Answer :
Option A) 2.00 eV
Explanation : The conversion of J to eV is done with the following formula;
Here, we have the value of particle in terms of Joules which is 3.2 X
So, on substituting we get,
= 3.2 X
X
= 1.99 eV so, it can be rounded off to 2.00 eV.
Answer:
6 Percent Composition. 1. Molar Mass of KBr K = 1(39.10) = 39.10 Br =1(79.90) =79.90 MM = 119.0 79.90 g 119.0 g = 0.6714 3. 0.6714 x 50.0g = 33.6 g Br 2.
Reactivity of non-metals depend on their ability to gain electrons. So, smaller is the size of a non-metal more readily it will attract electrons because then nucleus will be more closer to valence shell. ... Hence, Br is the non-metal which will be more reactive than At.
Ca + 2HCl = CaCl₂ + H₂
c=4.50 mol/l
v=2.20 l
n(HCl)=cv
m(Ca)/M(Ca)=n(HCl)/2
m(Ca)=M(Ca)cv/2
m(Ca)=40g/mol·4.50mol/l·2.20l/2=198 g
198 grams of Ca are needed
Using the Henderson-Hasselbalch equation on the solution before HCl addition: pH = pKa + log([A-]/[HA]) 8.0 = 7.4 + log([A-]/[HA]); [A-]/[HA] = 4.0. (equation 1) Also, 0.1 L * 1.0 mol/L = 0.1 moles total of the compound. Therefore, [A-] + [HA] = 0.1 (equation 2) Solving the simultaneous equations 1 and 2 gives: A- = 0.08 moles AH = 0.02 moles Adding strong acid reduces A- and increases AH by the same amount. 0.03 L * 1 mol/L = 0.03 moles HCl will be added, soA- = 0.08 - 0.03 = 0.05 moles AH = 0.02 + 0.03 = 0.05 moles Therefore, after HCl addition, [A-]/[HA] = 0.05 / 0.05 = 1.0 Resubstituting into the Henderson-Hasselbalch equation: pH = 7.4 + log(1.0) = 7.4, the final pH.
