The balanced equation for the reaction is as follows;
Ca(OH)₂ + 2HBr --> CaBr₂ + 2H₂O
stoichiometry of Ca(OH)₂ to HBr is 1:2
number of Ca(OH)₂ moles reacted - 0.10 mol/L x 0.1000 L = 0.010 mol
Number of HBr moles added - 0.10 mol/L x 0.4000 = 0.040 mol
1 mol of Ca(OH)₂ needs 2 mol of HBr for neutralisation
therefore 0.010 mol of Ca(OH)₂ needs - 0.010 x 2 = 0.020 mol of HBr to be neutralised
but 0.040 mol of HBr has been added therefore number of moles of HBr in excess - 0.040 - 0.020 = 0.020 mol
then pH of the medium can be calculated using the excess H⁺ ions
HBr is a strong acid therefore complete ionization
[HBr] = [H⁺]
[H⁺] = 0.020 mol / (100.0 + 400.0 mL)
= 0.020 mol / 0.5 L
= 0.040 mol/L
pH = -log[H⁺]
pH = - log [0.040 M]
pH = 1.40
pH of the medium is 1.40
<span>a. Tall prarie grass burns after being struck by lightning.</span>
Answer:
The activation energy for the decomposition = 33813.28 J/mol
Explanation:
Using the expression,
Wherem
is the activation energy
R is Gas constant having value = 8.314 J / K mol
Thus, given that, = ?
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (5 + 273.15) K = 278.15 K
T = (25 + 273.15) K = 298.15 K
So,
<u>The activation energy for the decomposition = 33813.28 J/mol</u>
Hydrogen is the limiting reactant because when doing a stoichiometry equation for the reactants, hydrogen will be used completely by having a smaller yield and oxygen will be excess (7 moles to be exact)
Answer:
The answer to your question is 1.51 moles of PBr₃
Explanation:
Data
moles of Br = 2.27
moles of PBr₃ = x
Balanced chemical reaction
2P + 3Br₂ ⇒ 2PBr₃
Reactant Element Products
2 P 2
6 Br 6
-Use proportions to find the answer
3 moles of Br₂ ------------------ 2 moles of PBr₃
2.27 moles of Br₂ ------------------ x
x = (2.27 x 2) / 3
x = 4.54 / 3
x = 1.51 moles