Those would be called indicators , simply google them and you should find many. I'll leave some examples "bromophenol blue" "Methyl red" and "phenol red".
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:
2H+(aq) + 2OH-(aq) → 2H2O(l)
Explanation:
Step 1: The balanced equation
2HCl(aq)+Ca(OH)2(aq) → 2H2O(l)+CaCl2(aq)
This equation is balanced, we do not have the change any coefficients.
Step 2: The netionic equation
The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will.
2H+(aq) + 2Cl-(aq) + Ca^2+(aq) + 2OH-(aq) → 2H2O(l) + Ca^2+(aq) + 2Cl-(aq)
After canceling those spectator ions in both side, look like this:
2H+(aq) + 2OH-(aq) → 2H2O(l)
Answer: No
Explanation: <em>Reactants</em> are the substances present at the beginning of a chemical reaction. In the burning of natural gas, for example, methane (CH4) and oxygen (O2) are the reactants in the chemical reaction. <em>Products </em>are the substances formed by a chemical reaction. In the burning of natural gas, carbon dioxide (CO2) and water (H2O) are the products formed by the reaction.
Answer:
It becomes a positive ion and its radius decreases
Explanation:
As per the Octet rule, Barium has 2 electrons in its outermost shell. When it loses the two electron it gains two positive charge i.e Ba2+. As the barium loses the two electron from its outermost shell, the outermost shell becomes vacant and thus is no more considered as a part of atomic geometry of the barium atom and since the outermost shell is considered negligible the radius of barium atom reduces automatically.
