Help me with both questions

4. Froath floatation process is used for*

lbvjy [14]

Answer:

Explanation:

Froath flotaion process is based on thie wttebality properteis like hydrophillicity and hydrophobicity.

Incase of sulphide ore processing,

The metallic sulphide particles of ore are preferentially wetted by oil and the gangue particles by water. By doing thise you are sperating required sulphide ore particels from unwanted gaunge , by doing this you are incresing the concentation of Sulfide in ore by removing the not required gaunge particles.

Hope this helps you to decide your annswer

5 0

3 years ago

Nitric oxide is formed in automobile exhaust when nitrogen and oxygen in air react at high temperatures.N2(g) + O2(g) 2NO(g)The

yanalaym [24]

Answer : The correct option is, (E) 7.8 atm

Explanation :

The partial pressure of $N_2$ = 8.00 atm

The partial pressure of $O_2$ = 5.00 atm

$K_p$ = 0.0025

The balanced equilibrium reaction is,

$N_2(g)+O_2(g)\rightleftharpoons 2NO(g)$

Initial pressure 8.00 5.00 0

At eqm. (8.00-x) (5.00-x) 2x

The expression of equilibrium constant $K_p$ for the reaction will be:

$K_p=\frac{(p_{NO})^2}{(p_{N_2})(p_{O_2})}$

Now put all the values in this expression, we get :

$0.0025=\frac{(2x)^2}{(8.00-x)\times (5.00-x)}$

By solving the terms, we get:

$x=0.15atm$

The equilibrium partial pressure of $N_2$ = (8.00 - x) = (8.00 - 0.15) = 7.8 atm

Therefore, the equilibrium partial pressure of $N_2$ is 7.8 atm.

3 0

4 years ago

Natural gas (CH4) has a molar mass of 16.0 g/mole. You started out the day with a tank containing 200.0 g of natural gas. At the

hodyreva [135]

Considering the definition of molar mass, the moles of gas used are 10.625 moles.

<h3>Definition of molar mass</h3>

The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.

<h3>Amount of moles used</h3>

Natural gas has a molar mass of 16.0 g/mole.

You started out the day with a tank containing 200.0 g of natural gas. So, you can apply the following rule of three: If by definition of molar mass 16 grams are contained in 1 mole, 200 grams are contained in how many moles?

$amount of moles at the beginning=\frac{200 gramsx1 mole}{16 grams}$

<u><em>amount of moles at the beginning= 12.5 moles</em></u>

At the end of the day, your tank contains 30.0 g of natural gas. So, you can apply the following rule of three: If by definition of molar mass 16 grams are contained in 1 mole, 30 grams are contained in how many moles?

$amount of moles at the end=\frac{30 gramsx1 mole}{16 grams}$

<u><em>amount of moles at the end= 1.875 moles</em></u>

The number of moles used will be the difference between the number of moles used initially and the contents at the end of the day.

moles used= amount of moles at the beginning - amount of moles at the end

moles used= 12.5 moles - 1.875 moles

<u><em>moles used= 10.625 moles</em></u>

Finally, the moles of gas used are 10.625 moles.

Learn more about molar mass:

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3 0

2 years ago

There is no connection between the study of chemistry and physics. A. true B. false

9966 [12]

B. False, have a connection between chemistry and physis

4 0

3 years ago

A particular reaction, A- products, has a rate that slows down as the reaction proceeds. The half-life of the reaction is found

Thepotemich [5.8K]

Explanation:

Half life of zero order and second order depends on the initial concentration. But as the given reaction slows down as the reaction proceeds, therefore, it must be second order reaction. This is because rate of reaction does not depend upon the initial concentration of the reactant.

a. As it is a second order reaction, therefore, doubling reactant concentration, will increase the rate of reaction 4 times. Therefore, the statement a is wrong.

b. Expression for second order reaction is as follows:

$\frac{1}{[A]} =\frac{1}{[A]_0} +kt$

the above equation can be written in the form of Y = mx + C

so, the plot between 1/[A] and t is linear. So the statement b is true.

c.

Expression for half life is as follows:

$t_{1/2}=\frac{1}{k[A]_0}$

As half-life is inversely proportional to initial concentration, therefore, increase in concentration will decrease the half life. Therefore statement c is wrong.

d.

Plot between A and t is exponential, therefore there is no constant slope. Therefore, the statement d is wrong

8 0

3 years ago