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Paladinen [302]

3 years ago

14

Two commonly used sulfonic acids are methanesulfonic acid (CH3SO3H) and trifluoromethanesulfonic acid (CF3SO3H). Which acid's co

njugate base is the better leaving group?

1 answer:

cluponka [151]3 years ago

4 0

Answer:trifluoromethanesulfonic acid (CF3SO3H).

Explanation:

The trifluoromethanesulfonic acid (CF3SO3H) has a halogen atom which stabilizes the leaving group by withdrawal of charge from the SO3- moiety. The methanesulfonic acid (CH3SO3H) contains an electron pushing group which tends to destabilize the charge centre. The better leaving group will be the stabilized anion which in this case is trifluoromethanesulfonic acid (CF3SO3H). This typifies the role of stabilizing factors in formation of chemical species.

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A gas at 20.0°C and 52.1 mL is heated to 93.5°C. What is the new volume?

lawyer [7]

$\frac{ V_{1} }{ T_{1} }$ = $\frac{ V_{2} }{ T_{2} }$
$V_{1}$ = 0.0521 L
$T_{1}$ = 293.15 K
$T_{2}$ = 366.65 K

Solve for $V_{2}$

$\frac{0.0521 L}{293.15 K}$ = $\frac { V_{2} }{366.65 K}$

$V_{2}$ = 0.06516 L or 65.2 mL

7 0

3 years ago

A student measures the boiling point of water in the lab over three trials and gets the following

maw [93]

I think both are accurate

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3 years ago

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Calculate the enthalpy for this reaction: 2C(s) + H2(g) ---> C2H2(g) ΔH° = ??? kJ Given the following thermochemical equation

nordsb [41]

Answer:

The enthalpy for given reaction is 232 kilo Joules.

Explanation:

$C_2H_2(g) + \frac{5}{2}O_2(g)\rightarrow 2CO_2(g) + H_2O(l), \Delta H^o_{1} = -1,123 kJ$ ...[1]

$C(s) + O_2(g)\rightarrow CO2(g), \Delta H^o_{2} = -340 kJ$ ..[2]

$H_2(g) + \frac{1}{2}O_2(g)\rightarrow H_2O(l) ,\Delta H^o_{3} = -211 kJ$ ..[3]

$2C(s) + H_2(g)\rightarrow C_2H_2(g),\Delta H^o_{4} =?$ ..[4]

2 × [2] + [3] - [1] ( Using Hess's law)

$\Delta H^o_{4}=2\times \Delta H^o_{2}+\Delta H^o_{3} - \Delta H^o_{1}$

$\Delta H^o_{4}=2\times (-340 kJ) + (-211 kJ) - (-1,123 kJ)$

$\Delta H^o_{4}=232 kJ$

The enthalpy for given reaction is 232 kilo Joules.

5 0

3 years ago

Why is nitro group called an ambident group?

IrinaK [193]

Answer:

An ambident ("both teeth") group is a group that can attach to another by either of two atoms. A nitro group can bond through either N or O. In organic chemistry, this property gives rise to functional group isomers.

Explanation:

4 0

3 years ago

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Butane (C4H10) has a heat of vaporization of 22.44 kJ/mol and a normal boiling point of -0.4 ∘C. A 250 mL sealed flask contains

erastova [34]

Given that:

The heat of vaporization = 22.44 kJ/mol = 22440 J/mol
normal boiling point which is the initial temperature = 0.4° C = (273 + (-0.4))K = 272.6 K
volume = 250 mL = 0.250 L
Mass of butane = 0.8 g
the final temperature = -22° C = (273 + (-22)) K = 251 K

The first step is to determine the vapor pressure at the final temperature of 251K by using the Clausius-Clapeyron equation. This is following by using the ideal gas equation to determine the numbers of moles of butane gas. After that, the mass of butane present in the liquid is determined by using the relation for the number of moles.

Using Clausius-Clapeyron Equation:

$\mathbf{In (\dfrac{P_2}{P_1} )= -\dfrac{\Delta H_{vap}}{R}(\dfrac{1}{T_2} - \dfrac{1}{T_1})}$

where;

P1 and P2 correspond to the temperature at T1 and T2.

∴

replacing the values into the given equation, we have;

$\mathbf{In \dfrac{P_2}{1\ atm} = -\dfrac{22440 \ J/mol}{8.314 \ J/mol.K}(\dfrac{1}{251 \ K} - \dfrac{1}{272.6 \ K})}$

$\mathbf{In \dfrac{P_2}{1\ atm} =-(0.852053785)}$

$\mathbf{P_2=0.427 \ atm}$

As such, at -22° C; the vapor pressure = 0.427 atm

Now, using the ideal gas equation:

PV = nRT

where:

P = Pressure
V = volume
n = number of moles of butane
R = universal gas constant
T = temperature

∴

Making (n) the subject of the formula:

$\mathbf{n = \dfrac{PV}{RT}}$

$\mathbf{n = \dfrac{0.427 atm \times 0.250 L}{(0.08206 \ L.atm/k.mol) \times 251}}$

$\mathbf{n =0.00518 mol}$

We all know that the standard molecular weight of butane = 58.12 g/mol

∴

Using the relation for the number of moles which is:

$\mathbf{number \ of \ moles = \dfrac{mass}{molar mass}}$

mass = 0.00518 mole × 58.12 g/mol

mass = 0.301 g

∴

The mass of butane in the flask = 0.301 g

But the mass of the butane present as a liquid in the flask is

= 0.8 g - 0.301 g

= 0.499 g

In conclusion, the mass of the butane present as a liquid in the flask is 0.499 g

Learn more about vapourization here:

brainly.com/question/17039550?referrer=searchResults

7 0

3 years ago