Answer:
2.387 mol/L
Explanation:
The reaction that takes place is:
- 2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O
First we <u>calculate how many moles of each reagent were added</u>:
- HCl ⇒ 200.0 mL * 3.85 M = 203.85 mmol HCl
- Ba(OH)₂ ⇒ 100.0 mL * 4.6 M = 460 mmol Ba(OH)₂
460 mmol of Ba(OH)₂ would react completely with (2*460) 920 mmol of HCl. There are not as many mmoles of HCl so Ba(OH)₂ will remain in excess.
Now we <u>calculate how many moles of Ba(OH)₂ reacted</u>, by c<em>onverting the total number of HCl moles to Ba(OH)₂ moles</em>:
- 203.85 mmol HCl *
= 101.925 mmol Ba(OH)₂
This means the remaining Ba(OH)₂ is:
- 460 mmol - 101.925 mmol = 358.075 mmoles Ba(OH)₂
There are two OH⁻ moles per Ba(OH)₂ mol:
- OH⁻ moles = 2 * 358.075 = 716.15 mmol OH⁻
Finally we <u>divide the number of OH⁻ moles by the </u><u><em>total</em></u><u> volume</u> (100 mL + 200 mL):
- 716.15 mmol OH⁻ / 300.0 mL = 2.387 M
So the answer is 2.387 mol/L
All of the above. If you are going to narrow it down, it would be high voltage and radioactivity.
Answer is C.
Converting temperatures to degrees K: 33 degrees C = 306 K and -55 = 218 degrees K.
By the ideal gas law:-
760 * 0.50 / 306 = P * 0.10 / 218
P = 760 * 0.50 * 218 / 306 * 0.10
= 2700 mm Hg answer
Answer:
Group 8 or Group 0
Explanation:
Group 8 or Group 0 are generally inert gases with Helium as the first member in that group. Their complete duplet (in the case of Helium) and Octet (in the case of Neon) configuration makes them very stable and chemically un-reactive.