Answer:
pH = 2.69
Explanation:
The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>
<em />
The reaction of HNO₂ with KOH is:
HNO₂ + KOH → NO₂⁻ + H₂O + K⁺
Moles of HNO₂ and KOH that react are:
HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>
KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>
That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:
NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻
HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂
It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:
pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]
pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]
<h3>pH = 2.69</h3>
Answer:
moles of water in 
of water.
Explanation:
Mass of water = 
Molar mass of water = M = 18 g/mol
Moles = n = 
So, there are moles of water in of water.
Answer:
483 nm corresponds to blue light hence the complex will appear orange.
Explanation:
Using the formula;
E= hc/λ
Where;
E = energy of the photon
h = Plank's constant (6.6*10^-34Js)
c = Speed of light (3*10^8 ms-1)
λ = wavelength
λ = hc/E
λ = 6.6*10^-34 * 3*10^8/4.10×10^−19
λ = 4.83 * 10^-7 or 483 nm
483 nm corresponds to blue light
Using the colour wheel approach, if a complex absorbs blue light, then it will appear orange.
I think it would be solubility but I’m not sure
Answer:
<h3>5.06282 × 10²⁴ molecules</h3>
Explanation:
The number of molecules of Ca2(SO3) can be found by using the formula
<h3>N = n × L</h3>
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question we have
N = 8.41 × 6.02 × 10²³
We have the final answer as
<h3>5.06282 × 10²⁴ molecules</h3>
Hope this helps you
