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Elemental sulfur is one of the products of the gas-phase reaction of nitric acid and hydrogen sulfide. The other

Goshia [24]

Answer:

The manufacturing processes for liquefied petroleum gas are designed so that the majority, if not all, of the sulfur compounds are removed. The total sulfur level is therefore considerably lower than for other crude oil-based fuels and a maximum limit for sulfur content helps to define the product more completely. The sulfur compounds that are mainly responsible for corrosion are hydrogen sulfide, carbonyl sulfide and, sometimes, elemental sulfur. Hydrogen sulfide and mercaptans have distinctive unpleasant odors. A control of the total sulfur content, hydrogen sulfide and mercaptans ensures that the product is not corrosive or nauseating. Stipulating a satisfactory copper strip test further ensures the control of the corrosion.

8 0

3 years ago

Which type of mass movement makes pattern of wrinkles, or terraces, on hillsides

kondor19780726 [428]

Erosion is the correct answer

8 0

4 years ago

Read 2 more answers

What two pieces of evidence did van Helmont use to reach his conclusion

Sav [38]

Explanation:

He weighed silver, dissolved it in acid, and then recovered all the original silver by reacting the solution with copper. He also showed, by using iron to recover the copper, that this displacement of one metal from its salt by using a second metal was not because of transmutation, as many had held.

6 0

2 years ago

How many liters of gas will be in the closed reaction flask when 36.0L of ethane (C2H6) is allowed to react with 105.0L of oxyge

Ivan

Answer:- Volume of the gas in the flask after the reaction is 156.0 L.

Solution:- The balanced equation for the combustion of ethane is:

$2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)$

From the balanced equation, ethane and oxygen react in 2:7 mol ratio or 2:7 volume ratio as we are assuming ideal behavior.

Let's see if any one of them is limiting by calculating the required volume of one for the other. Let's say we calculate required volume of oxygen for given 36.0 L of ethane as:

$36.0LC_2H_6(\frac{7LO_2}{2LC_2H_6})$

= 126 L $O_2$

126 L of oxygen are required to react completely with 36.0 L of ethane but only 105.0 L of oxygen are available, It means oxygen is limiting reactant.

let's calculate the volumes of each product gas formed for 105.0 L of oxygen as:

$105.0LO_2(\frac{4LCO_2}{7L O_2})$

= 60.0 L $CO_2$

Similarly, let's calculate the volume of water vapors formed:

$105.0L O_2(\frac{6L H_2O}{7L O_2})$

= 90.0 L $H_2O$

Since ethane is present in excess, the remaining volume of it would also be present in the flask.

Let's first calculate how many liters of it were used to react with 105.0 L of oxygen and then subtract them from given volume of ethane to know it's remaining volume:

$105.0LO_2(\frac{2LC_2H_6}{7LO_2})$