The formula of chalcopyrite is CuFeS₂.
The molar mass of chalcopyrite - 183.5 g/mol
The mass of Cu in 1 mol of chalcopyrite is - 63.5 g
When 183.5 g of chalcopyrite is mined, the mass of Cu obtained - 63.5 g
So if 63.g of Cu is needed mass of chalcopyrite required - 183.5 g
Then to gain 370 g of pure Cu - mass of chalcopyrite needed -183.5/63.5 x 370
therefore mass of chalcopyrite to be mined - 1069.21 g
Answer is: a) ionization energy and electronegativity.
1) The ionization energy (Ei) is the minimum amount of energy required to remove the valence electron, when element lose electrons, oxidation number of element grows (oxidation process).
Barium, potassium and arsenic are metals (easily lost valence electrons), chlorine is nonmetal (easily gain electrons).
Alkaline metals (far left in Periodic table) have lowest ionizations energy and easy remove valence electrons (one electron), earth alkaline metals (next right to alkaline metals) have higher ionization energy than alkaline metals, because they have two valence electrons.
Nonmetals are far right in the main group and they have highest ionization energy, because they have many valence electrons.
2) Electronegativity (χ) is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.
Atoms with higher electronegativity attracts more electrons towards it, electrons are closer to that atom.
Nonmetals hava higher electronegativity than metals and metalloids.
3) The atomic radius decreases across the periods because an increasing number of protons, because greater attraction between the protons and electrons.
Answer:
B. The cell membrane surrounds particles outside the cell.
Explanation:
The last step is when the cell membrane completely surrounds particles outside the cell.
This process is often known as endocytosis.
- endocytosis is the process whereby a cell ingests materials by engulfing them using the cell membrane.
- In this process, the cell membrane completely covers the food.
The correct answer for the question that is being presented above is this one: "<span>16.728 g."</span>
Given that
ΔHsolid = -5.66 kJ/mol.
This means that 5.66 kJ of heat is released when 1 mole of NH3 solidifies
When 5.57 kJ of heat is released
amount of NH3 solidifies = 5.57/5.66 = 0.984 moles
<span>molar mass of NH3 = 17 g/mole </span>
<span>1 mole of NH3 = 17 g </span>
So, 0.984 moles of NH3 = 17 X 0.984 = 16.728 g
Answer:
1.427x10^-3mol per L
Explanation:

I could use ⇌ in the math editor so I used ----
from the question each mole of Y(IO3)3 is dissolved and this is giving us a mole of Y3+ and a mole of IO3^3-
Ksp = [Y^3+][IO3-]^3
So that,
1.12x10^-10 = [S][3S]^3
such that
1.12x10^-10 = 27S^4
the value of s is 0.001427mol per L
= 1.427x10^-3mol per L
so in conclusion
the molar solubility is therefore 1.427x10^-3mol per L