Answer:
The concentration of monosodium phosphate is 0.1262M
Explanation:
The buffer of H₂PO₄⁻ / HPO₄²⁻ (Monobasic phosphate and dibasic phosphate has a pKa of 7.2
To determine the pH you must use Henderson-Hasselbalch equation:
pH = pKa + log [A⁻] / [HA]
<em>Where [A⁻] is molarity of the conjugate base of the weak acid, [HA].</em>
For H₂PO₄⁻ / HPO₄⁻ buffer:
pH = 7.2 + log [HPO₄⁻² ] / [H₂PO₄⁻]
As molarity of the dibasic phosphate is 0.2M and you want a pH of 7.4:
7.4 = 7.2 + log [0.2] / [H₂PO₄⁻]
0.2 = log [0.2] / [H₂PO₄⁻]
1.58489 = [0.2] / [H₂PO₄⁻]
[H₂PO₄⁻] = 0.1262M
<h3>The concentration of monosodium phosphate is 0.1262M</h3>
<em />
Answer:
The answer is "2%"
Explanation:
Equation:


Formula:
![Ka = \frac{[H^{+}][NO_2^{-}]}{[HNO_2]}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5BNO_2%5E%7B-%7D%5D%7D%7B%5BHNO_2%5D%7D)
Let
at equilibrium

therefore,
![[H^{+}] = 2.0\times 10^{-2} \ M = 0.02 \ M](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%202.0%5Ctimes%2010%5E%7B-2%7D%20%5C%20M%20%3D%200.02%20%5C%20M)
Calculating the % ionization:
![= \frac{([H^{+}]}{[HNO_2])} \times 100 \\\\= \frac{0.02}{1}\times 100 \\\\= 2\%\\\\](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%28%5BH%5E%7B%2B%7D%5D%7D%7B%5BHNO_2%5D%29%7D%20%5Ctimes%20100%20%5C%5C%5C%5C%3D%20%5Cfrac%7B0.02%7D%7B1%7D%5Ctimes%20100%20%5C%5C%5C%5C%3D%202%5C%25%5C%5C%5C%5C)
(a) The speed of light in the unknown substance is determined 1.82 x 10⁸ m/s.
(b) The light will bend away from the normal since speed of light in air is not equal to speed of light in the substance.
<h3>What is the speed of light?</h3>
The speed of light passing from air into the substance is calculated as follows;
refractive index = speed of light in air / speed of light in the substance
speed of light in the substance = speed of light in air/refractive index
speed of light in the substance = (3 x 10⁸) / (1.65)
speed of light in the substance = 1.82 x 10⁸ m/s
Thus, the light will bend away from the normal since speed of light in air is not equal to speed of light in the substance.
Learn more about speed of light here: brainly.com/question/104425
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Answer:
2.76 × 10⁻¹¹
Explanation:
I don’t have access to the ALEKS Data resource, so I used a different source. The number may be different from yours.
1. Calculate the free energy of formation of CCl₄
C(s)+ 2Cl₂(g)→ CCl₄(g)
ΔG°/ mol·L⁻¹: 0 0 -65.3
ΔᵣG° = ΔG°f(products) - ΔG°f(reactants) = -65.3 kJ·mol⁻¹
2. Calculate K

T = (25.0 + 273.15) K = 298.15 K
