0.50769... It is just the 100g divided by golds molar mass, which is 196.97
Answer:
ΔH0reaction = [ΔHf0 CO2(g)] - [ΔHf0 CO(g) + ΔHf0 O2(g)]
Explanation:
Chemical equation:
CO + O₂ → CO₂
Balanced chemical equation:
2CO + O₂ → 2CO₂
The standard enthalpy for the formation of CO = -110.5 kj/mol
The standard enthalpy for the formation of O₂ = 0 kj/mol
The standard enthalpy for the formation of CO₂ = -393.5 kj/mol
Now we will put the values in equation:
ΔH0reaction = [ΔHf0 CO2(g)] - [ΔHf0 CO(g) + ΔHf0 O2(g)]
ΔH0reaction = [-393.5 kj/mol] - [-110.5 kj/mol + 0]
ΔH0reaction = [-393.5 kj/mol] - [-110.5 kj/mol]
ΔH0reaction = -283 kj/mol
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
I would assume so.
Given

, we can simplify the fraction to

Both would obtain the same proportions, so I don't see why putting a half cup of sugar would make things any different.
Hope this is the answer you are looking for.
Answer: I HOPE THIS HELPS, HAVE A GREAT EARLY HALLOWEEN
Explanation:
PubChem CID: 1084
Molecular Formula: S2O3(2−) or O3S2-2
Synonyms: Thiosulphate THIOSULFATE ION sulfurothioate UNII-LLT6XV39PY Thiosulfate (S2O32-) More...
Molecular Weight: 112.13 g/mol