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3 years ago

15

Did changing the concentration of the solid effect the change in temperature of the reaction?

2 answers:

wolverine [178]3 years ago

7 0

I’m pretty sure it’s true

Debora [2.8K]3 years ago

5 0

The answer you are looking for is True

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Two chemical bonds unite three atoms into a molecule, what type of bonds are they

marishachu [46]

Answer:

They are covalent bond.

Explanation:

For example; Water which is formed by the combination of 2 hydrogen and 1 oxygen and both are non-metal element.

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3 years ago

What is the estimated effective nuclear charge, Z eff, experienced by an electron in a 3p orbital of a chlorine atom?

Lunna [17]

Because 3pd hope it helps

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3 years ago

The data were collected over five spring seasons. What is the difference between the number of frogs in the pond when the rainfa

Lunna [17]

Answer: they like deep water

Explanation: they do

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3 years ago

The fire department needs information on friction losses occurring between a water main and an open fire hydrant. At maximum mai

lisov135 [29]

Explanation:

The given data is as follows.

$P_{1}$ = 85 psig, $P_{2}$ = $P_{atm}$ = 15 psia

Q = 1620 gpm, d = 2.5 inch, l = 8 ft = 2.4384 m

According to Darey-Weisbach equation,

$h_{l} = \frac{4fl \nu^{2}}{2gD}$ ......... (1)

Value of 'f' will be decided on the basis of Reynold number.

As, it is known that $R_{l} = \frac{\rho \nu d}{\mu}$

where, $\mu_{water}$ = $10^{-3}$ kg/ms

As it is known that 1 gpm = $\frac{1}{3.67} m^{3}/hr$

So, $1 m^{3}/hr$ = 3.67 gpm

Therefore, Q = $1620 \times \frac{1}{3.67}$

= $441.4168 m^{3}/hr$

= 0.1226 $m^{3}/s$

In, 1 inch = 2.54 cm = 0.0254 m

Therefore, d = 2.5 \times 0.0254 = 0.0635 m

V = $\frac{Q}{\frac{\pi}{4}d^{2}}$

= $\frac{0.1226}{0.785 \times (0.0635)^{2}}$

= 38.73 m/s

Hence, we will calculate Reynold number as follows.

$R_{l}$ = $\frac{1000 \times 38.73 \times 0.0635}{10^{-3}}$

= 2459355

As $R_{l}$ > 2000 then, it means that flow is turbulent.

As, f = 0.079 $R^{-0.25}_{l}$

= 0.001994

Putting all the values into equation (1) formula as follows.

$h_{l} = \frac{4fl \nu^{2}}{2gD}$

= $\frac{4 \times 0.001994 \times 2.4384 \times (38.73)^{2}}{2 \times 9.81 \times 0.0635}$

= $1.04069 \times 10^{5} m$

Thus, we can conclude that friction loss from the main to the discharge point is $1.04069 \times 10^{5} m$ .

4 0

3 years ago

HELP ASAP! STUDY ISLAND

madreJ [45]

Explanation:

done here u gooooooooo

6 0

3 years ago