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3 years ago

Determine the hydronium ion concentration Big Points

Chemistry

1 answer:

nalin [4]3 years ago

8 0

Answer:

[H⁺] = 0.00013 M

[OH⁻] = 7.7 × 10⁻¹¹ M

Explanation:

Step 1: Calculate the concentration of H⁺ ions

HCl is a strong acid that dissociates according to the following equation.

HCl ⇒ H⁺ + Cl⁻

The molar ratio of HCl to H⁺ is 1:1. The concentration of H⁺ is 1/1 × 0.00013 M = 0.00013 M.

Step 2: Calculate the concentration of OH⁻ ions

We will use the ionic product of water equation.

Kw = 10⁻¹⁴ = [H⁺] × [OH⁻]

[OH⁻] = 10⁻¹⁴/[H⁺] = 10⁻¹⁴/0.00013 = 7.7 × 10⁻¹¹ M

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

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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

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$h_{l} = \frac{4fl \nu^{2}}{2gD}$ ......... (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 = $1620 \times \frac{1}{3.67}$

= $441.4168 m^{3}/hr$

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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.

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

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

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

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