The solubility of a substance in water is dependent on the temperature. Thus for
1 & 2: Temperature is the independent variable (the one that changes in the first place) and Solubility is a dependent variable (a variable that changes in response to changes in the independent variable.)
The graph: by convention you shall label the horizontal axis with the independent variable and the vertical axis with the dependent variable. For clarity's sake you shall use the finest scale possible that accommodates for all data provided for both axis. Plot the data points on the graph as if they are points on a cartesian plane.
My teacher made no detailed requirements on the phrasing on titles of solubility curve plots; however, like most other graphs in chemistry, the title shall specify the name of variables presented in this visualization. For instance, "the solubility of
under different temperatures" might do. You shall refer to your textbooks for such convention.
It is necessary to interpolate to find the solubility at a temperature not given in the table. Start by connecting all given data points with a smooth line; find the vertical line corresponding to temperature = 75 degree Celsius and determine the solubility at the intersection of the vertical line and the trend line. That point shall approximates the solubility of the salt at that temperature.
Answer: second answer
Explanation:
The thing is that every new nuclear cycle a new element forms and reduces an electron. I may be wrong but this is the most logical and scientifically correct answer
I know the answer to your question but if you want me to answer this question first, I need your help with my question "How do I solve number 12? Quick please I need help ASAP!!!"
What exactly are you asking
There are 2.32 x 10^6 kg sulfuric acid in the rainfall.
Solution:
We can find the volume of the solution by the product of 1.00 in and 1800 miles2:
1800 miles2 * 2.59e+6 sq m / 1 sq mi = 4.662 x 10^9 sq m
1.00 in * 1 m / 39.3701 in = 0.0254 m
Volume = 4.662 x 10^9 m^2 * 0.0254 m
= 1.184 x 10^8 m^3 * 1000 L / 1 m3
= 1.184 x 10^11 Liters
We get the molarity of H2SO4 from the concentration of [H+] given by pH = 3.70:
[H+] = 10^-pH = 10^-3.7 = 0.000200 M
[H2SO4] = 0.000100 M
By multiplying the molarity of sulfuric acid by the volume of the solution, we can get the number of moles of sulfuric acid:
1.184 x 10^11 L * 0.000100 mol/L H2SO4 = 2.36 x 10^7 moles H2SO4
We can now calculate for the mass of sulfuric acid in the rainfall:
mass of H2SO4 = 2.36 x 10^7 moles * 98.079 g/mol
= 2.32 x 10^9 g * 1 kg / 1000 g
= 2.32 x 10^6 kg H2SO4