<span>Answer:
Mass % KCL:
Add the grams of both compounds (31.0 g KCL + 225 g water) to find total mass and then divide the grams of KCL over the total mass, then multiply by 100: ( 31.0 g KCL / 31.0 g + 225 G) * 100%
Mole fraction KCL
Calculate the moles of KCL and water and add them to find the total moles (Moles of KCL + moles of water). Then, divide the number of KCL moles over the total moles.
moles of KCL/ moles kcl + moles water= mole fraction of KCL</span>
The less soluble salt : PbCl₂
<h3>Further explanation</h3>
Given
0.1 M NaCl
Required
The less soluble salt
Solution
If we see from the answer option, the salt that is more difficult to dissolve in NaCl is PbCl₂ because it has the same ion (Cl)
When PbCl₂ is dissolved in water, ionization will occur
PbCl₂ ⇒ Pb²⁺+ 2Cl⁻
So, when dissolved in NaCl, NaCl itself will be ionized
NaCl ⇒ Na⁺ + Cl⁻
Based on the principle of equilibrium, the addition of an ion (one of the ions is enlarged), the reaction will shift towards the ion that was not added. In addition to this Cl ion, the reaction will shift to the left so that the solubility of PbCl₂ will decrease (the reaction to the right decreases)
Answer:
Correct option is
A
Explanation:
The number of molecules of CO involves in the slowest step will be 0 because CO is not involve in the slowest step i.e. rate determing step.
In formation of a Type II Binary Compound, the metal atom present is<span>
NOT</span> found in either Group 1 or Group 2 on the periodic table. For the choices, Ba is under Group 2 on the periodic table, which makes it the atom not involved in formation of type II compounds. The answer is B.
