Answer:
B) Native fish species will begin to decrease in population size.
Hope it helps:D
Answer:
6.33×10¯²² g
Explanation:
From the question given above, the following data were obtained:
Number of atoms = 6 atoms
Mass of copper (Cu) =?
From Avogadro's hypothesis, we understood that:
6.02×10²³ atoms = 1 mole of Cu
But 1 mole of Cu = 63.5 g
Thus,
6.02×10²³ atoms = 63.5 g of Cu
Finally, we shall determine the mass of 6 atoms of copper. This can be obtained as illustrated below:
6.02×10²³ atoms = 63.5 g of Cu
Therefore,
6 atoms = (6 × 63.5) / 6.02×10²³
6 atoms = 6.33×10¯²² g of Cu
Therefore, the mass of 6 atoms of copper is 6.33×10¯²² g.
Because the intention is to boil the solution, the purpose of the solvent is to dissolve so it has a higher boiling point so ensure it stays in liquid form and doesn't evaporate into a gas
also, a pure solvent is made of 1 substance so it has 1 boiling point and the solution must evaporate/boil first
The s orbitals are not symmetrical in shape is a FALSE statement.
An s orbital is so symmetric, more specifically spherically symmetric that it looks the same from all directions.
- The atomic orbitals in the atoms of elements differ in shape.
In essence, the electrons they describe have varying probability distributions around the nucleus. The spherical symmetry of s orbitals is evident in the fact that all orbitals of a given shell in the hydrogen atom have the same energy.
- All s orbitals are spherically symmetrical. Put simply, an electron that occupies an s orbital can be found with the same probability at any orientation (at a distance) from the nucleus.
The s orbitals are therefore represented by a spherical boundary surface which is a surface which captures a high proportion of the electron density.
Read more:
brainly.com/question/5087295
Answer:
Explanation:
The air 9% mole% methane have an average molecular weight of:
9%×16,04g/mol + 91%×29g/mol = 27,8g/mol
And a flow of 700000g/h÷27,8g/mol = 25180 mol/h
In the reactor where methane solution and air are mixed:
In = Out
Air balance:
91% air×25180 mol/h + 100% air×X = 95%air×(X+25180)
Where X is the flow rate of air in mol/h = <em>20144 mol air/h</em>
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The air in the product gas is
95%×(20144 + 25180) mol/h = 43058 mol air× 21%O₂ = 9042 mol O₂ ×32g/mol = <em>289 kg O₂</em>
43058 mol air×29g/mol <em>1249 kg air</em>
Percent of oxygen is: 
=<em>0,231 kg O₂/ kg air</em>
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I hope it helps!
