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2 years ago

Define ""acidity"" of an aqueous solution. How do you compare the strength of acidity of solutions ?

Engineering

1 answer:

Vlad [161]2 years ago

7 0

Answer with Explanation:

The acidity of an aqueous solution is a term used to identify how acidic the solution is. An acidic solution is a solution in which the concentration of hydrogen ions is greater than the concentration of hydroxide ions. In the other case around if the concentration of hydrogen ions is lesser than the concentration of hydroxide ions the solution is termed to be basic or alkaline. For a solution with equal concentration of hydrogen and hydroxide ions the solution is termed to be neutral.

The acidity of solutions is compared on the basis of the concentration of the hydrogen ions reduced to log of base 10 to ease calculations. The comparison is made in terms of 'pH' value which is defined as

$pH=-log[H^+]$

where

$[H^+]$ is the hydrogen ion concentration of the solution in moles per liter of solution.

If the pH is < 7 the solution is acidic and the closer the pH value to 1 the higher is the acidity of the solution.

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2. One of the many methods used for drying air is to cool the air below the dew point so that condensation or freezing of the mo

REY [17]

Answer:

0.5°c

Explanation:

Humidity ratio by mass can be expressed as

the ratio between the actual mass of water vapor present in moist air - to the mass of the dry air

Humidity ratio is normally expressed in kilograms (or pounds) of water vapor per kilogram (or pound) of dry air.

Humidity ratio expressed by mass:

x = mw / ma (1)

where

x = humidity ratio (kgwater/kgdry_air, lbwater/lbdry_air)

mw = mass of water vapor (kg, lb)

ma = mass of dry air (kg, lb)

It can be as:

x = 0.005 (100) / [(100 - 100)]

x = 0.005 x 100 / (100 - 100)

x = 0.005 x 100 / 0

x = 0.5°c

So the temperature to which atmospheric air must be cooled in order to have humidity ratio of 0.005 lb/lb is 0.5°c

6 0

3 years ago

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the str

Murljashka [212]

The question is incomplete. The complete question is :

The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?

Solution :

Given data :

Diameter of the rod : 46 mm

Torque, T = 85 Nm

The polar moment of inertia of the shaft is given by :

$J=\frac{\pi}{32}d^4$

$J=\frac{\pi}{32}\times (46)^4$

J = 207.6 $mm^4$

So the shear stress at point A is :

$\tau_A =\frac{Tc_A}{J}$

$\tau_A =\frac{85 \times 10^3\times 12 }{207.6}$

$\tau_A = 4913.29 \ MPa$

Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.

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2 years ago

Which of the following is used in the electrical field?

weeeeeb [17]

Answer:

pliers

Explanation:

because that makes the most sense

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2 years ago

What is A roofed structure that is similar to a porch, but is detached from the house.

agasfer [191]

Answer:

a gazebo

Explanation:

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3 years ago

Read 2 more answers

Write the following decorators and apply them to a single function (applying multiple decorators to a single function): 1. The f

natita [175]

Answer:

Complete question is:

write the following decorators and apply them to a single function (applying multiple decorators to a single function):

1. The first decorator is called strong and has an inner function called wrapper. The purpose of this decorator is to add the html tags of and to the argument of the decorator. The return value of the wrapper should look like: return “” + func() + “”

2. The decorator will return the wrapper per usual.

3. The second decorator is called emphasis and has an inner function called wrapper. The purpose of this decorator is to add the html tags of and to the argument of the decorator similar to step 1. The return value of the wrapper should look like: return “” + func() + “.

4. Use the greetings() function in problem 1 as the decorated function that simply prints “Hello”.

5. Apply both decorators (by @ operator to greetings()).

6. Invoke the greetings() function and capture the result.

Code :

def strong_decorator(func):

def func_wrapper(name):

return "{0}".format(func(name))

return func_wrapper

def em_decorator(func):

def func_wrapper(name):

return "{0}".format(func(name))

return func_wrapper

@strong_decorator

@em_decorator

def Greetings(name):

return "{0}".format(name)

print(Greetings("Hello"))

Explanation:

5 0

3 years ago