Answer:
Download Ptable, it will show all of the elements, and you can click on them, and it will show all the information you need about them!
Explanation:
I hope this helps!!!!!
Answer:If we dissolve NaF in water, we get the following equilibrium:
text{F}^-(aq)+text{H}_2text{O}(l) rightleftarrows text{HF}(aq)+text{OH}^-(aq)
The pH of the resulting solution can be determined if the K_b of the fluoride ion is known.
20.0 g of sodium fluoride is dissolve in enough water to make 500.0 mL of solution. Calculate the pH of the solution. The K_b of the fluoride ion is 1.4 × 10 −11 .
Step 1: List the known values and plan the problem.
Known
mass NaF = 20.0 g
molar mass NaF = 41.99 g/mol
volume solution = 0.500 L
K_b of F – = 1.4 × 10 −11
Unknown
pH of solution = ?
The molarity of the F − solution can be calculated from the mass, molar mass, and solution volume. Since NaF completely dissociates, the molarity of the NaF is equal to the molarity of the F − ion. An ICE Table (below) can be used to calculate the concentration of OH − produced and then the pH of the solution.
Explanation:
Answer:
The answer to your question is: 85.458 amu
Explanation:
data
Rubidium-85 A = 84.9118 amu abundance = 72.15%
Rubidium - 87 A = 86.9092 amu abundance = 27.85%
Atomic weight = ?
Atomic weight = 84.9118(0.7215) + 86.9092(0.2785)
Atomic weight = 61.2538 + 24.2042
Atomic weight = 85.458 amu
Answer:
149.79
Explanation:
Formula
Joules = m * c * delta (t)
Givens
J = 28242
m = ?
c = 4.19
Δt = 63 - 18 = 45
Solution
28242 = m * 4.19 * 45
28242 = m * 188.55
m = 28242 / 188.55
m = 149.79
