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2 years ago

Which of the following compounds has the highest pKa value? A)CCl3CH2NH2

1 answer:

3 0

The lower the pKa of a Bronsted acid, the more easily it gives up its proton. The higher the pKa of a Bronsted acid, the more tightly the proton is held, and the less easily the proton is given up.

Here we need the highest pKa, so we need to see which compound will less likely to give proton or hydrogen ion.

Now, all Nitrogen contains a lone pair. But HALOGEN groups( F, Cl, only) being electronegative than NITROGEN [electronegativity of N=3, F=4 and Cl=3], pulls electron pair towards itself.

The more the lone pair of nitrogen is pulled, the more strong bond between N and H will become, which means less likely to give hydrogen ion.
means high Pka

C) option is the answer because it has 3 F very close to N.

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It requires 0.0780L of a 0.12 M HCl solution to completely neutralize 0.0280L of an unknown LiOH solution. What is the concentra

kari74 [83]

Answer:

C₂ = 0.334 M

Explanation:

Given data:

Volume of HCl = 0.0780 L

Concentration of HCl = 0.12 M

Volume of LiOH = 0.0280 L

Concentration of LiOH = ?

Solution:

Formula:

C₁V₁ = C₂V₂

C₁ = Concentration of HCl

V₁ = Volume of HCl

C₂ = Concentration of LiOH

V₂ = Volume of LiOH

Now we will put the values in formula.

C₁V₁ = C₂V₂

0.12 M × 0.0780 L = C₂ × 0.0280 L

0.00936 M.L = C₂ × 0.0280 L

C₂ = 0.00936 M.L/0.0280 L

C₂ = 0.334 M

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1 year ago

The specific heat of aluminum is 0.214 cal/g.oC. Determine the energy, in calories, necessary to raise the temperature of a 55.5

Natasha2012 [34]

For this problem, we use the formula for sensible heat which is written below:

Q= mCpΔT
where Q is the energy
Cp is the specific heat capacity
ΔT is the temperature difference

Q = (55.5 g)(0.214 cal/g·°C)(48.6°C- 23°C)
Q = 304.05 cal

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2 years ago

Calculate the wavelength of the photon emitted when an electron makes a transition from n=6 to n=3. You can make use of the foll

Angelina_Jolie [31]

Answer: The wavelength of light is $1.094\times 10^{-6}m$

Explanation:

To calculate the wavelength of light, we use Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = $1.097\times 10^7m^{-1}$

$n_f$ = Final energy level = 3

$n_i$ = Initial energy level = 6

Putting the values in above equation, we get:

$\frac{1}{\lambda }=1.097\times 10^7m^{-1}\left(\frac{1}{3^2}-\frac{1}{6^2} \right )\\\\\lambda =\frac{1}{914617m^{-1}}=1.094\times 10^{-6}m$

Hence, the wavelength of light is $1.094\times 10^{-6}m$

6 0

2 years ago

In a 0.01 M solution of HCl, Litmus will be

Nadya [2.5K]

In a 0.01 M solution of HCl, Litmus will be red. Litmus paper will turn into red in acidic conditions. Hydrochloric acid is an acid. Litmus is an indicator for acidity and alkalinity made from inchens.

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2 years ago

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How many moles of O2 should be supplied to burn 1 mol of C3H8 (propane) molecules in a camping stove

Shkiper50 [21]

The combustion of any hydrocarbon yields water and carbon dioxide. We will now construct a balanced equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Each mole of propane requires 5 moles of oxygen.

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1 year ago