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3 years ago

Why factory waste should be neutralized before disposing it into the water bodies?

Chemistry

1 answer:

Fudgin [204]3 years ago

6 0

Explanation:

Factory wastes are usually acidic and they pose serious environmental challenges.

If an untreated acidic waste is discharged into the river, the acidity of the river increases and this is toxic to life in the water.

On land, the acid can enter into the soil and prevent plant growth.

Acid discharge can enter the groundwater system and render it unfit for use.

Bases and a couple of additives are used to remove acidity from factory wastes before they are discharged.

Learn more:

Neutralization brainly.com/question/4455839

#learnwithBrainly

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10.0 mL of a HF solution was titrated with a 0.120 N solution of KOH; 29.6 mL of

Ilia_Sergeevich [38]

Answer:

0.355 N of HF

Explanation:

The titration reaction of HF with KOH is:

HF + KOH → H₂O + KF

Where 1 mole of HF reacts per mole of KOH

Moles of KOH are:

0.0296L × (0.120 equivalents / L) = 3.552x10⁻³ equivalents of KOH = equivalents of HF.

As volume of the titrated solution was 10.0mL, normality of HF solution is:

3.552x10⁻³ equivalents of HF / 0.010L = 0.355 N of HF

3 0

3 years ago

A hydrogen atom in an excited state emits a photon of frequency ν = 3.084 x 1015 s-1. If the electron returns to the ground stat

VLD [36.1K]

Answer: The photon was emitted from the 4th energy level

Explanation:

To calculate the wavelength of light, we use the equation:

$\lambda=\frac{c}{\nu}$

where,

$\lambda$ = wavelength of the light

c = speed of light = $3\times 10^8m/s$

$\nu$ = frequency of light = $3.084\times 10^{15}s^{-1}$

Putting the values in above equation, we get:

$\lambda=\frac{3\times 10^8m/s}{3.084\times 10^{15}s^{-1}}=9.73\times 10^{-8}m$

To calculate the initial level, we use the equation given by Rydberg's equation, which is:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2} \right )$

where,

$\lambda$ = Wavelength of radiation $9.73\times 10^{-8}m$

$R_H$ = Rydberg's Constant = $1.097\times 10^7m^{-1}$

$n_f$ = Final energy level = 1

$n_i$ = Initial energy level = ?

Putting the values, in above equation, we get

$\frac{1}{9.73\times 10^{-8}m}=1.097\times 10^7m^{-1}\left(\frac{1}{1^2}-\frac{1}{n_i^2} \right )$

$n_i=4$

Hence, the photon was emitted from the 4th energy level

4 0

3 years ago

What are the intervals on the fahrenheit scale called

Darya [45]

Fahrenheit temperature scale fâr´ənhīt˝ [key], temperature scale in which the temperature difference between two reference temperatures, the melting and boiling points of water, is divided into 180 equal intervals called degrees.

4 0

3 years ago

A certain heat engine operates between 800 K and 300 K. (a) What is the maximum efficiency of the engine? (b) Calculate the maxi

Sholpan [36]

Answer :

(a) The maximum efficiency of the engine is, 62.5 %

(b) The maximum work done is, 0.625 KJ.

(c) The heat discharge into the cold sink is, 0.375 KJ.

Explanation : Given,

Temperature of hot body $T_h$ = 800 K

Temperature of cold body $T_c$ = 300 K

(a) First we have to calculate the maximum efficiency of the engine.

Formula used for efficiency of the engine.

$\eta =1-\frac{T_c}{T_h}$

Now put all the given values in this formula, we get :

$\eta =1-\frac{300K}{800K}$

$\eta =0.625\times 100=62.5\%$

(b) Now we have to calculate the maximum work done.

Formula used :

$\eta =\frac{Q_h-Q_c}{Q_h}=\frac{w}{Q_h}$

where,

$Q_h$ = heat supplied by hot source = 1 KJ

$Q_c$ = heat supplied by hot source

w = work done = ?

Now put all the given values in this formula, we get :

$\eta =\frac{w}{Q_h}$

$0.625=\frac{w}{1KJ}$

$w=0.625KJ$

(c) Now we have to calculate the heat discharge into the cold sink.

Formula used :

$w=Q_h-Q_c$

$Q_c=Q_h-w$

$Q_c=1-0.625$

$Q_c=0.375KJ$

Therefore, (a) The maximum efficiency of the engine is, 62.5 %

(b) The maximum work done is, 0.625 KJ.

(c) The heat discharge into the cold sink is, 0.375 KJ.

4 0

4 years ago

What is the volume of 84 g of bromine gas (Br2) at STP?

Alla [95]

Answer : The correct option is, 11.77 L

Solution : Given,

Mass of bromine gas = 84 g

Molar mass of bromine gas = 159.8 g/mole

First we have to calculate the moles of bromine gas.

$\text{Moles of }Br_2=\frac{\text{ given mass of }Br_2}{\text{ molar mass of }Br_2}=\frac{84g}{159.8g/mole}=0.5256moles$

Now we have to calculate the volume of bromine gas at STP.

As we know that at STP,

1 mole of bromine gas contains 22.4 L volume of bromine gas

So, 0.5256 mole of bromine gas contains $22.4\times 0.5256=11.77L$ volume of bromine gas

Therefore, the volume of bromine gas at STP is, 11.77 L

3 0

4 years ago

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