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3 years ago

Please Help ASAP!!!!!!!!!!!!!!!!!!

Chemistry

1 answer:

eduard3 years ago

6 0

Answer:

Using the drop down Menu

Proton : Possessing a Charge of +1. It is found in the Nucleus and has a Mass of "A"

Neutron; Possesses a charge of 0. It is situated in the Nucleus and Has a Mass of "B"

Electron: Has a charge of -1. Found in the Orbitals of the atom and has a mass of "C"

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3 years ago

Which of the following bases are strong enough to deprotonate CH3CH2CH2C≡CH (pKa = 25), so that equilibrium favors the products?

seraphim [82]

Answer:

a, and f.

Explanation:

To be deprotonated, the conjugate acid of the base must be weaker than the acid that will react, because the reactions favor the formation of the weakest acid. The pKa value measures the strength of the acid. As higher is the pKa value, as weak is the acid. So, let's identify the conjugate acid and their pKas:

a. NaNH2 will dissociate, and NH2 will gain the proton and forms NH3 as conjugate acid. pKa = 38.0, so it happens.

b. NaOH will dissociate, and OH will gain the proton and forms H2O as conjugate acid. pKa = 14.0, so it doesn't happen.

c. NaC≡N will dissociate, and CN will gain a proton and forms HCN as conjugate acid. pKa = 9.40, so it doesn't happen.

d. NaCH2(CO)N(CH3)2 will dissociate and forms CH3(CO)N(CH3)2 as conjugate acid. pKa = -0.19, so it doesn't happen.

e. H2O must gain one proton and forms H3O+. pKa = -1.7, so it doesn't happen.

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3 years ago

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3 years ago

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How many half-lives are required for the concentration of reactant to decrease to 1.56% of its original value?4247.56.56

neonofarm [45]

Answer:

6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.

Explanation:

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given:

Concentration is decreased to 1.56 % which means that 0.0156 of $[A_0]$ is decomposed. So,

$\frac {[A_t]}{[A_0]}$ = 0.0156

Thus,

$\frac {[A_t]}{[A_0]}=e^{-k\times t}$

$0.0156=e^{-k\times t}$

kt = 4.1604

The expression for the half life is:-

Half life = 15.0 hours

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$\frac{4.1604}{t}=\frac {ln\ 2}{t_{1/2}}$

$t = 6\times t_{1/2}$

<u>6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.</u>

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