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
H atoms are found in the molecules of covalent compounds that dissolve in water. These compound are called acids (proton donors). They form acidic solutionst{acidic solutions} acidic solutions.
Because there are no ions moving about in solution with covalent molecules, they are characterised as non-electrolytes. Non-electrolyte solutions are those that do not conduct electricity.
Examples are hydrochloric acid (HCl), sulfuric acid (H2SO4), and nitric acid solutions (HNO3).
<h3>
What type of compounds dissociate in water to form ions?</h3>
When certain substances dissolve in water, they undergo either a physical or a chemical change that results in the formation of ions in solution. These substances are members of an important class of compounds known as electrolytes. Nonelectrolytes are substances that do not produce ions when dissolved.
A substance is known as a strong electrolyte if the physical or chemical process that generates the ions is essentially 100% efficient (all of the dissolved compound yields ions). A weak electrolyte is one in which only a small portion of the dissolved substance undergoes the ion-producing process.
learn more about Covalent refer
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Answer:
24.32 amu
Explanation:
From the question given above, the following data were obtained:
Isotope A (Mg–24):
Mass of A = 24 amu
Abundance (A%) = 79%
Isotope B (Mg–25):
Mass of B = 25 amu
Abundance (B%) = 10%
Isotope C (Mg–26):
Mass of C = 26 amu
Abundance (C%) = 11%
Average atomic mass of Mg =?
Average atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100] + [(Mass of C × C%)/100]
= [(24 × 79)/100] + [(25 × 10)/100] + [(26 × 11)/100]
= 18.96 + 2.5 + 2.86
= 24.32 amu
Thus, the average atomic mass of Mg is 24.32 amu
