A result of the common ion effect is that the selective precipitation of a metal ion, such as Ag+, is promoted by the addition of an appropriate counter ion (X-) that produces a compound (AgX) with a very low solubility.
Multiply 9.0g/cm3 by 0.31g = 2.79cm3
Answer:
The answer to your question is: 8.31 x 10 ²² atoms
Explanation:
butane C₄H₁₀
carbon atoms in 2 g of butane
MW C₄H₁₀ = (4 x 12) + (1 x10) = 48 + 10 = 58 g
58 g of butane ------------------- 48 g of C
2 g of butane ----------------- x
x = (2 x 48) / 58 = 1.655 g of C
1 mol C ------------------- 12 g
x ------------------ 1.655 g
x = (1.655 x 1) / 12 = 0.14 mol of C
1 mol of C ------------------- 6.023 x 10 ²³ atoms
0.14 mol of C ---------------- x
x = ( 0.14 x 6.023 x 10 ²³) / 1
x = 8.31 x 10 ²² atoms
Answer:
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