Answer is: A. 1.1 3 1023 NiCl2 formula units.
m(NiCl₂) = 24.6 g; mass of nickel(II) chloride.
M(NiCl₂) = 129.6 g/mol; molar mass of nickel(II) chloride.
n(NiCl₂) = m(NiCl₂) ÷ M(NiCl₂).
n(NiCl₂) = 24.6 g ÷ 129.6 g/mol.
n(NiCl₂) = 0.19 mol; amount of nickel(II) chloride.
Na = 6.022·10²³ 1/mol; Avogadro constant.
N(NiCl₂) = n(NiCl₂) · Na.
N(NiCl₂) = 0.19 mol · 6.022·10²³ 1/mol.
N(NiCl₂) = 1.13·10²³; number of formula units.
Answer:
SO2(g) + O2(g) SO3(g) (needs to be balanced. Balance it by placing a 1/2 in front of the 02.) This is a 5-sig-fig problem, so when you calculate your molar masses, you must use all of the sig figs available to you from the periodic table. That's how you can get this problem correct. Melissa Maribel likes to round the numbers from the periodic table, and usually that is ok. But for problems where you have many sig figs, your molar masses from the periodic table must have at least as many sig figs as your data. Therefore, for each oxygen atom, please use 15.9994 g/mol. For each sulfur atom, please use 32.066 g/mol. Thank you.
Answer:
Explanation:
Given parameters:
Hydroxide ion concentration = 5.5 x 10⁻¹⁰M
Unknown:
concentration of hydroxonium ion [H₃O⁺]= ?
pH = ?
Solution
To determine the concentration of the hydroxonium ion, we know that from the ionic product of water:
[H₃O⁺][OH⁻]= 10⁻¹⁴
[H₃O⁺] = 
[H₃O⁺] = 
[H₃O⁺] = 1.8 x 10⁻⁵M
To find the pH of the solution, we use the expression below:
pH = -log₁₀[H₃O⁺]
pH = -log₁₀ 1.8 x 10⁻⁵ = -(-4.74) = 4.7
Since the pH of the solution is 4.7, it is acidic.
She should have removed her contact lenses when rinsing.
