The energy of the photon from the calculation performed is 3.5 * 10^-16 J.
<h3>What is a photon?</h3>
A photon is a packet of light. It was derived from the theory of Albert Einstein. The energy of a photon is obtained from; E = hc/λ
- h= Plank's constant
- c = speed of light
- λ = wavelength
Now we have that;
E = 6.6 * 10^-34 * 3 * 10^8/562 * 10^-9
E = 3.5 * 10^-16 J
Missing parts:
Calculate the energy, in joules, of a photon of green light having a wavelength of 562nm?
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Answer:
yes
Explanation:
Although plants use photosynthesis to produce glucose, they use cellular respiration to release energy from the glucose.
Answer:
1.84 L
Explanation:
Using the equation for reversible work:
Where:
W is the work done (J) = -287 J.
Since the gas did work, therefore W is negative.
P is the pressure in atm = 1.90 atm.
However, work done is in joules and pressure is in atm. We can use the values of universal gas constant as a convenient conversion unit. R = 8.314 J/(mol*K); R = 0.0821 (L*atm)/(mol*K)
Therefore, the conversion unit is 0.0821/8.314 = 0.00987 (L*atm)/J
is the initial volume = 0.350 L
is the final volume = ?
Thus:
(-287 J)*0.00987 (L*atm)/J = -1.9 atm*( - 0.350) L
= [(287*0.00987)+(1.9*0.350)]/1.9 = (2.833+0.665)/1.9 =1.84 L
If Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).
<h3>
What is base dissociation constant?
</h3>
The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.
The dissociation reaction of hydrogen cyanide can be given as
HCN --- (H+) + (CN-)
Given,
The value of Ka for HCN is 2.8× 10^(-9)
The correlation between base dissociation constant and acid dissociation constant is
Kw = Ka × Kb
Kw = 10^(-14)
Substituting values of Ka and Kw,
Kb = 10^(-14) /{2.8×10^(-9) }
= 3.5× 10^(-6)
Thus, we find that if Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).
DISCLAIMER: The above question have mistake. The correct question is given as
Question:
Given that Ka for HBrO is 2. 8×10^−9 at 25°C. What is the value of Kb for BrO− at 25°C?
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