Sodium Sulfate
= Na2(SO4) meaning there are two ions of Na+ in one mole of Sodium Sulfate the M
stands for Molarity, defined as Molarity = (moles of solute)/(Liters of
solution), So if the Na2SO4 solution is 3.65M that means one Liter of has 3.65
moles of Na2SO4, the stoichiometry of Na2SO4 shows that there would be two Na+
ions in solution for every one Na2SO4.
Therefore if
3.65 moles of Na2SO4 was to dissolve, it would produce 7.3 moles of Na+, and
since this is still a theoretical solution, we can assume 1 L of solution.
Finally we find
[Na+] = 2*3.65 = 7.3M
Use the same
logic for parts b and c
Explanation:
Because they share a similar composition, similar chemicals affect their structures in positive or negative ways. For example, fluoride – a staple in many dental practices – strengthens both enamel and eggshells and helps protect them from acids. Acids weaken and break down both substances. Scientists find this particularly concerning given that the ocean is growing increasingly acidic. They fear this may weaken the eggs of some marine species and harm their chance of survival. Most dentists recommend limiting aggressively acidic foods and beverages such as soft drinks.
Answer:
Explanation:
Given parameters:
Mass of aluminium oxide = 3.87g
Mass of water = 5.67g
Unknown:
Limiting reactant = ?
Solution:
The limiting reactant is the reactant in short supply in a chemical reaction. We need to first write the chemical equation and convert the masses given to the number of moles.
Using the number of moles, we can ascertain the limiting reactants;
Al₂O₃ + 3H₂O → 2Al(OH)₃
Number of moles;
Number of moles = 
molar mass of Al₂O₃ = (2x27) + 3(16) = 102g/mole
number of moles =
= 0.04mole
molar mass of H₂O = 2(1) + 16 = 18g/mole
number of moles =
= 0.32mole
From the reaction equation;
1 mole of Al₂O₃ reacted with 3 moles of H₂O
0.04 mole of Al₂O₃ will react with 3 x 0.04 mole = 0.12 mole of H₂O
But we were given 0.32 mole of H₂O and this is in excess of amount required.
This shows that Al₂O₃ is the limiting reactant
Answer: So if you had 570 cm of ribbon, then 570%2F8.5=67.05 which means that about 67 students can do the experiment (round down to the nearest whole number).
Explanation: If you had 8.5 cm of ribbon, then only 8.5%2F8.5=1 student can do the experiment. If you had 17 cm of ribbon, then 17%2F8.5=2 students can do the experiment.