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denis-greek [22]
3 years ago
6

During reduction, which of the following is true? A. Electrons are gained, so the oxidation number increases. B. Electrons are g

ained, so the oxidation number decreases. C. Electrons are given off, so the oxidation number increases. D. Electrons are given off, so the oxidation number decreases.
Chemistry
1 answer:
Pani-rosa [81]3 years ago
6 0

Answer:

B. Electrons are gained, so the oxidation number decreases.

Explanation:

Reduction is the <em>gain of electrons</em>.

Oxidation number is the charge that an atom <em>appears</em> to have when we count its electrons in a specific way.

Electrons have a negative charge so, if an atom gains electrons, its charge (oxidation number) becomes more negative. The oxidation number decreases.

If electrons are given off, the atom is being oxidized. Loss of electrons is <em>oxidation</em>.

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6 0
2 years ago
If normal rainwater has a pH of about 6.50, what is the hydrogen ion concentration, [H+]?
IceJOKER [234]

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.000000316

Explanation:

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6 0
3 years ago
How many significant digits does the number 700 have?
OLga [1]

Answer: 700 has one significant figure which is 7.

Explanation: These are some rules for significant figures

•All non-zero digits are significant: 1,2,3,4,5,6,7,8,9

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•Leading zeros are not significant: There are two significant figures in 0.56.

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5 0
3 years ago
Read 2 more answers
When 40.5 g of Al and 212.7 g of Cl2 combine in the reaction: 2 Al (s) + 3 Cl2 (g) --&gt; 2 AlCl3 (s) c) How many moles of the e
sineoko [7]

Answer:

The answer to your question is 1.49 mol of Cl₂  

Explanation:

Data

mass of Al = 40.5 g

mass of Cl₂ = 212.7 g

moles of excess reactant = ?

- Balanced chemical reaction

                2 Al(s)  +  3Cl₂(g)  ⇒   2AlCl₃

Process

a) Calculate the molar mass of the reactants  

molar mass of Al = 2 x 26.98 = 53.96 g

molar mass of Cl₂ = 6 x 35.45 = 212.7 g

b) Calculate the theoretical proportion  Al/Cl₂ = 53.96/212.7 = 0.254

   Calculate the experimental proportion Al/Cl₂ = 40.5/212.7 = 0.19

As the experimental proportion is lower than the theoretical proportion we conclude that the limiting reactant is Aluminum.

c) Calculate the grams of excess reactant

                    53.96 g of Al ------------------ 212.7 g of Cl₂

                     40.5 g of Al -------------------  x

                      x = (40.5 x 212.7) / 53.96

                      x = 8614.35 / 53.96

                      x = 159.64 g of Cl₂

Excess Cl₂ = 212.7 - 159.64

                  = 53.057 g

d) Calculate the moles of Cl

                       35.45 g of Cl ----------------- 1 mol

                       53.057 g of Cl ---------------  x

                        x = (53.057 x 1)/35.45

                       x = 1.49 mol of Cl₂                    

6 0
3 years ago
Read 2 more answers
Consider the reaction 4PH3(g) → P4(g) + 6H2(g) At a particular point during the reaction, molecular hydrogen is being formed at
Svetllana [295]

Answer:

The rate at which P_4 is being produced is 0.0228 M/s.

The rate at which PH_3 is being consumed is 0.0912 M/s.

Explanation:

4PH_3\rightarrow P_4(g)+6H_2(g)

Rate of the reaction : R

R=\frac{-1}{4}\frac{d[PH_3]}{dt}=\frac{1}{6}\frac{d[H_2]}{dt}=\frac{1}{1}\frac{d[P_4]}{dt}

The rate at which hydrogen is being formed = \frac{d[H_2]}{dt}=0.137 M/s

R=\frac{1}{6}\frac{d[H_2]}{dt}

R=\frac{1}{6}\times 0.137 M/s=0.0228 M/s

The rate at which P_4 is being produced:

R=\frac{1}{1}\frac{d[P_4]}{dt}

0.0228 M/s=\frac{1}{1}\frac{d[P_4]}{dt}

The rate at which PH_3 is being consumed :

R=\frac{-1}{4}\frac{d[PH_3]}{dt}

0.0228 M/s\times 4=\frac{-1}{1}\frac{d[PH_3]}{dt}

\frac{-1}{1}\frac{d[PH_3]}{dt}=0.912 M/s

6 0
3 years ago
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