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Answer:
In 33.7 grams SnF2 we have 8.17 grams of F
Explanation:
Step 1: Data given
Mass of SnF2 = 33.7 grams
Molar mass of SnF2 = 156.69 g/mol
Molar mass of F = 19.00 g/mol
Step 2: Calculate moles of SnF2
Moles SnF2 = mass / molar mass
Moles SnF2 = 33.7 grams / 156.69 g/mol
Moles SnF2 = 0.215 moles
Step 3: Calculate moles F
For 1 mol SnF2 we have 2 moles F
For 0.215 moles SnF2 we have 2*0.215 = 0.430 moles F
Step 4: Calculate mass F
Mass F = moles F * molar mass F
Mass F = 0.430 moles * 19.00 g/mol
Mass F = 8.17 grams
In 33.7 grams SnF2 we have 8.17 grams of F
A possible cause of a large percentage of error in an
experiment where MgO is produced from the combustion of magnesium would be not all of the Mg has
completely reacted. <span>
I hope this helps and if you have any further questions, please don’t hesitate
to ask again. </span>
Your answer is B: The outer electron of Adam B has moved to a higher energy state
Answer:
oxidation- reduction
Explanation:
where gaining electronic reduces one element and losing them oxidize the other nitric acid is not only strong it is also a oxidizing agent
<h2>Oxidize: copper = Cu+2</h2>
