Assuming the other part of the question is;
A solid mixture consists of 44.2g of KNO3 (potassium nitrate) and 7.8g of K2SO4 (potassium sulfate). The mixture is added to 130. g of water. (Assume KNO3 has a solubility of 14 g solute/ 100 g water)
Answer;
26 g
Solution;
-X g of KNO3/ 130 g water= 14 g of KNO3/ 100 g water
X= 18.2 g (this is the amount of solute that dissolves at 0 degrees C in 130 g water).
if you have 44.3g of KNO3 in 130 g of water at 0 degrees celsius and only 18.2 can dissolve then
44.2g - 18.2g = 26 g left undisolved which gives you the amount that would crystallize
Well, water is composed H2 gas and O2 gas in a 2:1 ratio, and the bubbles that form at the bottom of the pan is just the water changing from a liquid to a gas, and im pretty sure its water vapor that comes out of the pan. I hope that helped??? :)
Answer:
The measured absorbance will be too large.
Explanation:
Fe³⁺(aq) + SCN⁻(aq) ⟶Fe(SCN)²⁺(aq)
A = log₁₀(I₀/I)
If the student orients the cuvette so that the path of the light is through the frosted sides of the cuvette, little light will be able to reach the detector.
The measured intensity (I) will be quite small, so the absorbance (A) will be unusually large.
Molarity of solution = 1.6 M
<h3>Further explanation</h3>
Given
40 g NaOH
6 L solution
Required
Steps to solve the problem of molarity
Solution
No additional information about the question.
If you want to make the solution above, then we just need to put the existing NaOH (40 g) into 6 L of water, then do the stirring (in a warm temperature above the hot plate will speed up the NaOH dissolving process)
But if you want to know the molarity of a solution, then
- 1. we calculate the moles of NaOH
MW(molecular weight) of NaOH=
Ar Na+ Ar O + Ar H
23 + 16 + 1 = 40 g/mol
so mol NaOH :
There are several ways of expressing concentration of solution. Few of them are listed below
1) mass percentage
2) volume percentage
3) Molarity
4) Normality
5) Molality
In most of the drugs, concentration is expressed either in terms of mass percentage or volume percentage. For, solid in liquid type systems, mass percentage is convenient way of expressing concentration, while for liquid in liquid type solutions, expressing concentration in terms of volume percentage is preferred. Present system is an example of liquid in liquid type solution
Here, concentration of H2O2 is given antiseptic = 3.0 % v/v
It implies that, 3ml H2O2 is present in 100 ml of solution
Thus, 400 ml of solution would contain 4 X 3 = 12 ml H2O2
