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LenaWriter [7]
3 years ago
11

xidation in redox reactions involves a. a loss of electrons and an increase in the oxidation number b. a gain of electrons and a

n increase in the oxidation number c. a loss of electrons and a reduction in the oxidation number d. a gain of electrons and a decrease in the oxidation number
Chemistry
1 answer:
Strike441 [17]3 years ago
4 0

Answer:

Option a is the right one

Explanation:

Redox reactions are defined as the reactions where one element is oxidized (so the oxidation state is increased); and another element is reduced (oxidation state decreases). These changes in the oxidations states are defined by a transference of electrons.

When the oxidation state decrease → reduction → the element gains electrons

When the oxidation state increase → oxidation → the element release electrons

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What is one source of CO?
Murljashka [212]

Answer:

Carbon monoxide (CO)—a colorless, odorless, tasteless, and toxic air pollutant—is produced in the incomplete combustion of carbon-containing fuels, such as gasoline, natural gas, oil, coal, and wood. The largest anthropogenic source of CO in the United States is vehicle emissions.

Explanation:

Hope this helps- good Luck! ^w

5 0
3 years ago
A sample of a pure compound that weighs 60.3 g contains 20.7 g Sb (antimony) and 39.6 g F (fluorine). What is the percent compos
Volgvan

Answer:

The percent composition of fluorine is 65.67%

Explanation:

Percent Composition is a measure of the amount of mass an element occupies in a compound. It is measured in percentage of mass.

That is, the percentage composition is the percentage by mass of each of the elements present in a compound.

The calculation of the percentage composition of an element is made by:

percent composition element A=\frac{total mass of element A}{mass of compound} *100

In this case, the percent composition of fluorine is:

percent composition of fluorine=\frac{39.6 g}{60.3 g} *100

percent composition of fluorine= 65.67%

<u><em>The percent composition of fluorine is 65.67%</em></u>

4 0
3 years ago
With the Haber process, how many grams of hydrogen will be needed to make 0.3820 pounds of ammonia? Needed to 3 decimal places f
masya89 [10]

30.576 g of hydrogen

Explanation:

First we need to covert the pound in grams.

if        1 pound is equal to 453.592 grams

then  0.3820 pounds is equal to X grams

X = (0.3820 ×  453.592) / 1 = 173.272 grams

So we have 173.272 grams of ammonia.

Now we look at the chemical reaction where hydrogen (H₂) reacts with nitrogen (N₂) to produce ammonia (NH₃):

3 H₂ + N₂ → 2 NH₃

number of moles = mass / molar weight

number of moles of ammonia = 173.272 / 17 = 10.192 moles

Now taking in account the chemical reaction, we formulate the following reasoning:

if         3 moles of hydrogen produce 2 moles of ammonia

then    Y moles of hydrogen produce 10.192 moles of ammonia

Y = (3 × 10.192) / 2 = 15.288 moles of hydrogen

mass = number of moles × molar weight

mass of hydrogen = 15.288 × 2 = 30.576 g

Learn more about:

problems with number of moles

brainly.com/question/13950919

brainly.com/question/13947602

#learnwithBrainly

4 0
3 years ago
At the heart of the stars, nuclear fusion merges _____ together in the nucleus of atoms to create new elements?
Goshia [24]

Answer:

lighter atoms especially hydrogen

Explanation:

At the heart of the stars, nuclear fusion merges lighter atoms especially hydrogen together in the nucleus of atoms to create new elements.

During nuclear fusion small atomic nuclei combines to form larger ones with the release of a large amount of energy.

The energy released provides the needed temperature for another set of hydrogen atoms to fuse. This process is in turn yields another set of helium atom which releases a lot of energy.

A chain reaction progresses which leads to the formation of new elements.

3 0
3 years ago
if i add water to 100 mL of a 0.75 M NaOH solution un til the final volume is 165mL what will the molarity of the diluted soluti
ehidna [41]
Answer is: <span>the molarity of the diluted solution 0,454 M.
</span>V₁(NaOH) = 100 mL ÷ 1000 mL/L = 0,1 L.
c₁(NaOH) = 0,75 M = 0,75 mol/L.
n₁(NaOH) = c₁(NaOH) · V₁(NaOH).
n₁(NaOH) = 0,75 mol/L · 0,1 L.
n₁(NaOH) = 0,075 mol
n₂(NaOH) = n₁(NaOH) = 0,075 mol.
V₂(NaOH) = 165 mL ÷ 1000 mL/L = 0,165 L.
c₂(NaOH) = n₂(NaOH) ÷ V₂(NaOH).
c₂(NaOH) = 0,075 mol ÷ 0,165 L.
c₂(NaOH) = 0,454 mol/L.
7 0
3 years ago
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