Methyle orange is the indicator that is used in sulfuric acid.
Answer:
162 kJ
Explanation:
The reaction given by the problem is:
- NBr₃(g) + 3 H₂O(g) → 3 HOBr(g) + NH₃(g) ∆H = +81 kJ
If we turn it around, we have:
- 3 HOBr(g) + NH₃(g) → NBr₃(g) + 3 H₂O(g) ∆H = -81 kJ
If we think now of HOBr and NH₃ as our reactants, then now we need to find out <u>which one will be the </u><em><u>limiting reactant</u></em> when we have 9 moles of HOBr and 2 moles of NH₃:
- When we have 1 mol NH₃, we need 3 mol HOBr. So when we have 2 moles NH₃, we need 6 moles HOBr. We have more than 6 moles HOBr so that's our <em>reactant in excess</em>, thus NH₃ is our limiting reactant.
-81 kJ is our energy change when there's one mol of NH₃ reacting, so we <u>multiply that value by two when there's two moles of NH₃ reacting</u>. The answer is 81*2 = 162 kJ.
Mol = # atoms ÷ Avagadaro's number (6.02×10^23)
mol = 2.4×10^24 ÷ 6.02×10^23
mol = 3.986 mol
Answer:
It takes 178.4 years to drop the activity from 15.0 Ci to 0.25 Ci
Explanation:
Step 1: Data given
Half-life time = 30.2 years
Initial activity = 15.0 Ci
Final activity = 0.250 Ci
Step 2: Calculate the time needed
N / No = e^(-0.693 * t / T1/2)
⇒ with N = The activity after dropped = 0.250 Ci
⇒ with N0 = the initial activity = 15.0
⇒ with t= the time (in years) needed to drop the activity from 15.0 to 0.250
⇒ with t1/2 = the half- life time is 30.2 years
0.250 / 15.0 = e^(-0.693 * t / 30.2)
ln(0.250/15.0) = (-0.693 * t / 30.2)
-4.09 * 30.2 = -0.693t
-123.65 = -0.693t
t= 178.4 years
It takes 178.4 years to drop the activity from 15.0 Ci to 0.25 Ci