<u>Answer:</u> The correct answer is 1.18 g.
<u>Explanation:</u>
We are given a chemical equation:

We know that at STP conditions:
22.4L of volume is occupied by 1 mole of a gas.
So, 2.21L of carbon dioxide is occupied by =
of carbon dioxide gas.
By Stoichiometry of the above reaction:
1 mole of carbon dioxide gas is produced by 1 mole of carbon
So, 0.0986 moles of carbon dioxide is produced by =
of carbon.
Now, to calculate the mass of carbon, we use the equation:

Moles of carbon = 0.0986 mol
Molar mass of carbon = 12 g/mol
Putting values in above equation, we get:

Hence, the correct answer is 1.18 g.
If you spill a chemical on you, you should call poison control or 911
Answer:

-) Oxidizing agent: 
-) Reducing agent: 
Explanation:
The first step is separate the reaction into the <u>semireactions</u>:
A.
B.
If we want to balance in <u>basic medium </u>we have to follow the rules:
1. We adjust the oxygen with 
2. We adjust the H with 
3. We adjust the charge with 
Lets balance the first semireaction A. :

Now, lets balance semireaction B:

Finally, we have to add the two semireactions:
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Cancel out the species on both sides:

Simplifying the equation :

The
is <u>oxidized</u> therefefore is the <u>reducing agent</u>. The
is<u> reduced</u> therefore is the <u>oxidizing agent</u>.
Answer : The protons and neutrons subatomic particles will be located at Z.
Explanation :
In the model of an atom, there are three subatomic particles. Protons, neutrons and electrons are the subatomic particles.
The protons and the neutrons subatomic particles are located inside the nucleus and the electrons subatomic particle are located around or outside the nucleus.
The protons are positively charged, electrons are negatively charged and neutrons are neutral that means it has no charge.
In the given dartboard, Z is the nucleus in which the protons and neutrons subatomic particles are present and x, w & y are the electrons because they are located around the nucleus.
Hence, the protons and neutrons subatomic particles will be located at Z.