Answer:
1.008moles of iodine
Explanation:
Hello,
This question requires us to calculate the theoretical yield of I₂ or number of moles that reacted.
Percent yield = (actual yield / estimated yield) × 100
Actual yield = 1.2moles
Estimated yield = ?
Percentage yield = 84%
84 / 100 = 1.2 / x
Cross multiply and solve for x
100x = 84 × 1.2
100x = 100.8
x = 100.8/100
x = 1.008moles
1.008 moles of I₂ reacted in excess of H₂ to give 1.2 moles of HI
The pool being cooler at the deep end. Also the air closer to the roof of a house being cooler.
The principle quantum number "n" represents the relative overall energy of each orbital, and the energy of each orbital increases as the distance from the nucleus increases. The sets of orbitals with the same "n" value are often referred to as electron shells or energy levels.
The name is Potassium bromide.
<em>Calculate the pH of the following substances formed during a volcanic eruption:
</em>
<em>• Acid rain if the [H +] is 1.9 x 10-5
</em>
<em>• Sulfurous acid if [H +] = 0.10
</em>
<em>• Nitric acid if [H +] = 0.11</em>
<em />
<h3>Further explanation </h3>
pH is the degree of acidity of a solution that depends on the concentration of H⁺ ions. The greater the value the more acidic the solution and the smaller the pH.
pH = - log [H⁺]
![\tt pH=-log[1.9\times 10^{-5}]\\\\pH=5-log1.9\\\\pH=4.72](https://tex.z-dn.net/?f=%5Ctt%20pH%3D-log%5B1.9%5Ctimes%2010%5E%7B-5%7D%5D%5C%5C%5C%5CpH%3D5-log1.9%5C%5C%5C%5CpH%3D4.72)
![\tt pH=-log[10^{-1}]\\\\pH=1](https://tex.z-dn.net/?f=%5Ctt%20pH%3D-log%5B10%5E%7B-1%7D%5D%5C%5C%5C%5CpH%3D1)
![\tt pH=-log[11\times 10^{-2}]\\\\pH=2-log~11=0.959](https://tex.z-dn.net/?f=%5Ctt%20pH%3D-log%5B11%5Ctimes%2010%5E%7B-2%7D%5D%5C%5C%5C%5CpH%3D2-log~11%3D0.959)
