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IgorLugansk [536]

3 years ago

5

Melting point 134-137C you range 125-146C is sample pure?

1 answer:

nika2105 [10]3 years ago

5 0

No because the deviation from the theoretical value is quite high and the temperature range of your sample is too wide.

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[24 points] A sample of soil has a total volume of 205 cm3. The soil mass when saturated is 361 g. A specific yield test was con

telo118 [61]

Answer:

Follows are the solution to the given question:

Explanation:

Dry Soil weight = solid soil weight = $284 \ grams$

solid soil volume = $205 \ cc$

saturated mass soil = $361 \ g$

The weight of the soil after drainage is = $295 \ g$

Water weight for soil saturation = $(361-284) = 77 \ g$

Water volume required for soil saturation = $\frac{77}{1} = 77 \ cc$

Sample volume of water: $= \frac{\text{water density}}{\text{water density input}}$

$= 361- 295 \\\\ = 66 \ cc$

Soil water retained volume = (draining field weight - dry soil weight)

$= 295 - 284 \\\\ = 11 \ cc.$

$\text{POROSITY}= \frac{\text{Vehicle volume}}{\text{total volume Soil}}$

$= \frac{77}{(205 + 77)} \\\\= \frac{77}{(282)} \\\\ = 27.30 \%$

(Its saturated water volume is equal to the volume of voids)

$\text{YIELD SPECIFIC} = \frac{\text{Soil water volume}}{\text{Soil volume total}}$

$= \frac{66}{(205+77)}\\\\= \frac{66}{(282)}\\\\=0.2340\\\\ = 0.23$

$\text{Specific Retention}= \frac{\text{Volume of soil water}}{\text{Total soil volume}}$

$= \frac{11}{282} \\\\= 0.0390 \\\\ = 0.04$

6 0

3 years ago

For the reaction below, if the rate of appearance of Br2 is 0.180 M/s, what is the rate of disappearance of

brilliants [131]

Answer:

–0.360 M/s

Explanation:

7 0

3 years ago

PLS PLS HELP!!<br> ILL MARK U BRAINLIEST IF U GET IT RIGHT.<br> THIS IS SCIENCE

elixir [45]

Answer:

rocks and magma

Explanation:

there you go lol ok

5 0

3 years ago

Read 2 more answers

Consider the second-order reaction:

kirza4 [7]

Answer:

Initial concentration of HI is 5 mol/L.

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

Explanation:

$2HI(g)\rightarrow H_2(g)+I_2(g)
$

Rate Law: $k[HI]^2
$

Rate constant of the reaction = k = $6.4\times 10^{-9} L/mol s$

Order of the reaction = 2

Initial rate of reaction = $R=1.6\times 10^{-7} Mol/L s$

Initial concentration of HI = $[A_o]$

$1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2$

$[A_o]=5 mol/L$

Final concentration of HI after t = [A]

t = $4.53\times 10^{10} s$

Integrated rate law for second order kinetics is given by:

$\frac{1}{[A]}=kt+\frac{1}{[A_o]}$

$\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}$

$[A]=0.00345 mol/L$

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

5 0

4 years ago

I was born in ……… west.<br><br>complete this sentense with article

Xelga [282]

Answer:

Article is

I was born in <em><u>t</u></em><em><u>h</u></em><em><u>e</u></em><em><u> </u></em> west .

4 0

3 years ago