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
To determine whether the amount of H2 in the lab is dangerous, we first need to know how much hydrogen gas is present in the room in units of percent by volume. For this particular problem, we cannot exactly determine since we do not know the total volume of the room. Hope this answers the question.
NaOH+HCl-> NaCl+H2O
1 mole of NaOH
1 mole of HCl.
To calculate volume of NaOH
CaVa/CbVb= Na/Nb
Where Ca=2M
Cb=1M
Va=200cm³
Vb=xcm³
Substitute into the equation.
2×200/1×Vb=1/1
400/Vb=1/1
Cross multiply
Vb×1=400×1
Vb=400cm³
To calculate the mass of sodium chloride, NaCl from the neutralization rxn.
Mole of NaCl=1
Molar mass of NaCl= 23+35.5=58.5
Mass=xgrammes.
Mass of NaCl=Number of moles × Molar mass.
Substitute
Mass of NaCl= 1×58.5
=58.5g
This is what I could come up with.
The balanced chemical reaction:
<span>Cu + 2AgNO3 = Cu(NO3)2 + 2Ag
</span>
We are given the amount of the reactants to be used for the reaction. These values will be the starting point of our calculations.
9.85 g Cu ( 1 mol Cu / 63.55 g Cu ) = 0.15 mol Cu
31.0 g AgNO3 ( 1 mol AgNO3 / 169.87 g AgNO3 ) = 0.18 mol AgNO3
The limiting reactant is AgNO3.
0.18 mol AgNO3 ( 1 mol Cu(NO3)2 / 2 mol AgNO3 ) (187.56 g / 1 mol) =16.88 g Cu(NO3)2
0.15 mol Cu - 0.18 mol AgNO3 ( 1 mol Cu / 2 mol AgNo3) = 0.06 mol Cu excess
<span>0.06 mol Cu ( 63.55 g Cu / 1 mol Cu ) = 3.81 g Cu excess</span>
