The highest atom economy
2CO + O₂ ⇒ 2CO₂
<h3>Further explanation</h3>
Given
The reaction for the production of CO₂
Required
The highest atom economy
Solution
In reactions, there are sometimes unwanted products that can be said to be a by-product or a waste product. Meanwhile, the desired product can be said to be a useful product, which can be shown as the atom economy
of the reaction
the higher the atomic economy value of a reaction, the smaller the waste/ byproducts produced, so that less energy is wasted
The general formula:
Atom economy = (mass of useful product : mass of all reactants/products) x 100
<em>or
</em>
Atom economy = (total formula masses of useful product : total formula masses of all reactants/products) x 100
So a reaction that only produces one product will have the highest atomic value, namely the reaction in option C
Answer:
Take a look at the attachment below
Explanation:
Take a look at the periodic table. As you can see, Rubidium is the closest element to Cesium, and happens to have the closest boiling point to Cesium, with only a difference of about 30 degrees.
Respectively, you would think that fluorine should have the least similarity to Cesium with respect to it's boiling point, considering it is the farthest away from the element out of the 4 given. This is not an actual rule, there are no fixed trends of boiling points in the periodic table, there are some but overall the trends vary. However in this case fluorine does have the least similarity to Cesium with respect to it's boiling point, a difference of about 1,546.6 degrees.
<em>Hope that helps!</em>
Answer : Option E) 50 grams.
Explanation : According to the solubility curves the compound 
to dissolve at 50 °C in 100 mL of water will need 50 grams of the compound. It is clearly indicated in the graph which is marked with red that at 50°C approximately 50.4 grams of the compound will be needed to dissolved in 100 mL of water to form a solution.
We assume that the method that made use of urea was able to recover all of the recoverable substance. The method in question is the method that makes use of water.
The total amount of substance is 43 mg/dl. The recovered amount is 25 mg/dl. The percent recovery is
(25 mg/dl / 43 mg/dl) * 100 = 58.14%
