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
Mixing of pure orbitals having nearly equal energy to form equal number of completely new orbitals is said to be hybridization.
For the compound, the electronic configuration of the atoms, carbon and hydrogen are:
Carbon (atomic number=6): In ground state=
In excited state:
Hydrogen (atomic number=1):
All the bonds in the compound is single bond(-bond) that is they are formed by head on collision of the orbitals.
The structure of the compound is shown in the image.
The Carbon-Hydrogen bond is formed by overlapping of s-orbital of hydrogen to p-orbital of carbon.
In order to complete the octet the required number of electrons for carbon is 4 and for hydrogen is 1. So, the electron in of hydrogen will overlap to the 2p^{3}-orbital of carbon.
Thus, the hybridization of Hydrogen is -hybridization and the hybridization of Carbon is -hybridization.
The hybridization of each atom is shown in the image.
Answer:
325
Explanation:
velocity = 3250 Hz x 0.1 m
This question requires the knowledge of density.
The density of ethyl alcohol = 789 kg m⁻³
The density of water = 1000 kg m⁻³
Density = Mass / Volume
By applying ethyl alcohol,
789 kg m⁻³ = Mass / 0.9 m³
Mass = 710.1 kg
hence the mass of 0.9 m³ ethyl alcohol is 710.1 kg.
Then by applying water,
1000 kg m⁻³ = 710.1 kg / Volume
Volume = 0.7101 m³
= 0.7 m³
hence the equal water volume is 0.7 m³
Move your rope up and down and that will create transverse waves.
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