Answer:
phenotype,phenotype,genotype,genotype,
genotype
Explanation:
phenotype is physical appearance and genotype is just like
yy Tt
Magnesium dichloride or Magnesium chloride
Answer:

Explanation:
Scientific notation is the way of writing numbers which are either large or small. The number is written in the scientific notation when the number is between 1 and 10 and then multiplied by the power of 10. Engineering notation is the same version of the scientific notation but the number can be between 1 and 1000 and in this exponent of the ten is divisible by three.
For example,
is to be written as
in engineering notation.
The given number:
0.00000009345 can be written as 
Answer upto 4 significant digits = 
Answer: The metalloids are a group of elements in the periodic table. They are located to the right of the post-transition metals and to the left of the non-metals. Metalloids have some properties in common with metals and some in common with non-metals.
Explanation:
Answer:
a) The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.
b) 0.0035 mole
c) 0.166 M
Explanation:
Phosphoric acid is tripotic because it has 3 acidic hydrogen atom surrounding it.
The equation of the reaction is expressed as:

1 mole 3 mole
The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.
b) if 10.00 mL of a phosphoric acid solution required the addition of 17.50 mL of a 0.200 M NaOH(aq) to reach the endpoint; Then the molarity of the solution is calculated as follows

10 ml 17.50 ml
(x) M 0.200 M
Molarity = 
= 0.0035 mole
c) What was the molar concentration of phosphoric acid in the original stock solution?
By stoichiometry, converting moles of NaOH to H₃PO₄; we have
= 
= 0.00166 mole of H₃PO₄
Using the molarity equation to determine the molar concentration of phosphoric acid in the original stock solution; we have:
Molar Concentration = 
Molar Concentration = 
Molar Concentration = 0.166 M
∴ the molar concentration of phosphoric acid in the original stock solution = 0.166 M