Answer: The average valence electron energy (AVEE) of this element =
1014.2 KJ/ mol or 1.0142mJ/mol.
Explanation:
The average valence electron energy = (number of electrons in s subshell x Ionization energy of that subshell) + (number of electrons in p subshell x Ionization energy of that subshell) / total number of electrons in both subshells of the valence shells.
The 5A elements are non-metals like Nitrogen and Phosphorus with the metallic character increasing as you go down the group, So a new 5A element will have characteristics of its group with 5 valence electron in its outermost shell represented as ns2 np3
Therefore the average valence electron energy (AVEE) of this element will be calculated as
The average valence electron energy = (2 x 1370 kJ/mol + 3 x 777 kJ/mol.) / 5
2740+2331/ 5 =5071/5
=1014.2 KJ/ mol or 1.0142mJ/mol.
Answer:
0.12M
Explanation:
A balanced equation for the reaction will go a great deal in obtaining our desired result. So, let us write a balanced equation for the reaction
HCl + NaOH —> NaCl + H2O
From the above equation,
nA (mole of the acid) = 1
nB (mole of the base) = 1
Data obtained from the question include:
Vb (volume of the base) = 30mL
Mb (Molarity of the base) = 0.1M
Va (volume of the acid) = 25mL
Ma (Molarity of the acid) =?
The molarity of the acid can be obtained as follow:
MaVa/MbVb = nA/nB
Ma x 25/ 0.1 x 30 = 1
Cross multiply to express in linear form
Ma x 25 = 0.1 x 30
Divide both side by 25
Ma = (0.1 x 30) / 25
Ma = 0.12M
The molarity of the acid is 0.12M
Answer:
I think the answer is forest
Maple syrup would be categorized under solutions. As it contains sugar which is very soluble in water, it dissolves completely in water depending on its solubility at that temperature making it impossible for the components to be filtered out. A solution is a homogeneous mixture.
Answer:
No, there is no evidence that the manufacturer has a problem with underfilled or overfilled bottles, due that according our results we cannot reject the null hypothesis.
Explanation:
according to this exercise we have the following:
σ^2 =< 0.01 (null hypothesis)
σ^2 > 0.01 (alternative hypothesis)
To solve we can use the chi-square statistical test. To reject or not the hypothesis, we have that the rejection region X^2 > 30.14
Thus:
X^2 = ((n-1) * s^2)/σ^2 = ((20-1)*0.0153)/0.01 = 29.1
Since 29.1 < 30.14, we cannot reject the null hypothesis.
