Answer:
In the presence of salt water and oxygen the coating will not corrode. As long as zinc coating is present and remains intact corrosion is not possible.
Explanation:
Here it is given that a tin is present so firstly tin is made of a chemical element
which belongs to carbon family in periodic table of group 14.
It is a silvery,soft, white metal with a bluish tinge.
Now the covering which is been done on the tin is Zinc.
so, zinc is known to be served as a sacrificial coater.
Their is an amazing reason behind zinc coating being so effective and intact i.e. Its own corrosive properties are such that it stops corrosion.
Their is a process which is known as a galvanic corrosion which refers to that "ZINC" defers to the metal to which it is protecting.
It is even more electrochemically active than iron itself.
Here, it is mentioned that zinc coating gets chipped but the coating remains intact. So, if the zinc is not removed from the tin's surface it will not get corroded when it is exposed to salt water and oxygen.
Answer:
pH 8.89
Explanation:
English Translation
If the MgCl₂ solution of 0.2 M has its pH raised by adding NH₄OH, the precipitate will begin to form at a pH of approximately.
Given the solubility product (Ksp) of Mg(OH)₂ = 1.2 x 10⁻¹¹
Assuming all of the salts involved all ionize completely
MgCl₂ ionizes to give Mg²⁺ and Cl⁻
MgCl₂ ⇌ Mg²⁺ + 2Cl⁻
1 mole of MgCl₂ gives 1 moles of Mg²⁺
Since the concentration of Mg²⁺ is the same as that of MgCl₂ = 0.2 M
Mg(OH)₂ is formed from 1 stoichiometric mole of Mg²⁺ and 2 stoichiometric moles of OH⁻
Ksp Mg(OH)₂ = [Mg²⁺][OH⁻]²
(1.2 x 10⁻¹¹) = 0.2 × [OH⁻]²
[OH⁻]² = (6×10⁻¹¹)
[OH⁻] = √(6×10⁻¹¹)
[OH⁻] = 0.000007746 M
p(OH) = - log [OH⁻] = - log (0.000007746)
pOH = 5.11
pH + pOH = 14
pH = 14 - pOH = 14 - 5.11 = 8.89
Hope this Helps!!!
Answer:
The answer is Steam is produced when water is heated. The color does not change, a precipice is not formed, and the odor does not change!
Answer: 41.5 mL
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
where,
n = moles of solute
= volume of solution in L
Given : 59.4 g of 
in 100 g of solution
moles of 
Volume of solution =
Now put all the given values in the formula of molality, we get
To calculate the volume of acid, we use the equation given by neutralisation reaction:
where,
are the molarity and volume of stock acid which is
are the molarity and volume of dilute acid which is
We are given:
Putting values in above equation, we get:
Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid
