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IgorLugansk [536]
2 years ago
9

A substance composed of two or more elements in a fixed, definite proportion is.

Chemistry
2 answers:
Arisa [49]2 years ago
7 0

Answer:

Compound

Explanation:

-compound is a chemical substance composed of many identical molecules composed of atoms from more than one element held together by chemical bonds.

emmasim [6.3K]2 years ago
3 0

\sf{\pmb{ANSWER  :}}

\sf{\pmb{Compound}}

Two or more elements are combined to form a compound with a definite proportion.

<h3>ex :</h3>

\green {{{\sf Na+Cl = NaCl}}}

https://brainly.in/app/profile/18799022

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Calculate the pH and fraction of dissociation ( α ) for each of the acetic acid ( CH 3 COOH , p K a = 4.756 ) solutions. A 0.002
marysya [2.9K]

Answer:

The degree of dissociation of acetic acid is 0.08448.

The pH of the solution is 3.72.

Explanation:

The pK_a=4.756

The value of the dissociation constant = K_a

pK_a=-\log[K_a]

K_a=10^{-4.756}=1.754\times 10^{-5}

Initial concentration of the acetic acid = [HAc] =c = 0.00225

Degree of dissociation = α

HAc\rightleftharpoons H^++Ac^-

Initially

c

At equilibrium ;

(c-cα)                                cα        cα

The expression of dissociation constant is given as:

K_a=\frac{[H^+][Ac^-]}{[HAc]}

1.754\times 10^{-5}=\frac{c\times \alpha \times c\times \alpha}{(c-c\alpha)}

1.754\times 10^{-5}=\frac{c\alpha ^2}{(1-\alpha)}

1.754\times 10^{-5}=\frac{0.00225 \alpha ^2}{(1-\alpha)}

Solving for α:

α = 0.08448

The degree of dissociation of acetic acid is 0.08448.

[H^+]=c\alpha = 0.00225M\times 0.08448=0.0001901 M

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3 years ago
How many mL will a 0.205 mole sample of He occupy at 3.00 atm and 200 K? Report your answer to the nearest mL.
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1.1214 mL will a 0.205-mole sample of He occupy at 3.00 atm and 200 K.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Using equation PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.

Given data:

P= 3.00 atm

V= ?

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R= 0.082057338 \;L \;atm \;K^{-1}mol^{-1}

T=200 K

Putting value in the given equation:

\frac{nRT}{P} =V

V= \frac{0.205 \;mole\;0.082057338 \;L \;atm \;K^{-1}mol^{-1} X 200}{3 \;atm}

V= 1.1214 mL

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