Molarity = number of mole of substance(n) / volume of solution (V).
n(CaCl2) = mass (CaCl2)/M(CaCl2)
M(CaCl2) = 40+2*35.5 = 111 g/mol
n(CaCl2) =39.5 g CaCl2*1 mol/111g
0.250 M = 39.5 g CaCl2*1 mol/111g*volume of solution (V).
volume of solution (V) = 39.5 g CaCl2*1 mol/(0.250 M*111g) = 1.42 L
It is less
effective to wash an insoluble precipitate with 15 ml of water once than it is
to wash the precipitate with 3 ml of water 5 times because commonly, when you
clean an <span>indissoluble
precipitate with water, the water will not be completely saturated with
contaminates. Therefore, the absorption of the contaminates would lower with
each wash, since if you only washed it once with a bigger amount or volume of
water, it’d become less contaminated with the wash water but it wouldn’t get
rinsed numerous times.</span>
Answer:
the double bond between c and o is shorter and weaker
Explanation:
this is because the bond between c and o involves unequal sharing of electrons whole c and c involves hybridization sp2 of orbitals and also catenation phenomenon in which carbon could form long chain with it's other carbon
Answer:
Altogether for both models; two red jellybeans, two white jellybeans, two black jellybeans and six blue jellybeans.
<em>Note: Since no specific color was stated for oxygen atoms, the answer assigns blue colored jellybeans to represent oxygen atoms.J</em>
Explanation:
Sodium bicarbonate, NaHCO₃ is a compound composed of one atom of sodium, one atom of hydrogen, one atom of carbon and three atoms of oxygen.
Since red jellybeans represent sodium atoms, white jellybeans represent hydrogen atoms, black jellybeans represent carbon atoms and blue jellybeans represent oxygen atoms, each of the two students will require the following number of each jellybean for their model of sodium carbonate: One red jellybean, one white jellybean, one black jellybean and three blue jellybeans.
Altogether for both models; two red jellybeans, two white jellybeans, two black jellybeans and six blue jellybeans.
Answer : The molar concentration of ethanol in the undiluted cognac is 8.44 M
Explanation :
Using neutralization law,
where,
= molar concentration of undiluted cognac = ?
= molar concentration of diluted cognac = 0.0844 M
= volume of undiluted cognac = 5.00 mL = 0.005 L
= volume of diluted cognac = 0.500 L
Now put all the given values in the above law, we get molar concentration of ethanol in the undiluted cognac.
Therefore, the molar concentration of ethanol in the undiluted cognac is 8.44 M
