The oxidation state of the compound Mn (ClO4)3 is to be determined in this problem. For oxygen, the charge is 2-, the total considering its number of atoms is -24. Mn has a charge of +3. TO compute for Mn, we must achieve zero charge overall hence 3+3x-24=0 where x is the Cl charge. Cl charge, x is +7.
The answer is 42 grams of NaF are there in 1 mole.
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What is a mole ?</h3>
A mole is defined as 6.022 × 10²³ atoms, molecules, ions, or other chemical units.
and the molar mass of a substance is defined as the mass of 1 mole of that substance, expressed in grams per mole.
It is equal to the mass of 6.022 × 10²³ atoms, molecules, or formula units of that substance.
Molar Mass , i.e. mass of 1 mole of NaF is sum of molar mass of Na and F
23 + 19
42 grams
Therefore 42 grams of NaF Sodium-fluoride are there in 1 mole.
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Answer:
The answer to your question is 1.1 moles of water
Explanation:
2Al(OH)₃ + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 6H₂O
0.45 mol 0.55 mol ?
Process
1.- Calculate the limiting reactant
Theoretical proportion
Al(OH)₃ / H₂SO₄ = 2/3 = 0.667
Experimental proportion
Al(OH)₃ / H₂SO₄ = 0.45 / 0.55 = 0.81
From the proportions, we conclude that the limiting reactant is H₂SO₄
2.- Calculate the moles of H₂O
3 moles of H₂SO₄ ---------------- 6 moles of water
0.55 moles of H₂SO₄ ----------- x
x = (0.55 x 6) / 3
x = 3.3 / 3
x = 1.1 moles of water
Sodium, hydrogen, carbon, and oxygen