1. Is charge 2. Is the first one
<span>Best Answer: Mn(ii) is catalyst
Stetep-1 is slow step
Steps-2,3 are fast steps
intermediates are Mn(iii)and Mn(iv)
since step -1 is slow rate depends on Ce(iv) and Mn(ii) only
not on Tl(i) as it is involved in fast step-3</span>
False. Due to:
Entropy increases with molecular size (mass of compound)
HBr has greater molecular size (molar mass) than HCl
The value of standard entropy of HCl(g) is 187 joules/ kelvin
The value of standard entropy of HBr(g) is 199 joules / kelvin
Answer:
65 g (C₂H₅)₂NH₂Br
Explanation:
It seems your question lacks the values required to solve the problem. However, an internet search tells me these are the values (if your values are different, keep those in mind when solving the problem, but the methodology remains the same):
" A chemistry graduate student is given 125 mL of a 0.90 M diethylamine ((C₂H₅)₂NH) solution. Diethylamine is a weak base with Kb=1.3×10⁻³. What mass of (C₂H₅)₂NH₂Br should the student dissolve in the (C₂H₅)₂NH solution to turn it into a buffer with pH=10.53? You may assume that the volume of the solution doesn't change the (C2H5)2NH2Br is dissolved in it. Be sure your answer has a unit symbol, and round it to 2 significant digits. "
To solve this problem we use the <em>Henderson-Hasselbach equation</em>:
- pH = pKa + log
We <u>calculate pKa from Kb</u>:
- Ka = Kw/Kb ⇒ Ka = 1x10⁻¹⁴/1.3×10⁻³ = 7.69x10⁻¹²
Now we possess all the required data to calculate the concentration of the bromide salt (C₂H₅)₂NH₂Br :
- pH = pKa + log
- 10.53 = 11.11 + log
- -0.58 = log
- 10⁻⁰°⁵⁸=
Now we use the total volume to<u> calculate the moles of (C₂H₅)₂NH₂Br required to have that molar concentration</u>:
- 3.42 M * 0.125 L = 0.4275 mol (C₂H₅)₂NH₂Br
Finally we calculate <u>the mass of (C₂H₅)₂NH₂Br</u>:
- 0.4275 mol (C₂H₅)₂NH₂Br * 154.05 g/mol = 65.86 g
Which rounding to 2 significant figures is 65 g (C₂H₅)₂NH₂Br
It depends on what kind of reaction it is. it could be 10 moles too, but it can be less, for example, if the products looked like this
CO2+2 H2O the answer would be:
x mol CO2 - 10 mol H20
1 mol CO2 - 2 mol H2O