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4 years ago

A(n) accepts electrons and becomes .

Chemistry

2 answers:

eimsori [14]4 years ago

6 0

Google says it's the first one.

Sever21 [200]4 years ago

6 0

Answer:

Oxidizing agent, Reduced

Explanation:

Oxidation is the loss of electrons and reduction is the gain of electrons.

An oxidizing agent oxidizes other elements but itself gets reduced. Therefore oxidizing agents accepts electrons and gets reduced.

A reducing agent reduces other elements but itself gets oxidized. Therefore reducing agents losses electrons and gets oxidized.

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Compare how fast the rate of CO2 increase is in the last few hundred years with the rate of increase (or decrease, for that matt

Tcecarenko [31]

The rate of CO2 increase is in the last few hundred years is 10 times more with the rate of increase (or decrease, for that matter) in the preceding 400,000 years.

There are many possible reasons for this cause , some primary factors are listed below:

Increase in population
increased emission of green house gases, as we all know auto mobile industry is growing rapidly and this vehicles releases harmful gases like CO2, CO ,etc. and increases carbon % , this CO2 is a main gas component in green house effect.
Deforestation, as the amount of plant decreases the CO2 present in atmosphere increases, plants uses CO2 and sun lite to make their food via photosynthesis.
Increased emission of Industrial flue gases, etc.

Learn more about green house effect here...

brainly.com/question/19521661

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7 0

2 years ago

Calculate the mass of nickel metal which will react with 25mL of 0.15mol/L (

Fittoniya [83]

Answer:

0.110625 g of Ni

Explanation:

The first step in solving this problem is to put down the accurate chemical reaction equation.

Ni(s) + 2HCl(aq) ---> NiCl2(aq) + H2(g)

Secondly, we obtain the amount of HCl that reacted from the information provided.

Volume of HCl (V)= 25 ml

Concentration of HCl (C)= 0.15 mol/L

Then, to find the number of moles of HCl (n);

n= CV

Substitution values

n= 25/1000 × 0.15

n= 3.75 ×10^-3 moles

Mass of 3.75 ×10^-3 moles of HCl = number of moles × molar mass

Molar mass of HCl= 36.5 gmol-1

Therefore;

Mass of HCl = 3.75 ×10^-3 moles × 36.5 gmol-1

Mass of HCl= 0.136875 g of HCl

Thirdly we determine the mass of Ni reacted;

If 1 mole of Ni reacted with 2 moles of HCl according to the reaction equation

Then x moles of Ni reacts with 3.75 ×10^-3 moles of HCl

x= 1 × 3.75 ×10^-3 moles/ 2

x= 1.875 × 10^-3 moles of Ni

Mass of Ni= 1.875 × 10^-3 moles of Ni × 59 gmol-1

Mass of Ni= 0.110625 g of Ni

6 0

4 years ago

Why is Anton Von Leeuwenhoek important in the cell theory?

vesna_86 [32]

Answer:

He was the first scientist to observe and describe bacteria and protozoa by looking at a drop of water from a pound under a microscope. He also was the one to build the first compound microscope.

Hope this helps :)

4 0

3 years ago

Sterling silver contains silver and copper metals. if a sterling silver chain contains 17.6g of silver and 2.40g of copper, what

Bogdan [553]

Answer:- 0.88

Solution:- Masses of silver and copper metals are given and we are asked to calculate the percentage of silver in the alloy.

mass percent of Ag = $(\frac{mass of Ag}{mass of alloy})*100$

Mass of Ag = 17.6 g

mass of Cu = 2.40 g

mass of alloy = 17.6 g + 2.40 g = 20.0 g

Let's plug in the values in the formula:

mass percentage of Ag = $(\frac{17.6}{20.0})*100$

mass percentage of Ag = 88%

So, the mass percentage of silver in sterling silver is 88% and in decimal form it is 0.88.

7 0

3 years ago

. Write the two resonance hybrids for the carbocation that would be formed by protonation at C-1 of 2-methyl-1,3-pentadiene. Wit

never [62]

Answer:

Answer is explained in the explanation section below.

Explanation:

This question is requires the diagram, which I have drawn and it is attached in the attachment below. Please for this answer, refer to the diagram attached in the attachment below.

Referring to the diagram, attached. As we know that, Allylic secondary carbocations are more stable than Allylic tertiary carbocations.

Hence,

C2 will have a more positive charge since a tertiary carbocation (C2) is more stable than a secondary carbocation (C4). Therefore, the resonance structure will favor the positive charge at C2.

8 0

3 years ago

A(n) ____ accepts electrons and becomes ____.

A(n) accepts electrons and becomes .