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3 years ago

You do an experiment in which you need 0.25 mole of tyrosine (C,H11 NO3).

Chemistry

1 answer:

Marina CMI [18]3 years ago

6 0

Answer:

B.90.5 g

Explanation:

not 100%sure hehe

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Answer:

Explanation:

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3 years ago

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Answer:

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3 years ago

Consider the reaction: A <=> B. Under standard conditions at equiliubrium, the concentrations of the compounds are [A] = 1

Katen [24]

Answer:

$Keq'>1\\\Delta G'$

Explanation:

Hello,

In this case, for the given reaction, the equilibrium constant turns out:

$Keq=\frac{[B]}{[A]}=\frac{0.5M}{1.5M} =1/3$

Nonetheless, we are asked for the reverse equilibrium constant that is:

$Keq'=\frac{1}{Keq}=3$

Which is greater than one.

In such a way, the Gibbs free energy turns out:

$\Delta G'=-RTln(Keq')\\$

Now, since the reverse equilibrium constant is greater than zero its natural logarithm is positive, therefore with the initial minus, the Gibbs free energy is less than zero, that is, negative.

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3 years ago

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Suppose that you add 26.7 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the adde

nadya68 [22]

Answer: The molar mass of the unknown compound is 200 g/mol

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=2.74^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for molecular compound)

$K_f$ = freezing point constant = $5.12^0C/m$

m= molality

$\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}$

Weight of solvent (benzene)= 0.250 kg

Molar mass of solute = M g/mol

Mass of solute = 26.7 g

$2.74^0C=1\times 5.12\times \frac{26.7g}{Mg/mol\times 0.250kg}$

$M=200g/mol$

Thus the molar mass of the unknown compound is 200 g/mol

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4 years ago

The elements located in the lower left corner of the Periodic Table are classified as

Rasek [7]

In the lower left corner of the Periodic table are elements of the Group 1 and they are called Alkali metals. The Alkali metals are highly reactive elements that readily lose their outmost electron to form cations with charge +1. The elements are: rubidium ( Rb ), caesium ( Cs ) and francium ( Fr ). Answer : 1 ) Metals
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3 years ago