Answer:
Yes
Explanation:
Calcium Chloride is an inorganic compound, a salt that can dissolve in water.
The scientist that performed the cathode ray experiment leading to the discovery of electrons is J.J. Thomson.
<h3>What is a cathode ray?</h3>
A cathode ray is a tube that contains negatively charged electrode( that is the cathode) which emits electrons when heated at a low pressure.
The cathode ray was used by the scientist, J.J. Thomson to find the ratio of charge to mass (e/m) of the electrons.
Therefore, the scientist that performed the cathode ray experiment leading to the discovery of electrons is J.J. Thomson.
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he required empirical formula based on the data provided is Na2CO3.H2O.
<h3>What is empirical formula?</h3>
The term empirical formula refers to the formula of a compound which shows the ratio of each specie present.
We have the following;
Mass of sodium = 37.07-g
Mass of carbonate = 48.39 g
Mass of water = 14.54-g
Number of moles of sodium = 37.07-g/23 g/mol = 2 moles
Number of moles of carbonate = 48.39 g/61 g/mol = 1 mole
Number of moles of water = 14.54/18 g/mol = 1 mole
The mole ratio is 2 : 1: 1
Hence, the required empirical formula is Na2CO3.H2O
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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
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