The pH a 0.25 m solution of C₆H₅NH₂ is equal to 3.13.
<h3>How do we calculate pH of weak base?</h3>
pH of the weak base will be calculate by using the Henderson Hasselbalch equation as:
pH = pKb + log([HB⁺]/[B])
pKb = -log(1.8×10⁻⁶) = 5.7
Chemical reaction for C₆H₅NH₂ is:
C₆H₅NH₂ + H₂O → C₆H₅NH₃⁺ + OH⁻
Initial: 0.25 0 0
Change: -x x x
Equilibrium: 0.25-x x x
Base dissociation constant will be calculated as:
Kb = [C₆H₅NH₃⁺][OH⁻] / [C₆H₅NH₂]
Kb = x² / 0.25 - x
x is very small as compared to 0.25, so we neglect x from that term and by putting value of Kb, then the equation becomes:
1.8×10⁻⁶ = x² / 0.25
x² = (1.8×10⁻⁶)(0.25)
x = 0.67×10⁻³ M = [C₆H₅NH₃⁺]
On putting all these values on the above equation of pH, we get
pH = 5.7 + log(0.67×10⁻³/0.25)
pH = 3.13
Hence pH of the solution is 3.13.
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Do the atomic mass times the number of atoms you have. That'll give you your answer. Hope it helps!
Answer:
Vertically Shrunk by a factor of 1/6
Explanation:
Parent Formula: f(x) = a(bx - c) + d
<em>a</em> - vertical shrink/stretch and x-reflections
<em>b</em> - horizontal shrink/stretch and y-reflections
<em>c</em> - horizontal movement left/right
<em>d</em> - vertical movement up/down
Since we are only modifying <em>a</em>, we are dealing with vertical shrink/stretch:
Since a < 1 (1/6 < 1), we are dealing with a vertical shrink of 1/6.
Since a > 0 (1/6 > 0), we do not have a reflection over the x-axis.
Answer:
A = 0.023 m
Explanation:
The relation between the frequency of a radiation and its wavelength is given by the following expression.
where,
c is the speed of light (it has a constant value of 3.00 x 108 m/s)
A is the wavelength of the radiation v is the frequency of the radiation
In this case, the frequency is 13 GHz = 13 x
10° Hz = 13 x 1o° s-
The wavelength associated with this frequency is:
A = c/v = (3.00 x 10° m/s)/(13 x 10° s-") = 0.023
<span>In ionic equations, the chemicals are written as the ions which are dissolved in water.
AgNO</span>₃<span>(aq) reacts with K</span>₂<span>SO</span>₄<span>(aq) and produces Ag</span>₂<span>SO</span>₄<span>(s) and KNO</span>₃<span>(aq). The balanced
complete ionic reaction is
2Ag</span>⁺<span>(aq) + NO</span>₃⁻<span>(aq) +2K</span>⁺<span>(aq) + SO</span>₄²⁻<span>(aq) → Ag</span>₂<span>SO</span>₄<span>(s) +2K</span>⁺<span>(aq) + NO</span>₃⁻<span>(aq)
K</span>⁺<span>(aq) and NO</span>₃⁻<span>(aq) present in both sides. Hence, we can cut off them to find net ionic equation.
So, net ionic equation is
2Ag⁺(aq) + SO₄</span>²⁻<span>(aq) → Ag₂SO₄(s)</span>
