Answer : The correct option is, 
Explanation :
Amino acid : The acid that contains two functional groups that are carboxylic group,
and ammine group,
.
When the two or more that two amino acids join together with the help of peptide bond, they produces polypeptide chain or protein.
The bond present between the two amino acid is called a peptide bond.
The peptide bond is a chemical bond that is formed between the two molecules when the nitrogen of one amino acid react with the carbon of another amino acid by releasing a water molecule. This is a dehydration synthesis or condensation reaction.
From this we conclude that, only two functional groups carboxylic group,
and ammine group,
are present in all amino acids.
Hence, the correct option is, 
The given reaction is:
3Fe + 4H2O → Fe3O4 + 4H2
Given:
Mass of Fe = 354 g
Mass of H2O = 839 g
Calculation:
Step 1 : Find the limiting reagent
Molar mass of Fe = 56 g/mol
Molar mass of H2O = 18 g/mol
# moles of Fe = mass of Fe/molar mass Fe = 354/56 = 6.321 moles
# moles of H2O = mass of h2O/molar mass of H2O = 839/18 = 46.611 moles
Since moles of Fe is less than H2O; Fe is the limiting reagent.
Step 2: Calculate moles of Fe3O4 formed
As per reaction stoichiometry:
3 moles of Fe form 1 mole of Fe3O4
Therefore, 6.321 moles of Fe = 6.321 * 1/ 3 = 2.107 moles of Fe3O4
Step 4: calculate the mass of Fe3O4 formed
Molar mass of Fe3O4 = 232 g/mol
# moles = 2.107 moles
Mass of Fe3O4 = moles * molar mass
= 2.107 moles * 232 g/mol = 488.8 g (489 g approx)
Answer: a) pH = 13.00 : basic
b)
: basic
c) pOH = 5.00 : basic
d)
: acidic
Explanation:
pH or pOH is the measure of acidity or alkalinity of a solution.
pH is calculated by taking negative logarithm of hydrogen ion concentration.
![pH=-\log [H_3O^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH_3O%5E%2B%5D)
Acids have pH ranging from 1 to 6.9 and bases have pH ranging from 7.1 to 14.Neutral substances have pH of 7.
a) pH = 13.00
As pH is more than 7, the solution is basic.
b) ![[H_3O^+]=1.0\times 10^{-12}](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D1.0%5Ctimes%2010%5E%7B-12%7D)
Putting in the values:
![pH=-\log[1.0\times 10^{-12}]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5B1.0%5Ctimes%2010%5E%7B-12%7D%5D)

As pH is more than 7, the solution is basic.
c) pOH = 5.00


As pH is more than 7, the solution is basic.
d) ![[OH^-]=1.0\times 10^{-9}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.0%5Ctimes%2010%5E%7B-9%7D)
Putting in the values:
![pOH=-\log[1.0\times 10^{-9}]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5B1.0%5Ctimes%2010%5E%7B-9%7D%5D)



As pH is less than 7, the solution is acidic.
<em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em>
<em>Sol</em><em>ution</em><em>,</em>
<em>Mass</em><em>=</em><em>5</em><em>0</em><em>0</em><em>g</em><em> </em><em>=</em><em>5</em><em>0</em><em>0</em><em>/</em><em>1000</em><em>=</em><em>0</em><em>.</em><em>5</em><em> </em><em>kg</em>
<em>gravity</em><em>(</em><em>g</em><em>)</em><em>=</em><em>9</em><em>.</em><em>8</em><em>m</em><em>/</em><em>s^</em><em>2</em>
<em>Now</em><em>,</em>
<em>Weight</em><em>=</em><em> </em><em>m</em><em>*</em><em>g</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>0</em><em>.</em><em>5</em><em>*</em><em>9</em><em>.</em><em>8</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>4</em><em>.</em><em>9</em><em> </em><em>N</em>
<em>So</em><em> </em><em>the</em><em> </em><em>weight</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>block</em><em> </em><em>is</em><em> </em><em>4</em><em>.</em><em>9</em><em> </em><em>Newton</em>
<em>hope</em><em> </em><em>it</em><em> </em><em>helps</em><em>.</em><em>.</em><em>.</em>
<em>Good</em><em> </em><em>luck</em><em> on</em><em> your</em><em> assignment</em>
<em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em>