Answer:
(B) 0.038 M
Explanation:
Kc = [H2][I2]/[HI]^2
Let the equilibrium concentration of H2 be y M
From the equation of reaction, mole ratio of H2 to I2 formed is 1:1, therefore equilibrium concentration of I2 is also y M
Also, from the equation of reaction, mole ratio of HI consumed to H2 formed is 2:1, therefore equilibrium concentration of HI is (1 - 2y) M
1.6×10^-3 = y×y/(1 - 2y)^2
y^2/1-4y+4y^2 = 0.0016
y^2 = 0.0016(1-4y+4y^2)
y^2 = 0.0016 - 0.0064y + 0.0064y^2
y^2-0.0064y^2+0.0064y-0.0016 = 0
0.9936y^2 + 0.0064y - 0.0016 = 0
The value of y must be positive and is obtained by using the quadratic formula
y = [-0.0064 + sqrt(0.0064^2 - 4×0.9936×-0.0016)] ÷ 2(0.9936) = 0.0736 ÷ 1.9872 = 0.038 M
Answer:
[H₃O⁺] = [F⁻] = 2.2 x 10⁻² M. & [OH⁻] = 4.55 x 10⁻¹³.
Explanation:
- For a weak acid like HF, the dissociation of HF will be:
<em>HF + H₂O ⇄ H₃O⁺ + F⁻.</em>
[H₃O⁺] = [F⁻].
<em>∵ [H₃O⁺] = √Ka.C,</em>
Ka = 6.8 x 10⁻⁴, C = 0.710 M.
∴ [H₃O⁺] = √Ka.C = √(6.8 x 10⁻⁴)(0.710) = 2.197 x 10⁻² M ≅ 2.2 x 10⁻² M.
<em>∴ [H₃O⁺] = [F⁻] = 2.2 x 10⁻² M.</em>
<em></em>
∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.
<em>∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺]</em> = 10⁻¹⁴/(2.2 x 10⁻²) = <em>4.55 x 10⁻¹³.</em>
a wave that the medium is displaced perpendicular to the direction of the wave, such as waves on a string.
Explanation:
A transverse wave is a wave that the medium is displaced perpendicular to the direction of the wave, such as waves on a string.
A wave is a disturbance that transmits energy from one place to another.
There are two major types of waves:
- Transverse waves are waves that are propagated perpendicularly to the direction of the wave.
- A wave in string going up and down is a transverse wave.
- Electromagnetic waves are transverse waves.
- Seismic s-waves are transverse waves.
- Sound waves are longitudinal waves propagated parallel to their source.
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