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:
Aluminum Gallium Reaction- Gallium reacts with Aluminum to create an Aluminum alloy that crumbles at the touch. Gallium attacks many metals including Aluminum & Steel by diffusing into the grain boundaries making them extremely brittle. Gallium easily alloys with many metals in low quantities
Explanation:
I believe the answer is C. an unstable nucleus emits radiation and is transformed into the nucleus of one or more elements. i hope this helps!
Answer:
6.02 x 1023 number of atoms in one mole.
Explanation:
i cant show work because i dont want to be loud in 3 in the moring
Cm^3 = mL
1.11 g/cm^3 = 1.11 g/mL
Density (g/mL) multiplied by volume (mL) will give us the mass (g)
1.11 g/mL * 387 mL = 429.57 g
And since we have 3 significant figures, we have 430. g.
