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Fantom [35]
3 years ago
6

g Oxidation refers to ________ Select one: a. an increase in oxidation number. b. a decrease in oxidation number. c. an increase

in mass. d. an increase in the atomic number. e. a gain in the number of protons.
Chemistry
1 answer:
mihalych1998 [28]3 years ago
4 0

Answer:

The correct alternative is "Option a".

Explanation:

Oxidation has become a mechanism whereby the physicochemical properties transform attributed to the formation of O₂.

  • The connection involving magnesium as well as O₂ requires the oxidation of the component named magnesium.
  • Even before exposed to the air, silicon is oxidized as well as generates silicon dioxide.

Other possibilities are not connected to the scenario in question. So Choice A is the best option.

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Suppose that, from measurements in a microscope, you determine that a certain layer of graphene covers an area of 2.60μm2. conve
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3 years ago
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1.12g H2 is allowed to react with 9.60 g N2, producing 1.23 g NH3.
andriy [413]

Answer:

A. m_{NH_3}^{theo} =1.50gNH_3

B. Y=82.2\%

Explanation:

Hello!

In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:

N_2+3H_2\rightarrow 2NH_3

Thus we proceed as follows:

A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:

n_{NH_3}^{by \ H_2}=1.12gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.370molNH_3\\\\  n_{NH_3}^{by \ N_2}=1.23gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.0878molNH_3

Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:

m_{NH_3}^{theo}=0.0878mol*\frac{17.04gNH_3}{1molNH_3} =1.50gNH_3

B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:

Y=\frac{1.23gNH_3}{1.50gNH_3} *100\%\\\\Y=82.2\%

Best regards!

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Answer:

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Explanation:

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