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3 years ago

6

Does any one here have morsin solutions?

1 answer:

miskamm [114]3 years ago

8 0

Answer:

THE ANSWERS IS A MODERN ANSWERS IS THE BIOLOGY WAY

Explanation:

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Can someone help me answer C

Digiron [165]

because B has more Surface are.

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3 years ago

Please hurry! classify the following as a

Talja [164]

Answer:

pure substance.

Explanation:

i hope that helps

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2 years ago

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What is the molecularity of this elementary step?

dybincka [34]

The correct answer is b. bimolecular

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4 years ago

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What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig

Mashutka [201]

Answer:

$P_H =2.86$

$c=1.4\times 10^{-4}$

Explanation:

first write the equilibrium equaion ,

$C_3H_6O_3$ ⇄ $C_3H_5O_3^{-} +H^{+}$

assuming degree of dissociation $\alpha$ =1/10;

and initial concentraion of $C_3H_6O_3$ =c;

At equlibrium ;

concentration of $C_3H_6O_3 = c-c\alpha$

$[C_3H_5O_3^{-} ]= c\alpha$

$[H^{+}] = c\alpha$

$K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}$

$\alpha$ is very small so $1-\alpha$ can be neglected

and equation is;

$K_a = {c\alpha \times \alpha}$

$[H^{+}] = c\alpha$ = $\frac{K_a}{\alpha}$

$P_H =- log[H^{+} ]$

$P_H =-logK_a + log\alpha$

$K_a =1.38\times10^{-4}$

$\alpha = \frac{1}{10}$

$P_H= 3.86-1$

$P_H =2.86$

composiion ;

$c=\frac{1}{\alpha} \times [H^{+}]$

$[H^{+}] =antilog(-P_H)$

$[H^{+} ] =0.0014$

$c=0.0014\times \frac{1}{10}$

$c=1.4\times 10^{-4}$

6 0

4 years ago

A standard 1.00 kg mass is to be cut from a bar of steel having an equilateral triangular cross section with sides equal to 2.50

Elden [556K]

Answer:

The section of the bar is 2.92 inches.

Explanation:

Mass of the steel cut ,m = 1.00 kg = 1000 g

Volume of the steel bar = V = Area × height

Height of the of the section of bar = h

Area of Equilateral triangular = $\frac{\sqrt{3}}{4}a^2$

a = 2.50 inches

Cross sectional area of the steel mass = A

$A=\frac{\sqrt{3}}{4}(2.50 inches)^2=2.71 inches^2$

$V = 2.71 inches^2\times h$

Density of the steel = d = $7.70 g/cm^3$

$1cm^3 = 0.0610237 inches^3$

$d=\frac{m}{v}$

$\frac{7.70 g}{0.0610237 inches^3}=\frac{ 1000 g}{2.71 inches^2\times h}$

$h=\frac{ 1000 g\times 0.0610237 inches^3}{2.71 inches^2\times 7.70 g}$

h = 2.92 inches

The section of the bar is 2.92 inches.

3 0

3 years ago