Because water is polar and oil is nonpolar, their molecules are not
attracted to each other. The molecules of a polar solvent like water are
attracted to other polar molecules, such as those of sugar. This explains
why sugar has such a high solubility in water. Ionic compounds, such
as sodium chloride, are also highly soluble in water. Because water
molecules are polar, they interact with the sodium and chloride ions.
In general, polar solvents dissolve polar solutes, and nonpolar solvents
dissolve nonpolar solutes. This concept is often expressed as “Like
dissolves like.”
So many substances dissolve in water that it is sometimes called
the universal solvent. Water is considered to be essential for life
because it can carry just about anything the body needs to take in
or needs to get rid of.
The reduction of a less active metal by a more active one is called metal displacement reactions. For example:
Fe + CuSO4 → FeSO4 + Cu
<h3>What is metal displacement reaction? </h3>
Displacement reactions is a reaction which includes a metal and the compound of a other metal. A more reactive metal will push or displace out a less reactive metal from its compound in this displacement reaction. The metal which is less reactive left uncombined after the reaction.
As we know that, electrons are the basis of the chemical reactions. If chemical compound or element A is more easily oxidized than B, then according to the terms of the activity series, the elements which are more easily oxidized can react with more chemicals, since they are able to act as a reducing agents for more chemicals.
Since, Metal ions are positively charged ions as they lose electrons. Some metals give up their electrons more readily than others and become more reactive.
Thus, we concluded that the reduction of a less active metal by a more active one is called metal displacement reactions. For example:
Fe + CuSO4 → FeSO4 + Cu
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The correct answer would b C
To determine the molar mass of the unknown gas, we use Graham's Law of Effusion where it relates the effusion rates of two gases with their molar masses. It is expressed as r1/r2 = √M2/M1. We calculate as follows:
Let 1 = argon gas 2 = unknown gas
r2 = 0.91r1r1/r2 = 1/0.91
1/0.91 = √M2/M1 = √M2/40M2 = 48.30 g/mol
Change in the amount of volume left inside the basketball for the remaining air coming in