Answer:
Pyrophoricity is a property of metals and oxides of lower oxidation states, including radioactive ones, in which they spontaneously ignite during or after stabilization.
Answer:
Homogeneous - With a uniform appearance
Heterogeneous - With visible differences in the mixture
Explanation:
Referring to mixtures, homogeneous looks the same throughout, while heterogeneous has particles or whatnot.
Basically:
Homo means same
Hetero means different
Answer:
0.0084
Explanation:
The mole fraction of BaCl₂ (X) is calculated as follows:
X = moles BaCl₂/total moles of solution
Given:
moles of BaCl₂ = 0.400 moles
mass of water = 850.0 g
We have to convert the mass of water to moles, by using the molecular weight of water (Mw):
Mw of water (H₂O) = (2 x 1 g/mol)+ 16 g/mol = 18 g/mol
moles of water = mass of water/Mw of water = 850.0 g/(18 g/mol) = 47.2 mol
The total moles of the solution is given by the addition of the moles of solute (BaCl₂) and the moles of solvent (water):
total moles of solution = moles of BaCl₂ + moles of water = 0.400 + 47.2 mol = 47.6 mol
Finally, we calculate the mole fraction:
X = 0.400 mol/47.6 mol = 0.0084
Answer:
One extraction: 50%
Two extractions: 75%
Three extractions: 87.5%
Four extractions: 93.75%
Explanation:
The following equation relates the fraction q of the compound left in volume V₁ of phase 1 that is extracted n times with volume V₂.
qⁿ = (V₁/(V₁ + KV₂))ⁿ
We also know that V₂ = 1/2(V₁) and K = 2, so these expressions can be substituted into the above equation:
qⁿ = (V₁/(V₁ + 2(1/2V₁))ⁿ = (V₁/(V₁ + V₁))ⁿ = (V₁/(2V₁))ⁿ = (1/2)ⁿ
When n = 1, q = 1/2, so the fraction removed from phase 1 is also 1/2, or 50%.
When n = 2, q = (1/2)² = 1/4, so the fraction removed from phase 1 is (1 - 1/4) = 3/4 or 75%.
When n = 3, q = (1/2)³ = 1/8, so the fraction removed from phase 1 is (1 - 1/8) = 7/8 or 87.5%.
When n = 4, q = (1/2)⁴ = 1/16, so the fraction removed from phase 1 is (1 - 1/16) = 15/16 or 93.75%.
Answer:
D. Increase in UV radiation
Explanation:
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