Answer:
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Explanation:
Answer : The initial quantity of sodium metal used is 17.25 grams.
Solution : Given,
Volume of 
gas = 8.40 L
Molar mass of Na metal = 23 g/mole
The Net balanced chemical reaction is,
At STP, 22.4 L of volume is occupied by 1 mole of gas
so, 8.40 L of volume is occupied by = 
= 0.375 moles of gas
Now from the above reaction, we conclude that
1 mole of gas produced by the 2 moles of Na metal
0.375 moles of gas produced = 
of Na metal
The quantity of Na metal used = Moles of Na metal × Molar mass of Na metal = 0.75 moles × 23 g/mole = 17.25 grams
Therefore, the initial quantity of sodium metal used is 17.25 grams.
Answer:
0.246 kg
Explanation:
There is some info missing. I think this is the original question.
<em>A chemist adds 370.0mL of a 2.25 M iron(III) bromide (FeBr₃) solution to a reaction flask. Calculate the mass in kilograms of iron(III) bromide the chemist has added to the flask. Be sure your answer has the correct number of significant digits.</em>
<em />
We have 370.0 mL of 2.25 M iron(III) bromide (FeBr₃) solution. The moles of FeBr₃ are:
0.3700 L × 2.25 mol/L = 0.833 mol
The molar mass of iron(III) bromide is 295.56 g/mol. The mass corresponding to 0.833 moles is:
0.833 mol × 295.56 g/mol = 246 g
1 kilogram is equal to 1000 grams. Then,
246 g × (1 kg/1000 g) = 0.246 kg
Answer:
liquid vibrate, move about, and slide past each other. solid vibrate (jiggle) but generally do not move from place to place.
Explanation:
