[Ar] 3d10 4s2 4p5 is the electron configuration of bromine
Answer:
US₂
Explanation:
Uranium sulfide (US₂)
Uranium atomic symbol = U
Sulfur atomic symbol = S
Uranium valency = +4
Sulfur valency = -2
So;
Uranium sulfide (US₂)
Answer: They are close to each other by 41.03 m^3
Explanation:
From Ideal gas equation, PV = nRT
Where n is negligible
R is gas constant = 8.314 J/mol.k
T = 30 + 273 = 303K
P = 1.02 * 103351.5 = 103351.5 Pascal
Then;
PV = RT
V = P/RT
V = 103351.5/(8.314*303)
V = 41.03m^3
32 electrons. as the orbitals get father away from the nucleus, they hold more electrons.
Answer:
a. 3; b. 5; c. 10; d. 12
Explanation:
pH is defined as the negative log of the hydronium concentration:
pH = -log[H₃O⁺] (hydronium concentration)
For problems a. and b., HCl and HNO₃ are strong acids. This means that all of the HCl and HNO₃ would ionize, producing hydronium (H₃O⁺) and the conjugate bases Cl⁻ and NO₃⁻ respectively. Further, since all of the strong acid ionizes, 1 x 10⁻³ M H₃O⁺ would be produced for a., and 1.0 x 10⁻⁵ M H₃O⁺ for b. Plugging in your calculator -log[1 x 10⁻³] and -log[1.0 x 10⁻⁵] would equal 3 and 5, respectively.
For problems c. and d. we are given a strong base rather than acid. In this case, we can calculate the pOH:
pOH = -log[OH⁻] (hydroxide concentration)
Strong bases similarly ionize to completion, producing [OH⁻] in the process; 1 x 10⁻⁴ M OH⁻ will be produced for c., and 1.0 x 10⁻² M OH⁻ produced for d. Taking the negative log of the hydroxide concentrations would yield a pOH of 4 for c. and a pOH of 2 for d.
Finally, to find the pH of c. and d., we can take the pOH and subtract it from 14, giving us 10 for c. and 12 for d.
(Subtracting from 14 is assuming we are at 25°C; 14, the sum of pH and pOH, changes at different temperatures.)
