A) deposition is the processes where particles of rock or laid down in sections with heavier sediments building up first
Answer:
oxygen
Explanation:
The given isotope has 8 protons and 8 electrons, so the atomic number of the given isotope is 8, which is the atomic number of oxygen.
To know the answer, we determine the polarities of the solvent we have. Water is polar and gasoline is nonpolar. We remember that like dissolves like. Therefore, the answers are:
<span>table sugar = water
motor oil = gasoline
rubber from tire marks = gasoline
adhesive residue from a packing tape = gasoline</span>
Answer:
1.51 X 10^23 ions
Explanation:
The number of ions in 17.1 gm of aluminum sulphate Al2 (SO4)3 =….. [Molar mass of Al2 (SO4)3 = 342 gm]
in one molecule of Al2(SO4)3 there are 5 ions 2 aluminum and 3 sulfate ions
in 2 molecules there are 2X5= 10 ions
in 10 molecules there are 10X5 = 50 ions
molar mass of Al2(SO4)3 = (2 X 26.98) +( 3 X 32.1) + (3 X 4 X 16.0 ) =342.gms = 17.1/342 =0.0500 moles
1 mole =6.02 X 10^23 molecules ( see Avogadros number)
0.0500 moles = 0.0500 X 6.02 X 10^23 molecules =
0.301 X 10^23 molecules = 3.01 X 10^22 molecules
We determined that each molecule of Al2(SO4)3 has 5 ions
so 3.01 X10^22 molecules have 5 X 3.01 X 10^22 ions =
15.05 X 10^22 ions = 1.51 X 10^23 ions
Answer:
The mass of oxygen needed to be pumped in the aquarium is 10 g
Explanation:
Here the value of the degree of solubility of oxygen in water is sought
The solubility of oxygen in fresh water is approximately 1.22 × 10⁻³ mol·dm⁻³ which is equivalent to approximately 40 mg/L
The volume of water the aquarium holds = 250 L
The mass of oxygen required = The solubility of oxygen in water × (The volume of water in the aquarium)
Therefore, the mass of oxygen needed to be pumped in the aquarium = 40 mg/L × 250 L = 10,000 mg
1,000 mg = 1 g
∴ 10,000 mg = 10 g
The mass of oxygen needed to be pumped in the aquarium = 10,000 mg = 10 g.
