4 0

Answer:

0.058 g of Al₂O₃

Explanation:

Al₂O₃ → Aluminum oxide

The molar mass is 101.96 g/mol. Therefore 1 mol (NA molecules) are contained in 101.96 grams

NA = Avogadro's number → 6.02×10²³

We make a rule of three to determine the grams we were asked for.

6.02×10²³ molecules are contained in 101.96 g

3.45×10²⁰ molecules will be contained in (3.45×10²⁰ . 101.96 g ) / NA = 0.058 g

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If 1.80 moles of NaCl was dissolved in enough water to make 3.60 L of solution, what is the Molarity?
sweet-ann [11.9K]

Answer:

0.5M

Explanation:

The equation for molarity is:

  • M = \frac{mol}{liters} ; where the "M" stands for molarity, the "mol" stands for moles of solute and the "liters" means the volume in liters of solution.

We are given that there are:

  • 1.80 moles of NaCl (the moles of solute)
  • 3.60 Liters of solution (the volume in liters of solution)

Now we just plug those numbers into the formula and get our answer:

  • M= \frac{1.80mol}{3.60L}= 0.5M

After doing the math and dividing the moles of solute by the liters of solution, we get that the molarity of the solution is 0.5M.

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3 years ago
Be sure to answer all parts. Consider the following balanced redox reaction (do not include state of matter in your answers): 2C
timurjin [86]

Answer:

The specie which is oxidized is:- CrO_2^-

The specie which is reduced is:- ClO^-

Explanation:

Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

X\rightarrow X^{n+}+ne^-

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

X^{n+}+ne^-\rightarrow X

For the given chemical reaction:

2CrO_2^- + 6ClO^- + 2H_2O\rightarrow 2CrO_4^{2-} + 3Cl_2 + 4OH^-

The half cell reactions for the above reaction follows:

Oxidation half reaction:  CrO_2^- + 2H_2O + 4OH^-\rightarrow CrO_4^{2-} + 4H_2O + 3e^-

Reduction half reaction:  2ClO^- + 4H_2O + 2e^-\rightarrow Cl_2 + 2H_2O + 4OH^-

Thus, the specie which is oxidized is:- CrO_2^-

The specie which is reduced is:- ClO^-

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Explanation:

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Explanation:

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