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34kurt
4 years ago
15

During the reaction, 3.50 μmol of HCl are produced. Calculate the final pH of the reaction solution. Assume that the HCl is comp

letely neutralized by the buffer.
Engineering
1 answer:
lyudmila [28]4 years ago
8 0

Answer:

The pH of the solution will be equal to 5.46

Explanation:

The dissociation reaction of HCl is equal to:

HCl → H+ + Cl-

To solve the exercise we must first convert the µmoles to moles using the following conversion factor:

3.5µmoles x \frac{1 mol}{1x10^{6} umol} = 3.5x10^{-6}moles

Assuming a liter of solution, we can calculate the molar concentration by:

M = \frac{Number of moles}{Liter of solution}

Replacing:

M = \frac{3.5x10^{-6}moles }{1 L} = 3.5x10^{-6}moles/L

As this acid dissociates completely, the concentration of protons and chloride will be equal to 3.5x10^{-6}moles/L

The pH will be equal to:

pH = -log[H+]

Replacing:

pH = -log[3.5x10^{-6}] = 5.46

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The speed of sound is 1150 ft/s convert to mile/h
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Answer:

784.090909mph

Explanation:

1ft/s=0.681818 mph

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3 years ago
The input power for a thermostat is wired to the
masya89 [10]

Answer:

A. R

Explanation:

There are basically two wires that supply input power to the thermostat namely the C wire which is the common wire and the R wire .The G wire simply completes the input power circuit from the R-leg of the power supply.On the another hand the Y wire also completes the circuit  from the compressor fan contactor to the R-leg of the power supply . While the W wire completes the path to the heater contactor coil.

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4 years ago
Please calculate the current for the circuit below
Aleks04 [339]

Answer:

The answer is "I = 0.0085106383 \ A"

Explanation:

Given:

R= 470  \ \Omega \\\\V= 4 \ v

Formula:

\to V=IR\\\\\to I = \frac{V}{R}\\\\

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3 0
3 years ago
a) A total charge Q = 23.6 μC is deposited uniformly on the surface of a hollow sphere with radius R = 26.1 cm. Use ε0 = 8.85419
dusya [7]

Answer:

(a) E = 0 N/C

(b) E = 0 N/C

(c) E = 7.78 x10^5 N/C

Explanation:

We are given a hollow sphere with following parameters:

Q = total charge on its surface = 23.6 μC = 23.6 x 10^-6 C

R = radius of sphere = 26.1 cm = 0.261 m

Permittivity of free space = ε0 = 8.85419 X 10−12 C²/Nm²

The formula for the electric field intensity is:

E = (1/4πεo)(Q/r²)

where, r = the distance from center of sphere where the intensity is to be found.

(a)

At the center of the sphere r = 0. Also, there is no charge inside the sphere to produce an electric field. Thus the electric field at center is zero.

<u>E = 0 N/C</u>

(b)

Since, the distance R/2 from center lies inside the sphere. Therefore, the intensity at that point will be zero, due to absence of charge inside the sphere (q = 0 C).

<u>E = 0 N/C</u>

(c)

Since, the distance of 52.2 cm is outside the circle. So, now we use the formula to calculate the Electric Field:

E = (1/4πεo)[(23.6 x 10^-6 C)/(0.522m)²]

<u>E = 7.78 x10^5 N/C</u>

4 0
3 years ago
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