Answer: A. Cilla Is Correct.
Answer: The methodology is the science that addresses these issues.
Explanation:
There are several steps to exploring to obtain valid scientific results. The scientific method is one of those elements. The scientific method involves specific steps that need to be taken to begin the scientific process adequately. By the scientific method, we also mean the hypothesis or assumption that needs to be put in the paper to try to defend it later and in fact. By scientific process, we suggest the following steps we take after evaluating a hypothesis. The scientific method is the very making of a particular scientific work, that is, its central part.
Explanation:
It is given that,
The time period of artificial satellite in a circular orbit of radius R is T. The relation between the time period and the radius is given by :
The radius of the orbit in which time period is 8T is R'. So, the relation is given by :
So, the radius of the orbit in which time period is 8T is 4R. Hence, this is the required solution.
The volume of the 0.279 M Ca(OH)₂ solution required to neutralize 24.5 mL of 0.390 M H₃PO₄ is 51.4 mL
<h3>Balanced equation </h3>
2H₃PO₄ + 3Ca(OH)₂ —> Ca₃(PO₄)₂ + 6H₂O
From the balanced equation above,
- The mole ratio of the acid, H₃PO₄ (nA) = 2
- The mole ratio of the base, Ca(OH)₂ (nB) = 3
<h3>How to determine the volume of Ca(OH)₂ </h3>
- Molarity of acid, H₃PO₄ (Ma) = 0.390 M
- Volume of acid, H₃PO₄ (Va) = 24.5 mL
- Molarity of base, Ca(OH)₂ (Mb) = 0.279 M
- Volume of base, Ca(OH)₂ (Vb) =?
MaVa / MbVb = nA / nB
(0.39 × 24.5) / (0.279 × Vb) = 2/3
9.555 / (0.279 × Vb) = 2/3
Cross multiply
2 × 0.279 × Vb = 9.555 × 3
0.558 × Vb = 28.665
Divide both side by 0.558
Vb = 28.665 / 0.558
Vb = 51.4 mL
Thus, the volume of the Ca(OH)₂ solution needed is 51.4 mL
Learn more about titration:
brainly.com/question/14356286
Answer:
1 . 
2. 
Explanation:
The more stable the ionic compound, the more is it lattice energy.
- The more the charge on the cation and the anion, the greater is the lattice energy.
- The less the size of the cation and the anion, the greater is the lattice energy.
Scandium oxide (
) is an oxide in which 
behaves as cation and 
behaves as anion.
The compounds which has higher lattice energy than scandium oxide are:
1 .
This is because the charge are same on the cation and the anion as in the case of the Scandium oxide but the size of the cation 
is smaller than . Thus, this corresponds to higher lattice energy.
2.
This is because the charge on the cation 
is greater than that of and also the size of the cation is smaller than . Thus, this corresponds to higher lattice energy.
