The mass of 2.15 mol of hydrogen sulphide (H₂S) will be 73.272 gm and the mass of 3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) will be 1.82 gm
<h3>
What is Mole ?</h3>
A mole is a very important unit of measurement that chemists use.
A mole of something means you have 6.023 x 10 ²³ of that thing.
- For 2.15 mol of hydrogen sulphide (H₂S) :
1 mole hydrogen sulphide (H₂S) = 34.08088 grams
Therefore,
2.15 mol of hydrogen sulphide (H₂S) = 34.08088 grams x 2.15 mol
= 73.272 gm
- For 3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) ;
1 mol of lead(II) iodide, (PbI₂) = 461.00894 grams
Therefore,
3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) = 461.00894 grams x 3.95 × 10⁻³ mol
= 1.82 gm
Hence,The mass of 2.15 mol of hydrogen sulphide (H₂S) will be 73.272 gm and the mass of 3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) will be 1.82 gm
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<h2>Question:</h2>
A precipitate is a solid that sometimes forms when two liquids combine.
<h2>Answer:</h2>
<u>A</u><u>.</u><u> </u><u>True</u><u> </u>
<h2>Explanation:</h2>
- <u>Because</u><u> </u><u>the</u><u> </u><u>Precipitate</u><u> </u><u>it's</u><u> </u><u>forms</u><u> </u><u>solid</u><u> </u><u>when</u><u> </u><u>two</u><u> </u><u>liquids</u><u> </u><u>combine</u><u> </u><u>to</u><u> </u><u>precipitate</u><u>.</u><u> </u>
<h2><u>#CARRYONLEARNING</u><u> </u></h2><h2><u>#STUDYWELL</u><u> </u></h2>
It is harder to remove an electron from fluorine than from carbon because the size of the nuclear charge in fluorine is larger than that of carbon.
The energy required to remove an electron from an atom is called ionization energy.
The ionization energy largely depends on the size of the nuclear charge. The larger the size of the nuclear charge, the higher the ionization energy because it will be more difficult to remove an electron from the atom owing to increased electrostatic attraction between the nucleus and orbital electrons.
Since fluorine has a higher size of the nuclear charge than carbon. More energy is required to remove an electron from fluorine than from carbon leading to the observation that; it is harder to remove an electron from fluorine than from carbon.
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Answer:
E) NaF and SrO
Explanation:
The ionic bonding occurs between atoms with a great difference in electronegativity. This usually happens between a metal and a non-metal.
<em>In which pair do both compounds exhibit predominantly ionic bonding? </em>
A) KCl and CO₂. NO. C and O are non-metals and present covalent bonding.
B) SO₂ and BaF₂. NO. S and O are non-metals and present covalent bonding.
C) F₂ and N₂O. NO. Both compounds contain non-metals and present covalent bonding.
D) N₂O₃ and Rb₂O. NO. N and O are non-metals and present covalent bonding.
E) NaF and SrO. YES. Na and Sr are metals while F and O are non-metals.