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3 years ago

11

What fraction of the earth did the former continental glaciers cover?

2 answers:

bija089 [108]3 years ago

5 0

<span>i think one third part of the earth </span>

8090 [49]3 years ago

4 0

Answer:

About 30-35%

Explanation:

Glaciers are defined as the large moving blocks of ice that are formed by the gathering and solidification of snow on the high elevated mountains. The poles and near-polar region are suitable locations for the formation of glaciers as the high latitude areas experience a very cold type of climate.

During the last glaciation or ice age, the maximum portion of the earth that was occupied with glaciers was about 30-35% of the total landmass.

At present, glaciers cover only 10% of the earth's total landmass that includes the glaciers, large ice caps, and the extensive sheets of ice that are present in Antarctica as well as Greenland.

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Read 2 more answers

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

Alex73 [517]

Answer:

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

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

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

8 0

3 years ago