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ratelena [41]
3 years ago
5

A blank is 2 or more substances that are together in the same place but are not chemically combined.

Chemistry
1 answer:
hammer [34]3 years ago
7 0

A mixture is a material system made up of two or more different substances, which are mixed but not combined chemically. A mixture refers to the physical combination of two or more substances in which the identities of the individual substances are retained.

Hope this helps!

Please give brainliest!

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increases my factor of 10

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represent rutheniums first 36 electrons using the chemical symbol for krypton written inside the brackets
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Answer:

[kr] hd^75s^1

Explanation:

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Read 2 more answers
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

The pH of the solution ;

pH=-\log[H^+]

=-\log[0.0001901 M]=3.72

3 0
3 years ago
Calculate the volume of 1M NaOH required to neutralize 200 cc of 2M HCI. What mass of sodium
Nezavi [6.7K]

NaOH+HCl-> NaCl+H2O

1 mole of NaOH

1 mole of HCl.

To calculate volume of NaOH

CaVa/CbVb= Na/Nb

Where Ca=2M

Cb=1M

Va=200cm³

Vb=xcm³

Substitute into the equation.

2×200/1×Vb=1/1

400/Vb=1/1

Cross multiply

Vb×1=400×1

Vb=400cm³

To calculate the mass of sodium chloride, NaCl from the neutralization rxn.

Mole of NaCl=1

Molar mass of NaCl= 23+35.5=58.5

Mass=xgrammes.

Mass of NaCl=Number of moles × Molar mass.

Substitute

Mass of NaCl= 1×58.5

=58.5g

This is what I could come up with.

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Is this a question? if so please elaborate.
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