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Alexandra [31]
4 years ago
8

What is the approximate volume injected into a GC? 10-50 μL 500-1000 μL 1-10 μL 100-500 μL 50-100 μL What technique is used to a

djust for injection volume inaccuracy? Standard Addition Nothing, the syringe is accurate External Standard Standardization Internal Standard
Chemistry
1 answer:
DiKsa [7]4 years ago
4 0

Answer:

1-10 μL

Explanation:

Gas Chromatography (GC) is a method of separating the constituents of a complex mixture by taking advantage of their different rates of movement in a solvent over an adsorbent medium. Gas Chromatography (GC) is a means of separation and analysis that utilises fractional separation. It is based on the principle that if a fluid containing a number of substances is allowed to pass through an adsorbent medium , the different substances in the fluid may travel at different rates and be separated.

The approximate volume injected through an inlet by using a split injection into a Gas Chromatography (GC) contains a sample with an approximate value of analyte which varies from  1-10 μL.

In order to adjust for injection volume inaccuracy, the technique used is known as  Internal Standard.

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If 3.0 atm of pure HN3(g) is decomposed initially, what is the final total pressure in the reaction container? What are the part
lawyer [7]

This is an incomplete question, here is a complete question.

Hydrogen azide, HN₃, decomposes on heating by thefollowing unbalanced reaction:

HN_3(g)\rightarrow N_2(g)+H_2(g)

If 3.0 atm of pure HN₃ (g) is decomposed initially,what is the final total pressure in the reaction container? Whatare the partial pressures of nitrogen and hydrogen gas? Assume thatthe volume and temperature of the reaction container are constant.

Answer : The partial pressure of N_2 and H_2 gases are, 4.5 atm and 1.5 atm respectively.

Explanation :

The given unbalanced chemical reaction is:

HN_3(g)\rightarrow N_2(g)+H_2(g)

This reaction is an unbalanced chemical reaction because in this reaction number of hydrogen and nitrogen atoms are not balanced on both side of the reaction.

In order to balance the chemical equation, the coefficient '2' put before the HN_3 and the coefficient '3' put before the N_2 then we get the balanced chemical equation.

The balanced chemical reaction will be,

2HN_3(g)\rightarrow 3N_2(g)+H_2(g)

As we are given:

The pressure of pure HN_3 = 3.0 atm

p_{Total}=2\times p_{HN_3}=2\times 3.0atm=6.0atm

From the reaction we conclude that:

Number of moles of N_2 = 3 mol

Number of moles of H_2 = 1 mol

Now we have to calculate the mole fraction of N_2 and H_2

\text{Mole fraction of }N_2=\frac{\text{Moles of }N_2}{\text{Moles of }N_2+\text{Moles of }H_2}=\frac{3}{3+1}=0.75

and,

\text{Mole fraction of }H_2=\frac{\text{Moles of }H_2}{\text{Moles of }N_2+\text{Moles of }H_2}=\frac{1}{3+1}=0.25

Now we have to calculate the partial pressure of N_2 and H_2

According to the Raoult's law,

p_i=X_i\times p_T

where,

p_i = partial pressure of gas

p_T = total pressure of gas  = 6.0 atm

X_i = mole fraction of gas

p_{N_2}=X_{N_2}\times p_T

p_{N_2}=0.75\times 6.0atm=4.5atm

and,

p_{H_2}=X_{H_2}\times p_T

p_{H_2}=0.25\times 6.0atm=1.5atm

Thus, the partial pressure of N_2 and H_2 gases are, 4.5 atm and 1.5 atm respectively.

8 0
3 years ago
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