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spayn [35]
3 years ago
6

What safety precautions are needed when running a brine electrolysis plant

Chemistry
1 answer:
nikdorinn [45]3 years ago
4 0
Good ventilation as a product of it is pure Cl2  gas 
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How many grams of FeCo3 will be produced from 57.2g FeCl2
Evgesh-ka [11]

Answer:

             287.30 g of FeCO₃

Solution:

The Balance Chemical Equation is as follow,

                           FeCl₂ + Na₂CO₃    →    FeCO₃ + 2 NaCl

Step 1: Calculate Mass of FeCl₂ as,

                            Molarity  =  Moles ÷ Volume

Solving for Moles,

                            Moles  =  Molarity × Volume

Putting Values,

                            Moles  =  2 mol.L⁻¹ × 1.24 L

                           Moles  =  2.48 mol

Also,

                            Moles  =  Mass ÷ M.Mass

Solving for Mass,

                            Mass  =  Moles × M.Mass

Putting Values,

                            Mass  =  2.48 mol × 126.75 g.mol⁻¹

                            Mass =  314.34 g of FeCl₂

Step 2: Calculate Mass of FeCO₃ formed as,

According to equation,

          126.75 g (1 mole) FeCl₂ produces  =  115.85 g (1 mole) FeCO₃

So,

               314.34 g of FeCl₂ will produce  =  X g of FeCO₃

Solving for X,

                     X =  (314.34 g × 115.85 g) ÷ 126.75 g

                     X =  287.30 g of FeCO₃

<h2>brainlyest pleas</h2>
4 0
2 years ago
Propane gas, C3H8, is sometimes used as a fuel. In order to measure its energy output as a fuel a 1.860 g sample was combined wi
lisov135 [29]

Answer:

The heat of the reaction is 105.308 kJ/mol.

Explanation:

Let the heat released during reaction be q.

Heat gained by water: Q

Mass of water ,m= 1kg = 1000 g

Heat capacity of water ,c= 4.184 J/g°C

Change in temperature = ΔT = 26.061°C - 25.000°C=1.061 °C

Q=mcΔT

Heat gained by bomb calorimeter =Q'

Heat capacity of bomb calorimeter ,C= 4.643 J/g°C

Change in temperature = ΔT'= ΔT= 26.061°C - 25.000°C=1.061 °C

Q'=CΔT'=CΔT

Total heat released during reaction is equal to total heat gained by water and bomb calorimeter.

q= -(Q+Q')

q = -mcΔT - CΔT=-ΔT(mc+C)

q=-1.061^oC(1000 g\times 4.184J/g^oC+4.643 J/^oC )=-4,444.15J=-4.444 kJ

Moles of propane =\frac{1.860 g}{44 g/mol}=0.0422 mol

0.0422 moles of propane on reaction with oxygen releases 4.444 kJ of heat.

The heat of the reaction will be:

\frac{4.444 kJ}{0.0422 mol}=105.308 kJ/mol

3 0
3 years ago
Compare and contrast the Bohr model and the electron cloud models of the atom.
bogdanovich [222]

Here we have to compare the Bohr atomic model with electron cloud model.

In the Bohr's atomic model the electrons of an element is assumed to be particle in nature. Which was unable to explain the deBroglie' hypothesis or the uncertainty principle and has certain demerits.

The uncertainty principle reveals the wave nature of the electrons or electron clod model. The Bohr condition of a stable orbits of the electron can nicely be explained by the electron cloud model, the mathematical form of which is λ = nh/mv, where, λ = wavelength, n is the integral number, h = Planck's constant, m = mass of the electron and v = velocity of the electron.

The integral number i.e. n is similar to the mathematical form of Bohr's atomic model, which is mvr = nh/2π. (r = radius of the orbit).

Thus, the electron cloud model is an extension of the Bohr atomic model, which can explain the demerits of the Bohr model. Later it is revealed that the electron have both particle and wave nature. Which is only can explain all the features of the electrons around a nucleus of an element.    

5 0
2 years ago
Read 2 more answers
Which physical property do plasmas and gases have in common?
sp2606 [1]
<span>They both have charged particles
They have the same attractive forces between particles
They have the same space between particles
They create magnetic and electric fields</span>
8 0
3 years ago
Read 2 more answers
copper forms two oxides. On heating 1 g of each in hydrogen 0.888 g and 0.798 g of the metal was obtained. Show that the results
Mashutka [201]

Answer:

For the first oxide, 1 g gives 0.888 g of copper.

Dividing by 0.888 tells us that 1.126 g gives 1 g of copper so has 0.126 g of oxygen.

For the second oxide, 1 g gives 0.798 g of copper.

Dividing by 0.798 tells us that 1.253 g gives 1 g of copper so has 0.253 g of oxygen.

So 1 g of copper combines with either 0.126 g or 0.253 g of oxygen.

Within the limits of experimental error, 0.253 is twice 0.126, confirming the law of multiple proportion.

6 0
2 years ago
Read 2 more answers
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