Answer:
4 × 10 g
Explanation:
Step 1: Write the balanced equation
2 H₂(g) + O₂(g) ⇒ 2 H₂O(I)
Step 2: Calculate the moles corresponding to 4 g of H₂
The molar mass of H₂ is 2.02 g/mol.
4 g × 1 mol/2.02 g = 2 mol
Step 3: Calculate the moles of H₂O produced from 2 moles of H₂
The molar ratio of H₂ to H₂O is 2:2. The moles of H₂O produced are 2/2 × 2 mol = 2 mol.
Step 4: Calculate the mass corresponding to 2 moles of H₂O
The molar mass of H₂O is 18.02 g/mol.
2 mol × 18.02 g/mol = 4 × 10 g
The theoretical yield of NaBr given that 2.36 moles of FeBr₃ reacts is 7.08 moles
<h3>Balanced equation </h3>
2FeBr₃ + 3Na₂S → Fе₂S₃ + 6NaBr
From the balanced equation above,
2 moles FeBr₃ reacted to produce 6 moles of NaBr
<h3>How to determine the theoretical yield of NaBr</h3>
From the balanced equation above,
2 moles FeBr₃ reacted to produce 6 moles of NaBr
Therefore,
2.36 moles FeBr₃ will react to produce = (2.36 × 6) / 2 = 7.08 moles of NaBr
Therefore,
Thus, the theoretical yield of NaBr is 7.08 moles
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In order to deprotonate an acid, we must remove protons in order to achieve a more stable conjugate base. For this example, we can use the relationship between carboxylic acid and hydroxide.
Deprotonation is the removal of a proton from a specific type of acid in reaction to its coming into contact with a strong base. The compound formed from this reaction is known as the conjugate base of that acid. The opposite process is also possible and is when a proton is added to a special kind of base. This is a process referred to as protonation, which forms the conjugate acid of that base.
For the example we have chosen to give, the conjugate base is the carboxylate salt. This would be the compound formed by the deprotonated carboxylic acid. The base in question was strong enough to deprotonate the acid due to the greater stability offered as a conjugated base.
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Answer:
P2= 125.26 Kpa
Explanation:
V1= 489.6 ml=0.4896L
V2= 750 ml= 0.750L
V1= 180 KPa= 180000 Pa
P2= ?
T1= 10 = 10 + 273.15 = 283.15K
T2= 28.7+273.15= 301.85K
180000Pa* 0.4896L/ 283.15K * 301.85K/0.75L
P2= 12526.28553
P2= 125.26 KPa
