Answer:
The answer to your question is the mass of solute = 53.5 g
Explanation:
Data
mass of solution = 482 g
mass of solute = ?
mass percent = 11.1 %
Mass percent is a unit of concentration. It measures the mass of the solute divided by the total mass of the solution
Process
1.- Write the formula
Mass percent = mass of solute / mass of solution x 100
-Solve for mass of solute
mass of solute = Mass percent x mass of solution / 100
2.- Substitution
mass of solute = 11.1 x 482 / 100
3.- Simplification
mass of solute = 5350.2 / 100
4.- Result
mass of solute = 53.5g
Answer:
C. Cs < K < P
Explanation:
Electronegativity is the basically the tendency of an atom to attract electrons. The pauling scale is used to quantify electronegativity, the most electronegative element; fluorine has a value of 4.0 and francium, the least electronegative element has a value of 0.7
The electronegative values of the following elements are given as;
K = 0.82
Cs =0.79
P = 2.19
Arranging in order of increasing electronegativity, we have;
Cs < K < P. The correct option is;
C. Cs < K < P
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:
E = 0.062 V
Explanation:
(a) See the attached file for the answer
(b)
Calculating the voltage (E) using the formula;
E = - (2.303RT/nf)log Cathode/Anode
Where,
R = 8.314 J/K/mol
T = 35°C = 308 K
F- Faraday's constant = 96500 C/mol,
n = number of moles of electron = 2
Substituting, we have
E = -(2.303 * 8.314 *308/2*96500) *log (0.03/3)
= -0.031 * -2
= 0.062V
Therefore, the voltmeter will show a voltage of 0.062 V
