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2 years ago

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What is the empirical formula of a carbon-oxygen compound, given that a 95.2 g sample of the compound contains 40.8 g of carbon

and the rest oxygen?

1 answer:

algol [13]2 years ago

4 0

Answer:

95.2 - 40.8 = 54.4 g of oxygen

number of moles = mass (g)/ Mr

no. of moles of carbon = 40.8/12 = 3.4

no. of moles of oxygen = 3.4

divide both by smallest value which is 3.4 and you’ll get 1 mole of carbon and 1 mole of oxygen therefore the empirical formula is CO

Explanation:

hope this helps :)

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A soft silvery metal has two naturally occurring isotopes: mass 84.9118, accounting for 72.15% and mass 86.9092, accounting for

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Answer: The atomic weight of the metal would be 85.47.

Explanation:

Mass of isotope 1 of metal = 84.9118

% abundance of isotope 1 of metal = 72.15% = $\frac{72.15}{100}$

Mass of isotope 2 of metal= 86.9092

% abundance of isotope 2 of metal = 27.85% = $\frac{27.85}{100}$

Formula used for average atomic mass of an element :

$\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})$

$A=\sum[(84.9118)\times \frac{72.15}{100})+(86.9092)\times \frac{27.85}{100}]]$

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A flash distillation chamber operating at 101.3 kpa is separating an ethanol water mixture the feed mixture contains z weight et

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The answer is $[\frac{LX_1}{46} + \frac{L(1-x)}{18}]\times0.454$ mol/hr

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Explanation:

For flash distillation

F = V+L

$\frac{V}{F} + \frac{L}{F} = 1$

$\frac{F}{V} -\frac{L}{V} = 1$

Fz = Vy+Lx

Y = $\frac{F}{V}\times Z - \frac{L}{V}\times X$ let, $\frac{V}{F} = F$

$y = \frac{Z}{F} -[ \frac{1}{F} -1]\times X$

Highlighted reading

F = 299; $\frac{V}{F}$ = 0.85 ; z = 0.36

y = $\frac{0.36}{0.85} - (-0.15)\times X$

= 0.423 + 0.15x ------------(i)

$y^{*}$ = -43.99713 $x^{6}$ + 148.27274 $x^{5}$ - 195.46 $x^{4}$ +127.99 $x^{3}$ -43.3 $x^{2}$ + 7.469 $x^{}$ + 0.02011

At equilibrium, $y^{*}$ = y

0.423+0.15 $x^{}$ = $y^{*}$

-43.99713 $x^{6}$ + 148.27274 $x^{5}$ - 195.46 $x^{4}$ +127.99 $x^{3}$ -43.3 $x^{2}$ + 7.319 $x^{}$ -0.403

F(x) for Newton's Law

Let $x_{0} = 0$

$x_{1}$ = $\frac{0-[{-0.403}]}{7.319}$

= 0.055

$x_{2}$ = $\frac{{0.055}-{f(0.055)} {{{{{{{}}}}}}}}{f^{'} (0.055)}$

= $\frac{{(0.055)}-(-0.11)}{3.59}$

= 0.085

$x^{3}$ = $\frac{{0.085}-(0.024)}{2.289}$

= 0.095

$x^{4}$ = $\frac{{0.095}-(-0.0353)}{-1.410}$

= 0.07

From This x and y are found from equation (i) and L and V are obtained from $\frac{V}{F}$ and F values

$[\frac{LX_1}{46} + \frac{L(1-x)}{18}]\times0.454$ mol/hr

$[\frac{Vy_1}{46}+\frac{V(1-y)}{18}]\times0.454$ mol/hr

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