Answer:
The difference in temperature is significant means that the lower-boiling liquid finishes distilling at a temperature that is too low for the higher-boiling liquid to be in vapor form yet.
Explanation:
The temperature will rise as the vapor of lower-boiling liquid rushes into the distillation head. However once the lower-boiling liquid is done distilling, there is a temperature drop because while the lower temperature liquid is done distilling, the temperature is still too low for the higher-boiling liquid to be rushing in as a vapour, so the temperature drops.
While staying in the same period, if we move from left to right across the period, the atomic radius decreases. The reason is, in a period the number of shells remain the same and the number of electrons and protons increase as we move across the period to the right. The increased electrons and protons attract each other with greater force and hence the atomic size decreases.
So the element on the left most will have the largest atomic radius. So the correct ans is Potassium. Potassium will have the largest atomic size among Potassium, Calcium and Scandium.
Answer:
The molarity of urea in this solution is 6.39 M.
Explanation:
Molarity (M) is <em>the number of moles of solute in 1 L of solution</em>; that is
To calculate the molality, we need to know the number of moles of urea and the volume of solution in liters. We assume 100 grams of solution.
Our first step is to calculate the moles of urea in 100 grams of the solution,
using the molar mass a conversion factor. The total moles of 100g of a 37.2 percent by mass solution is
60.06 g/mol ÷ 37.2 g = 0.619 mol
Now we need to calculate the volume of 100 grams of solution, and we use density as a conversion factor.
1.032 g/mL ÷ 100 g = 96.9 mL
This solution contains 0.619 moles of urea in 96.9 mL of solution. To express it in molarity, we need to calculate the moles present in 1000 mL (1 L) of the solution.
0.619 mol/96.9 mL × 1000 mL= 6.39 M
Therefore, the molarity of the solution is 6.39 M.
Answer:
The nitrogens are both sp3 hybridized. Their bonds are formed by sp overlaps. The carbon and oxygen are sp2 hybridized. The double bond with oxygen is produced by a sp2 overlap to form the sigma component and a probital overlap to form the pi component. The bonds with hydrogen are formed by sp2 overlaps.
Explanation:
1. A mixture of ammonium chloride, sand, and zinc chloride should be separated by sublimation.
2. A mixture of zinc chloride and silver chloride should be separated through crystallization.
<h3>What is a separation technique?</h3>
A separation technique can be defined as a technique that is typically used to separate or convert two (2) or more mixture and solution of chemical substances into distinct product such as chemical compounds or elements.
<h3>The types of
separation technique.</h3>
In Chemistry, there are various types of separation technique used for the separation of mixtures or solutions and these include:
In this scenario, the most effective and efficient means to separate a mixture of ammonium chloride, sand, and zinc chloride is by sublimation from solid to gas state.
On the other hand, the best means to separate a mixture of zinc chloride and silver chloride is through crystallization.
Read more on crystallization here: brainly.com/question/4980962
