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>
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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:
430 kPa
Explanation:
Use the formula PV/T = PV/T where the left side of the equation refers to the initial values and the right side refers to the final values: (170.2 kPa)(3.5 L)/(298 K) = (P)(1.35 L)/(293 K). The P represents the unknown pressure. Solving this equation gives us a value for P of 433.855 kPa, but because of significant figures (the two from 3.5 L), we round to 430 kPa.
Answer:
2C2H2 + 5O2 → 4CO2 + 2H2O
Explanation:
They are most likely formed by cold fronts!
