Substances that donate H⁺ in water are Brønsted–Lowry acids, and substances that accept H⁺ are Brønsted–Lowry bases.
H₃PO₄ and H₂PO₄⁻ can both act as Brønsted–Lowry acids because each can donate a proton. However, H₃PO₄ is a stronger acid than H₂PO₄⁻ since it carries no negative charge and more readily donates H⁺.
I⁻ and Ba(OH)₂ act as bases.
I⁻ can accept a proton to form HI.
Ba(OH)₂ provides OH⁻ ions, which readily accept H⁺ to form water.
H₂PO₄⁻ is amphiprotic—it can also act as a base by accepting a proton to form H₃PO₄.

Hi Mam,

I was wondering how to identify the Bronsted Lowry Acids and Bases?

Thank you  for the help!

Your method is correct however the value is for two moles of C4H10.In the balanced equation it is 2 moles of butane on combustion gives −5756 kJ of energy.
You need to take this ratio-
 580 g * (1mol C₄H₁₀ / 58.12g  ) * (−5756 kJ / 2 moles C₄H₁₀)

Not sure why answer choice C is wrong given my work

 

To show water and NO as the product
I will take equation 1 as such and will multiply eq 2  with 3/2 and eq3 as such and then add them together
Final equation will be
8NH3 + 13O2 ---> 4NO + 4HNO3 + 10H2O
∆H = -905 -3/2(112)-138 = -1211kJ
Rxn is exothermic

Born–Haber Cycle Steps:

Sublimation of Mg(s) → Mg(g)→ +150 kJ/mol

1st Ionization: Mg(g) → Mg⁺(g)→  +735 kJ/mol

2nd Ionization: Mg⁺(g) → Mg²⁺(g) → +1445 kJ/mol

Bond dissociation: F₂ →  2 F atoms  →  +154 kJ/mol

Electron affinity: 2 F(g) → 2 F⁻(g)  → 2 × (−328) = −656 kJ/mol

Lattice formation: Mg²⁺ + 2F⁻ → MgF₂(s) → −2913 kJ/mol
Total Enthalpy (ΔHf°):
Δ 𝐻 𝑓∘=
=(150+735+1445+154)−(656+2913)=2484−3569= −1085 kJ/mol
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I want practice and a better understanding on how to compute these types of questions.

I can't figure out the formula for this problem, NO should cancel out.

 This sequence converts the epoxide into an allylic bromide (alkene) by:
(acidic epoxide opening → make a good leaving group → E2 elimination).
Step 1: HBr (acidic epoxide opening)
The epoxide oxygen is protonated first.
Br⁻ attacks the more substituted epoxide carbon (the carbon bearing the Me), because the opening has carbocation-like character under acidic conditions.
Ring opens anti, giving a bromohydrin:
Br ends up on the Me-bearing carbon
OH ends up on the adjacent carbon
Step 2: TsCl, pyridine
Converts the alcohol into a tosylate:
–OH → –OTs (excellent leaving group)
Now you have a bromo-tosylate on adjacent carbons.
Step 3: tBuOK / heat (E2)
E2 elimination occurs, and it will preferentially expel OTs (better leaving group than Br).
The tosylate-bearing carbon has β-carbons on both sides, but:

The Me/Br-bearing carbon has no β-H (it is fully substituted: ring bonds + Me + Br)
Therefore elimination cannot form the C=C back to that carbon.
So the base must remove the β-H on the other adjacent carbon, giving the alkene one position away.
✅ Final product: 1-bromo-1-methylcyclohex-2-ene
(Br and Me on the same carbon; double bond between the next two carbons).

 
For this sequence, the blank box should show the deprotected alcohol plus a deuterohydrin (OH + D) added across the original double bond.

Step 1: TMSCl, pyridine
Protect the existing alcohol: ROH → ROTMS (configuration at that carbon is unchanged).

Step 2: peracid (mCPBA-type)
Epoxidation of the C=C happens from the less hindered face, i.e. opposite the bulky OTMS substituent.
Since the starting OH is drawn behind (dash), the protected OTMS is also "behind", so the epoxide oxygen is formed mainly on the front face.

Step 3–4: LAD, then H₂O
LiAlD₄ opens the epoxide by backside attack (anti opening).
Attack occurs at the less hindered epoxide carbon (the one farther from the OTMS side), so:

D ends up on that attacked carbon (back face)

The epoxide oxygen becomes OH on the other carbon (front face after workup)
So D and the new OH are trans (anti).

Step 5: TBAF
Removes the TMS group → restores the original OH (still drawn as dash, same stereochemistry as the start).