Answer : The original concentration of copper (II) sulfate in the sample is, 
Explanation :
Molar mass of Cu = 63.5 g/mol
First we have to calculate the number of moles of Cu.
Number of moles of Cu = 
Now we have to calculate the number of moles of 
Number of moles of Cu = Number of moles of 
Number of moles of
= 
Now we have to calculate the molarity of 

Now put all the given values in this formula, we get:

To change mol/L into g/L, we need to multiply it with molar mass of 
Molar mass of
= 159.609 g/mL
Concentration in g/L = 
Thus, the original concentration of copper (II) sulfate in the sample is, 
The answer is organisms, oxidation & water. Organisms such as plants have their roots penetrate the cracks in the rocks. As the roots grow and become bigger they put mechanical pressure on the rocks and the cracks increase in size and the rocks are split. Additionally, organisms secrete organic acids, as wastes, that chemically react with minerals in the rocks hence causing weathering.
Oxidation is a chemical reaction that changes the property of minerals by makes them prone to weathering. An example is the oxidation of iron in the rocks to iron III oxide which is brittle.
Water participates in chemical reaction by dissolving minerals, such as limestone (CaCO3) from the rock hence leaving the rock porous and more predisposed to physical weathering agents.
Answer:
The nucleus contains protons, which have a positive charge
Explanation:
Answer:
Keq = 19
ΔG° = -7.3kJ/mol
Explanation:
Based on the chemical reaction:
Glucose 1-phosphate ⇄ Glucose 6-phosphate
The equilibrium constant, Keq is defined as:
Keq = [Glucose 6-phosphate] / [Glucose 1-phosphate]
<em>Where [] are equilibrium concentrations of each substance</em>
<em />
Replacing:
Keq = [0.19M] / [0.01M]
Keq = 19
Now, standard free energy change, ΔG° is defined as:
ΔG° = -RT ln K
<em>Where R is gas constant 8.314J/molK</em>
<em>T is absolute temperature (25°C + 273.15K = 298.15K)</em>
<em>and K is equilibrium constant = 19</em>
<em />
Replacing:
ΔG° = -8.314J/molK*298.15K ln 19
ΔG° = -7299J/mol
ΔG° = -7.3kJ/mol