Answer: 2 kg/L or 2 g/mL
Explanation:
litres are a cubic measure. “Cubic litres” is redundant at best, potentially confusing.
Answer:
The change in enthalpy in the combustion of 3 moles of methane = -2406 kJ
Explanation:
<u>Step 1: </u>The balanced equation
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) ΔH = -802 kJ
<u>Step 2:</u> Given data
We notice that for 1 mole of methane (CH4), we need 2 moles of O2 to produce : 1 mole of CO2 and 2 moles of H20.
The enthalpy change of combustion, given here as Δ
H
, tells us how much heat is either absorbed or released by the combustion of <u>one mole</u> of a substance.
In this case: we notice that the combustion of 1 mole of methane gives off (because of the negative number), 802.3 kJ of heat.
<u>Step 3: </u>calculate the enthalpy change for 3 moles
The -802 kj is the enthalpy change for 1 mole
The change in enthalpy for 3 moles = 3* -802 kJ = -2406 kJ
The change in enthalpy in the combustion of 3 moles of methane = -2406 kJ
V( H₂SO₄) = 25.0 mL in liters = 25.0 / 1000 = 0.025 L
M(H₂SO₄) = ?
V(NaOH) = 50.0 mL = 50.0 / 1000 = 0.05 L
M(NaOH) = 0.150 M
number of moles NaOH :
n = M x V
n = 0.150 x <span> 0.05
</span>
n = 0.0075 moles of NaOH
H₂SO₄(aq) + 2 NaOH(aq) = Na₂SO₄(aq) + 2 H₂O(l)
1 mole H₂SO₄ ---------- 2 mole NaOH
? mole H₂SO₄ ---------- 0.0075 moles NaOH
moles = 0.0075 * 1 / 2
= 0.00375 moles of H₂SO₄
M(H₂SO₄) = n / V
M = 0.00375 / <span> 0.025
</span>= 0<span>.15 M
</span>
hope this helps!
Answer:
Here's what I get
Explanation:
The reaction is an E1 elimination of an alcohol to form an alkene. It has three steps:
1. Protonation
The alcohol is protonated with aqueous sulfuric acid to convert it into a better leaving group.
2. Loss of the leaving group
A water molecule leaves in a unimolecular process to form a stable 3° carbocation.
3. Loss of an α-hydrogen
A water molecule removes an α-hydrogen, forming 2-methylpropene and regenerating the original hydronium ion.
