Answer:
NH3(aq) + HNO3(aq) → NH4NO3(aq) Calculate the volume of an acid (1.5 M HNO3) needed to neutralize the 1.5 M HNO3.
Explanation:
I think the given is 3 g sample of NaHCO3. then if it will be reacted with an acid, it will produce H2CO3.
so the reaction NaHCO3 + HCl --> NaCl + H2CO3
mas of H2CO3 = 3 g NaHCO3 ( 1 mol NaHCO3 / 84 g ) ( 1 mol H2CO3 / 1 mol NaHCO3) ( 62.03 g / 1 mol )
mass of H2CO3 = 2.22 g H2CO3
FeBr₃ ⇒ limiting reactant
mol NaBr = 1.428
<h3>Further explanation</h3>
Reaction
2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr
Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)
211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :
186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :
Coefficient ratio from the equation FeBr₃ : Na₂S = 2 : 3, so mol ratio :
So FeBr₃ as a limiting reactant(smaller ratio)
mol NaBr based on limiting reactant (FeBr₃) :
<span>CH</span>₃<span>CH</span>₂<span>COOH + H</span>₂<span>O </span>↔ <span> CH</span>₃<span>CH</span>₂<span>COO</span>⁻<span> + H</span>₃<span>O</span>⁺<span>
</span>
pH = 0.5 pKa + 0.5 pCa
0.5 pCa = pH - 0.5 pKa
= 4.2 - (0.5 * (-log 1.34 x 10⁻⁵)) = 1.76
pCa = 3.53
Ca = antilog - 3.52 = 3 x 10⁻⁴
where Ca is the acid concentration
Answer:
B. 1.65 L
Explanation:
Step 1: Write the balanced equation
2 SO₂(g) + O₂(g) ⇒ 2 SO₃(g)
Step 2: Calculate the moles of SO₂
The pressure of the gas is 1.20 atm and the temperature 25 °C (298 K). We can calculate the moles using the ideal gas equation.
P × V = n × R × T
n = P × V / R × T
n = 1.20 atm × 1.50 L / (0.0821 atm.L/mol.K) × 298 K = 0.0736 mol
Step 3: Calculate the moles of SO₃ produced
0.0736 mol SO₂ × 2 mol SO₃/2 mol SO₂ = 0.0736 mol SO₃
Step 4: Calculate the volume occupied by 0.0736 moles of SO₃ at STP
At STP, 1 mole of an ideal gas occupies 22.4 L.
0.0736 mol × 22.4 L/1 mol = 1.65 L
