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3 years ago

A neutral particle formed when atoms share electrons.

Chemistry

1 answer:

Yakvenalex [24]3 years ago

6 0

Answer:

ion

ionic bond

compound

convalent bond

electrons

molecule

Explanation:

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Many ions are said to have a "noble gas configuration" because?

Yakvenalex [24]

Because they follow the octet rule and the outer most electrons, the valence electron shell is full.

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3 years ago

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A student has to design a titration experiment to determine the concentration of an unknown acid. Answer questions a–f.

Paraphin [41]

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3 years ago

In the laboratory, a general chemistry student measured the pH of a 0.529 M aqueous solution of phenol (a weak acid), C6H5OH to

Artyom0805 [142]

Answer:

The dissociation constant of phenol from given information is $9.34\times 10^{-11}$ .

Explanation:

The measured pH of the solution = 5.153

$C_6H_5OH\rightarrow C_6H_5O^-+H^+$

Initially c

At eq'm c-x x x

The expression of dissociation constant is given as:

$K_a=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OOH]}$

Concentration of phenoxide ions and hydrogen ions are equal to x.

$pH=-\log[x]$

$5.153=-\log[x]$

$x=7.03\times 10^{-6} M$

$K_a=\frac{x\times x}{(c-x)}=\frac{x^2}{(c-x)}=\frac{(7.03\times 10^{-6} M)^2}{ 0.529 M-7.03\times 10^{-6} M}$

$K_a=9.34\times 10^{-11}$

The dissociation constant of phenol from given information is $9.34\times 10^{-11}$ .

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4 years ago

What hypothesis can you draw about penicillin

astra-53 [7]

Penicillin V is an antibiotic in the penicillin group of drugs. It fights bacteria in your body. Penicillin V is used to treat many different types of infections caused by bacteria, such as ear infections,. Penicillin V may also be used for other purposes not listed in this medication guide.

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3 years ago

For parts a & b below, derive only the initial value problem set up.

Otrada [13]

Answer:

Explanation:

From the given information:

Let y(t) be the amount of sugar in tank A at any time.

Then, the rate of change of sugar in the tank is given by:

$\dfrac{dy}{dt}= (rate \ in ) - ( rate \ out )$

The rate of the sugar coming into the tank is 0

$\text{rate of sugar going out }= \dfrac{y(t) \ pound}{100}\times \dfrac{1 }{min} \\ \\ = \dfrac{y}{100} \ pound/min$

$So; \\ \\ \\ \dfrac{dy}{dt} = 0 - \dfrac{y}{100} \\ \\ \implies \dfrac{dy}{y }= -\dfrac{dt}{100 } \\ \\ \implies dx|y| = -\dfrac{t}{100}+C \\ \\ \implies e*{In|y|}= e^{-\dfrac{t}{100}+C} \\ \\ \implies y = Ce^{-\dfrac{t}{100}}$

Initial amount of sugar = 25 Pounds

Now; y(0) = 25

25 = Ce⁰

C = 25

So; y(t) = 25 $e^{-\dfrac{t}{100}$

Thus, the amount of sugar at any time t = $\mathbf{25 e^{^{-\dfrac{t}{100}}}}$

B) For tank B :

$\dfrac{dy}{dt}= 1 - \dfrac{y}{50 } \\ \\ \dfrac{dy}{dt}= \dfrac{50-y}{50} \\ \\ \dfrac{dy}{50-y }= \dfrac{dt}{50}$

8 0

3 years ago