The atom positions in a general molecule of formula (not shape class) AXn that has shape square pyramidal at the corers of square and one at the above center of the square.
<h3>What is square pyramidal?</h3>
The square pyramidal is a shape geometry of the hybridization in which it consists of one lone pair and 5 bond pairs of electrons that repel each other and due to which the geometry changes from octahedral to square pyramidal.
As atoms are located at the four corners of the planer and one atom at the above center of the planner which is repelled by 4 atoms present at the corner of the planer.
Therefore, the atom positions in a general molecule of formula (not shape class) AXn that has a shape square pyramidal at the corners of the square and one at the above center of the square.
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Answer:
209.98 g of NaOH
Explanation:
We are given;
- Volume of HCl as 3 L
- Molarity of HCl as 1.75 M
We are required to calculate the mass of NaOH required to completely neutralize the acid given.
First, we write a balanced equation for the reaction between NaOH and HCl
That is;
NaOH + HCl → NaCl + H₂O
Second, we determine the number of moles of HCl
Number of moles = Molarity × Volume
= 1.75 M × 3 L
= 5.25 moles
Third, we use the mole ratio to determine the moles of NaOH
From the reaction,
1 mole of NaOH reacts with 1 mole of HCl
Therefore;
Moles of NaOH = Moles of HCl
= 5.25 moles
Fourth, we determine the mass of NaOH
Molar mass of NaOH = 39.997 g/mol
Mass of NaOH = 5.25 moles × 39.997 g/mol
= 209.98 g
Thus, 209.98 g of NaOH will completely neutralize 3L of 1.74 M HCl
Newton's third law of motion states that for every action force, there is an equal and opposite reaction force so that means that the wall is pushing you with the same amount of force that you put on it.
Answer:
The entropy decreases.
Explanation:
The change in the standard entropy of a reaction (ΔS°rxn) is related to the change in the number of gaseous moles (Δngas), where
Δngas = n(gaseous products) - n(gaseous reactants)
- If Δngas > 0, the entropy increases
- If Δngas < 0, the entropy decreases.
- If Δngas = 0, there is little or no change in the entropy.
Let's consider the following reaction.
2 H₂(g) + O₂(g) ⟶ 2 H₂O(l)
Δngas = 0 - 3 = -3, so the entropy decreases.
