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3 years ago

Discuss the environmental impacts or gas fracking.

Chemistry

1 answer:

Allushta [10]3 years ago

5 0

Answer:

gas fracking is a process of drilling into the surface of the earth and injecting the rocks with a mixture of chemicals, sand and water at high pressure t release gas.

Explanation:

environmental impacts of gas fracking.

1. during fracking a chemical called methane is released into the atmosphere which is a greenhouse gas and thus pollutes the air

2. Fracking uses gallons of water. Thus, during this process a lot of water is wasted and that directly causes the shortage of clean and hygienic water available for the surrounding residents.

3. It is also a cause for soil degradation and soil pollution. fracking is a high risk potential for oil spill which degrades the soil. It also harms the vegetation around the fracking site.

4. Due to high pressure used during fracking, it may also cause natural disasters like earthquakes.

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Pls help, How is stoichiometry used to calculate amounts of substances in a chemical reaction?

Elden [556K]

Answer:

Thus, to calculate the stoichiometry by mass, the number of molecules required for each reactant is expressed in moles and multiplied by the molar mass of each to give the mass of each reactant per mole of reaction. The mass ratios can be calculated by dividing each by the total in the whole reaction.

Explanation: Stoichiometry is the field of chemistry that is concerned with the relative quantities of reactants and products in chemical reactions. For any balanced chemical reaction, whole numbers (coefficients) are used to show the quantities (generally in moles ) of both the reactants and products.

7 0

3 years ago

Find the molality of the solution if 42 grams of lithium chloride (LiCl) are dissolved in 3.6 kg of water.

omeli [17]

The molality is calculated using the following rule:
molality = number of moles of solute / kg of solvent

From the periodic table:
molar mass of lithium = 6.941 gm
molar mass of chlorine = 35.453 gm

molar mass of LiCl = 6.941 + 35.453 = 42.394 gm
number of moles found in 42 gm = mass / molar mass = 42 / 42.394 = 0.99

molality = 0.99 / 3.6 = 0.275 m

3 0

3 years ago

Copper(II) sulfide is oxidized b molecular oxygen to produce gaseous sulfur trioxide and solid copper (II) oxide. The gaseous pr

Tasya [4]

Explanation:

Copper(II) sulfide reacts with oxygen gas to give solid copper(II) oxide and sulfur trioxide gas.

The reaction is given as:

$CuS+2O_2\rightarrow CuO(s)+SO_3(g)$

When 1 mol copper(II) sulfide react with 2 moles of oxygen gas it gives 1 mol of solid copper(II) oxide and 1 mol of sulfur trioxide gas

The gas formed in above reaction that is sulfur trioxide reacts with water to give sulfuric acid or hydrogen sulfate.

The reaction is given as:

$SO_3(g)+H_2O(l)\rightarrow H_2SO_4(aq)$

1 mol of sulfur trioxide gas reacts with 1 mol of liquid water to produce 1 molo of liquid hydrogen sulfate or sulfuric acid

3 0

3 years ago

Determine the theoretical yield of HCl if 60.0 g of BC13 and 37.5 g of H20 are reacted according to the following balanced react

Pie

Answer : The theoretical yield of HCl is, 56.1735 grams

Explanation : Given,

Mass of $BCl_3$ = 60 g

Mass of $H_2O$ = 37.5 g

Molar mass of $BCl_3$ = 117 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $HCl$ = 36.5 g/mole

First we have to calculate the moles of $BCl_3$ and $H_2O$ .

$\text{Moles of }BCl_3=\frac{\text{Mass of }BCl_3}{\text{Molar mass of }BCl_3}=\frac{60g}{117g/mole}=0.513moles$

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{37.5g}{18g/mole}=2.083moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$BCl_3(g)+3H_2O(l)\rightarrow H_3BO_3(s)+3HCl(g)$

From the balanced reaction we conclude that

As, 1 mole of $BCl_3$ react with 3 mole of $H_2O$

So, 0.513 moles of $BCl_3$ react with $3\times 0.513=1.539$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $BCl_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $HCl$ .

As, 1 mole of $BCl_3$ react to give 3 moles of $HCl$

So, 0.513 moles of $BCl_3$ react to give $3\times 0.513=1.539$ moles of $HCl$

Now we have to calculate the mass of $HCl$ .

$\text{Mass of }HCl=\text{Moles of }HCl\times \text{Molar mass of }HCl$