Answer:
2.6 ×10^-42
Explanation:
From
∆G= -RTlnK
∆G= -237.2 KJmol-1 or -237.2×10^3 Jmol-1
R= 8.314 Jmol-1K-1
T= 25°C + 273= 298K
-237.2×10^3= 8.314 × 298 × ln K
ln K= -237.2×10^3/2477.572
K = 2.6 ×10^-42
Ionic bonds involve a cation and an anion. The bond is formed when an atom, typically a metal, loses an electron or electrons, and becomes a positive ion, or cation. Another atom, typically a non-metal, is able to acquire the electron(s) to become a negative ion, or anion.
One example of an ionic bond is the formation of sodium fluoride, NaF, from a sodium atom and a fluorine atom. In this reaction, the sodium atom loses its single valence electron to the fluorine atom, which has just enough space to accept it. The ions produced are oppositely charged and are attracted to one another due to electrostatic forces.
If 10 g of NaOH is diluted to 500 ml with water then the concentration expressed in percent is 0.5 mol/L .
Calculation ,
Given mass in gram = 10 g
Number of moles = given mass /molar mass = 10 g / 40 g/mol = 0.25mole
Given volume in ml = 500 ml
Given volume in liter = 0.5 L
Putting the value of mass and volume in equation i we get concentration expressed in percent .
C = number of moles ×100/ volume in liter = 0.25mole ×100/ 0.5 L
C = 0.5 mol/L
Concentration of solution in terms of percentage can be expressed in two ways
1) percentage by mass
2) percentage by volume
Hence, for liquid solutions, concentration is expressed in terms of percentage by volume.
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The answer would be A) sand, it is not soluble in water
Answer:
2 KClO3 (s) = 2 KCl (s) + 3 O2 (g)
2.5 g x g
Explanation:
x g O2 = 2.5 g KClO3 x (1 mol KClO3) x (3 mol O2) x (32 g O2) = 0.98 g O2
(122.5 g KClO3) (2 mol KClO3) (1 mol O2)
2 KClO3 (s) 2 KCl (s) + 3 O2 (g)
2.5 g x g
x g KCl = 2.5 g KClO3 x (1 mol KClO3) x (2 mol KClO3) x (74.5 g KCl) = 1.52 g KCl
(122.5 g KClO3) (2 mol KClO3) (1 mol KCl)
2 KClO3 (s) 2 KCl (s) + 3 O2 (g)
x mol 10 mol
x mol KClO3 = 10 mol O2 x (2 mol KClO3) = 6.7 mol KClO3
(3 mol O2)
