Answer is: <span>a. c6h14 and c10h20.
This pair will </span>most likely form a homogeneous solution because they are both nonpolar substances and "li<span>ke dissolves like".
Other pairs will not form homogeneous solution because nonpolar substances have low solubility in polar or ionic substances (for example LiBr is ionic and C</span>₅H₁₂ is nonpolar).
Answer:
146.3g NaCl (mol NaCl/58.44g NaCl) = 2.50 mol NaCl
1.5M NaCl = 1.5 mol NaCl / 1 L = 2.5 mol NaCl / x L, solve for x
x L = 2.5 mol NaCl / 1.5 mol NaCl = 1.66 L
It gives the answer and all the working.
To put it another way:
Dividing the amount required by the molar mass
we quickly see that 2.5 moles are required.
One litre of 1.5 molar solution gives 1.5 moles
we need a further mole, which is 2/3 of 1.5 so 2/3 of a litre.
Answer:

Explanation:
6.023 × 10²³ atoms of Ca are in 1 mol of Ca

Answer:
The flame test is used to visually determine the identity of an unknown metal or metalloid ion based on the characteristic color the salt turns the flame of a Bunsen burner.