Answer: C. Electrochemical cells involve oxidation-reduction reactions.
Explanation: Oxidation occurs at the anode, and reduction occurs at the cathode.
Rocks and minerals are usually composed of two or more minerals
<span>1 mole of calcium carbonate reacts with 1 mole of sulfuric acid and produces 1 mole of calcium sulfate.
3.1660 g of CaCO3 is how many moles of calcium carbonate? 3.1660 / 100.0869 = 0.031633 moles.
3.2900 g of H2S04 is how many moles of sulfuric acid? 3.2900 / 98.079 = 0.033544 moles.
</span><span>The lesser of the two is 0.031633 moles.
Therefore, 0.031633 moles of calcium carbonate will combine with 0.031633 moles of sulfuric acid to produce 0.031633 moles of calcium sulfate.
Molecular weight of calcium sulfate is 136.14 g/mol.
Therefore, 0.031633 moles of calcium sulfate will weight 0.031633 x 136.14 g/mol = 4.3065 grams.</span>
Explanation:
1)

Mass of NaOH = m
MOlar mass of NaOH = 40 g/mol
Volume of NaOH solution = 1.00 L
Molarity of the solution= 1.00 M


A student can prepare the solution by dissolving the 40. grams of NaOH in is small volume of water and making that whole volume of solution to volume of 1 L.
Upto two significant figures mass should be determined.
2)
(dilution equation)
Molarity of the NaOH solution = 
Volume of the solution = 
Molarity of the NaOH solution after dilution = 
Volume of NaOH solution after dilution= 


A student can prepare NaOH solution of 1.00 M by diluting the 0.500 L of 2.00 M solution of NaOH with water to 1.00 L volume.
Upto three significant figures volume should be determined.