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3 years ago

15

I NEED HELP FAST, BRAINLIEST! :) Why is water a good choice of fluid in a heat exchanger?

2 answers:

nadezda [96]3 years ago

4 0

Answer:

Water is a good heat transfer fluid as it has a high thermal capacity and low viscosity. It is cheaper to use because its application in direct steam generation saves cost in the heat exchanger. However, it is unstable and difficult to manage at high temperatures.

Explanation:

hope this helps

maw [93]3 years ago

4 0

Answer:

Because it has a high thermal capacity and low viscosity. :)

Explanation:

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What is the density of an 84.7 g sample of an unknown substance if the sample occupies 49.6cm cubed

Ahat [919]

Answer:

1.70 g.cm⁻³

Solution:

Data Given;

Mass = 84.7 g

Volume = 49.6 cm³

Density = ?

Formula Used;

Density = Mass ÷ Volume

Putting values,

Density = 84.7 g ÷ 49.6 cm³

Density = 1.70 g.cm⁻³

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3 years ago

Which of these is NOT a lipid?

Minchanka [31]

Answer: RNA

Explanation:

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2 years ago

Calculate the dissociation constant of nh4oh(aq) if the degree of dissociation of 0.006 mol/kg solution is 0.053 and the activit

Anastasy [175]

The dissociation equation will be

NH4OH ---> NH4+ + OH-

Initial 0.006 0 0

Change -0.006 X 0.053 +0.006 X 0.053 -0.006 X 0.053

Equlibrium 0.006 -0.006 X 0.053 0.006 X 0.053 0.006 X 0.053

Ka = [NH4+] [ OH-] / [NH4OH] = (0.006 X 0.053)^2 / 0.006 -0.006 X 0.053

Ka = 1.78 X 10^-5

7 0

3 years ago

Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =

Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Wherem

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314 J / K mol

Thus, given that, $E_a$ = ?

$k_2=4.0\times 10^3s^{-1}$

$k_1=1.5\times 10^3s^{-1}$

$T_1=5\ ^0C$

$T_2=25\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (5 + 273.15) K = 278.15 K

T = (25 + 273.15) K = 298.15 K

$T_1=278.15\ K$

$T_2=298.15\ K$

So,

$\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)$

$E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}$

$E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}$

$E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}$

$E_a=33813.28\ J/mol$

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

8 0

3 years ago

At what temperature do the gas and liquid phases become<br> indistinguishable from each other?

sergiy2304 [10]

Answer:

100 C or 212 F

Explanation:

That is the boiling point for liquid so both gas and liquid are presented.

5 0

3 years ago

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