The kinetic energy of the products is equal to the energy liberated which is 92.2 keV. But let's convert the unit keV to Joules. keV is kiloelectro volt. The conversion that we need is: 1.602×10⁻¹⁹ <span>joule = 1 eV
Kinetic energy = 92.2 keV*(1,000 eV/1 keV)*(</span>1.602×10⁻¹⁹ joule/1 eV) = 5.76×10²³ Joules
From kinetic energy, we can calculate the velocity of each He atom:
KE = 1/2*mv²
5.76×10²³ Joules = 1/2*(4)(v²)
v = 5.367×10¹¹ m/s
Explanation:
The given compounds are oxyacids and in these compounds more is the electronegativity of the central atom more will be its acidic strength.
This is because more is the electronegativity of the central atom more will be the polarity of OH bond. As a result, the compound can readily lose
ion.
Also, more is the electronegativity of central atom more will be the stability of conjugate base formed.
Thus, we can conclude that given compounds are arranged in increasing acid strength as follows.
HOI <
<
< HOF
Answer:
The pH of the buffer is 7.0 and this pH is not useful to pH 7.0
Explanation:
The pH of a buffer is obtained by using H-H equation:
pH = pKa + log [A⁻] / [HA]
<em>Where pH is the pH of the buffer</em>
<em>The pKa of acetic acid is 4.74.</em>
<em>[A⁻] could be taken as moles of sodium acetate (14.59g * (1mol / 82g) = 0.1779 moles</em>
<em>[HA] are the moles of acetic acid (0.060g * (1mol / 60g) = 0.001moles</em>
<em />
Replacing:
pH = 4.74 + log [0.1779mol] / [0.001mol]
<em>pH = 6.99 ≈ 7.0</em>
<em />
The pH of the buffer is 7.0
But the buffer is not useful to pH = 7.0 because a buffer works between pKa±1 (For acetic acid: 3.74 - 5.74). As pH 7.0 is out of this interval,
this pH is not useful to pH 7.0
<em />
Answer:
Equilibrium constant of the given reaction is 
Explanation:
....
....
The given reaction can be written as summation of the following reaction-


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Equilibrium constant of this reaction is given as-
![\frac{[NOBr]^{2}}{[N_{2}][O_{2}][Br_{2}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNOBr%5D%5E%7B2%7D%7D%7B%5BN_%7B2%7D%5D%5BO_%7B2%7D%5D%5BBr_%7B2%7D%5D%7D)
![=(\frac{[NOBr]}{[NO][Br_{2}]^{\frac{1}{2}}})^{2}(\frac{[NO]^{2}}{[N_{2}][O_{2}]})](https://tex.z-dn.net/?f=%3D%28%5Cfrac%7B%5BNOBr%5D%7D%7B%5BNO%5D%5BBr_%7B2%7D%5D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%29%5E%7B2%7D%28%5Cfrac%7B%5BNO%5D%5E%7B2%7D%7D%7B%5BN_%7B2%7D%5D%5BO_%7B2%7D%5D%7D%29)

