Answer:
5.0 × 10²⁴ molecules
Explanation:
Step 1: Write the balanced double displacement reaction
2 NaOH + CuSO₄ ⇒ Na₂SO₄ + Cu(OH)₂
Step 2: Calculate the moles corresponding to 5.0 × 10²⁴ molecules of Na₂SO₄
We will use Avogadro's number: there are 6.02 × 10²³ molecules in 1 mole of molecules.
5.0 × 10²⁴ molecule × 1 mol/6.02 × 10²³ molecule = 8.3 mol
Step 3: Calculate the moles of CuSO₄ required to produce 8.3 moles of Na₂SO₄
The molar ratio of CuSO₄ to Na₂SO₄ is 1:1. The moles of CuSO₄ required are 1/1 × 8.3 mol = 8.3 mol.
Step 4: Calculate the molecules corresponding to 8.3 moles of CuSO₄
We will use Avogadro's number.
8.3 mol × 6.02 × 10²³ molecule/1 mol = 5.0 × 10²⁴ molecule
The volume of the 0.15 M LiOH solution required to react with 50 mL of 0.4 M HCOOH to the equivalence point is 133.3 mL
<h3>Balanced equation </h3>
HCOOH + LiOH —> HCOOLi + H₂O
From the balanced equation above,
The mole ratio of the acid, HCOOH (nA) = 1
The mole ratio of the base, LiOH (nB) = 1
<h3>How to determine the volume of LiOH </h3>
- Molarity of acid, HCOOH (Ma) = 0.4 M
- Volume of acid, HCOOH (Va) = 50 mL
- Molarity of base, LiOH (Mb) = 0.15 M
- Volume of base, LiOH (Vb) =?
MaVa / MbVb = nA / nB
(0.4 × 50) / (0.15 × Vb) = 1
20 / (0.15 × Vb) = 1
Cross multiply
0.15 × Vb = 20
Divide both side by 0.15
Vb = 20 / 0.15
Vb = 133.3 mL
Thus, the volume of the LiOH solution needed is 133.3 mL
Learn more about titration:
brainly.com/question/14356286
Answer:
The veins that carry oxygenated bloof back into the heart are the pulmonary arteries.
Explanation:
Oxygen-rich blood flows from the lungs back into the left atrium (LA), or the left upper chamber of the heart, through four pulmonary veins. Oxygen-rich blood then flows through the mitral valve (MV) into the left ventricle (LV), or the left lower chamber.
The question is missing the graphics required to answer which I have attached as an image.
There are four different representations of the orientation of water molecules around chloride anion. Let's first analyze the water molecule.
We have H-O-H as the structure of water. The oxygen atom is more electronegative than the hydrogen atoms, which results in a partial positive charge on the hydrogen atoms and a partial negative charge on the oxygen atom.
The chloride anion is a negative charge. Therefore, the water molecules should orient themselves with the hydrogen atoms facing the chlorine atom as the partial positive charge on the hydrogen atoms will be attracted to the negative charge of the chlorine atom.
The correct representation is shown in graph 3 which shows all hydrogen atoms facing the chlorine anion.