Answer:
The war of the currents, sometimes called battle of the currents, was a series of events surrounding the introduction of competing electric power transmission systems in the late 1880s and early 1890s.
Explanation:
Answer:
molecules Cl- = 3.01 E23 molecules
Explanation:
molecules Cl- = 0.50 mol CL- * ( 6.02 E23 molecules/ mol ) = 3.01 E 23 molecules
Argon is a suitable choice for light bulbs because it's inert. Compared to a reactive gas like oxygen, the metal filimant would burn up in a reactive enviroment, which is why a noble gas is used.
Answer is: Ksp for silver sulfide is 8.00·10⁻⁴⁸.
Reaction
of dissociation: Ag₂S(s) → 2Ag⁺(aq) + S²⁻(aq)<span>.
</span>s(Ag₂S) = s(S²⁻) = 1.26·10⁻¹⁶ M.
s(Ag⁺) = 2s(Ag₂S) = 2.52·10⁻¹⁶ M; equilibrium concentration of silver cations.
Ksp = s(Ag⁺)² · s(S²⁻).
Ksp = (2.52·10⁻¹⁶ M)² · 1.26·10⁻¹⁶ M.
Ksp = 6.35·10⁻³² M² · 1.26·10⁻¹⁶ M.
Ksp = 8.00·10⁻⁴⁸ M³.
Answer:
15.75 grams of HNO3 was used and dissolved in 2.5 liters of solvent, to make a 0.10 M solution
Explanation:
Step 1: Data given
Nitric acid = HNO3
Molar mass of H = 1.01 g/mol
Molar mass of N = 14.0 g/mol
Molar mass O = 16.0 g/mol
Number of moles nitric acid (HNO3) = 0.25 moles
Molairty = 0.10 M
Step 2: Calculate molar mass of nitric acid
Molar mass HNO3 = Molar mass H + molar mass N + molar mass (3*O)
Molar mass HNO3 = 1.01 + 14.0 + 3*16.0
Molar mass HNO3 = 63.01 g/mol
Step 3: Calculate mass of solute use
Mass HNO3 = moles HNO3 * molar mass HNO3
Mass HNO3 = 0.25 moles * 63.01 g/mol
Mass HNO3 = 15.75 grams
15.75 grams of HNO3 was used and dissolved in 2.5 liters of solvent, to make a 0.10 M solution
