NaOH but you didnt give options
Answer:On these combined scales of pH and pH it can be shown that because for water when pH = pH = 7 that pH + pH = 14. This relationship is useful in the inter conversion of values. For example, the pH at a 0.01 M solution of sodium hydroxide is 2, the pH of the same solution must be 14-2 = 12.
Explanation:
Answer:
Total worth of gold in the ocean = $5,840,000,000,000,000
Explanation:
As stated in the question above, 4.0 x 10^-10 g of gold was present in 2.1mL of ocean water.
Therefore, In 1 L of ocean water there will be,
(4.0 x 10^-10)/0.0021
= 1.9045 x 10^-7 g of gold per Liter of ocean water.
So in 1.5 x 10^-21 L of ocean water, there will be
(1.9045 x 10^-7) * (1.5 x 10^-21)
= 2.857 x 10^14 g of gold in the ocean.
1 gram of gold costs $20.44, that is 20.44 dollars/gram. The total cost of the gold present in the ocean is
20.44 * (2.857 x 10^14)
= $5,840,000,000,000,000
Answer:
Explanation:
From the net ionic equation
Ba2+(aq) + SO42-(aq) ==> BaSO4(s) we see that 1 mole Ba2+ reacts with 1 mole SO42- to -> 1 mol BaSO4
Find moles of Ba2+ used: 0.250 moles/L x 0.0323 L = 0.008075 moles Ba2+
Find moles SO42- present: 0.008075 moles Ba2+ x 1 mol SO42-/1 mol Ba2+ = 0.008075 mol SO42-
Find mass of Na2SO4 present: 0.008075 mol SO42- x 1 mol Na2SO4/1 mol SO42- x 142.04 Na2SO4/mole = 1.14698 g = 1.15 g Na2SO4 (to 3 significant figures)
Answer:
c) 3,3-dimethylpent-1-yne
