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SOVA2 [1]
3 years ago
8

Understanding how much of a product is produced in a reaction is referred to as ________ understanding the reaction?

Chemistry
1 answer:
777dan777 [17]3 years ago
7 0
Hello!

Understanding how much of a product is produced in a reaction is referred to as Stoichiometrically understanding the reaction.

Stoichiometry is the calculation of the quantitative relationships between reactants and products in a chemical reaction. The first to talk about stoichiometry was Jeremias Benjamin Ritcher, who said that "Stoichiometry is the science that measures the quantitative proportions or mass ratios of chemical elements that are involved in a chemical reaction".

To calculate how much of a product is produced in a reaction, Stoichiometry is used, applying the law of conservation of mass. That means that the amount of product can be calculated from the amounts of reactants if they are known. 
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Answer:

An Arrhenius Base

Explanation:

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"What quantitative measurement and qualitative observations can be made about the interactions of matter and energy?" can someon
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Quantitative measurements are numerical values, they involve amounts and units like measuring things. Qualitative observations appeal to the five senses, like what does the interaction look and sound like

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C2H4(g) + H2(g) ----> C2H6(g)
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3 years ago
Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79
BARSIC [14]

Answer :  The mole fraction and partial pressure of CH_4,CO_2 and He gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

Explanation : Given,

Moles of CH_4 = 1.79 mole

Moles of CO_2 = 1.20 mole

Moles of He = 3.71 mole

Now we have to calculate the mole fraction of CH_4,CO_2 and He gases.

\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267

and,

\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179

and,

\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554

Thus, the mole fraction of CH_4,CO_2 and He gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of CH_4,CO_2 and He gases.

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  = 5.78 atm

X_i = mole fraction of gas

p_{CH_4}=X_{CH_4}\times p_T

p_{CH_4}=0.267\times 5.78atm=1.54atm

and,

p_{CO_2}=X_{CO_2}\times p_T

p_{CO_2}=0.179\times 5.78atm=1.03atm

and,

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

p_{He}=0.554\times 5.78atm=3.20atm

Thus, the partial pressure of CH_4,CO_2 and He gases are, 1.54, 1.03 and 3.20 atm respectively.

4 0
3 years ago
If you add oil to water and shake the two liquids together what will you form
dalvyx [7]
There will be oil bubbles because oil is more dense than water so therefore they would not mix
4 0
3 years ago
Read 2 more answers
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