You need to find the formula mass of NaCl first, which you find the amu of both sodium and chlorine on the periodic table and you add them both. Then you divide the 175.5g by the formula mass which I got 58 g to get 3.026 moles of NaCl. 4 sig figs because of the 175.5. I recommend doing it on your own to make sure you get the right answer. You divide the 175.5g of NaCl by the formula mass.
Hello,
You balanced the equation and then you have to set up what is called a psychometric equation for each reactant so that you get the same unit for each in order to compare the two. so for (calcium hydroxide) = 78.5 g ÷ (40.1 + 32 + 2) = 1.0593 mol * Note: I just broke it down in my own science way..u can just get the molar mass of calcium hydroxide which is approx 74 g/mol
Hope this helps
Based on the fgiven K_{a} values of the acids, the equilibrim concentrations of and is dtermined using ICE tables.
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What are acid dissociatio constants ?</h3>
An acid dissociation constant, of an acid is a measure of the strength of the acid in solution. The larger the value of an acid, the stronger the acid, therefore, acid dissociation constants are usually apply to only weak acids because strong acids have exceedingly large values.
It is written as a qoutient of the equilibrium concentrations of the aqueous species in the acid solution.
For the weak acids such as and , their equilibrim concentrations are determined using their values and an ICE table.
The pH of solutions are estimated using indicators such as methyl orange, methyl red and phenolphthalein. The colors of indicators change according to the pH of the solution
Acidic solutions have pH less than 7 while alkaline solutions have pH greater than 7.
Therefore, acids have low pH and weak acids have low values.
Learn more about acid dissociation constant at: brainly.com/question/3006391
Sedimentary rocks form layers.
Answer:
2000 mL
Explanation:
Given data:
Number of moles KMnO₄ = 6 mol
Molarity of solution = 3M
Volume of solution in mL = ?
Solution:
Formula:
Molarity = number of moles of solute / volume of solution in L
3 M = 6 mol / volume of solution in L
volume of solution in L = 6 mol/ 3 M
volume of solution in L = 2 L
Now we convert the L into mL:
1 L = 1000 mL
2 L × 1000 mL / 1 L
2000 mL