Firstly calculate the grams in the last 8 percent before moving onto the pyrite section.
50.8x0.08=4.064g
We know that iron ore in this case has 92 percent pyrite which contains 46.5 percent iron so we do 50.8x0.92=46.736g from this we need to find 46.5 percent of the iron content in the 92 percent pyrite section then add this answer to the 8 percent of iron ore we found at the start.46.736x0.465=21.73224g
21.73224g+4.064=25.79624g of iron ore 25.8g(3sf)
Answer: The given statement is true.
Explanation:
A catalyst is defined as the substance that lowers the activation energy of a chemical reaction without itself getting consumed so that there will be increase in rate of reaction.
As catalyst lowers the activation energy so, reactant molecules with lower energy also participates in the reaction. Hence, more collisions occur due to which there is rapid formation of products take place through another path.
Therefore, we can conclude that the statement a catalyst increases the rate of a reaction without being consumed. It accomplishes this by providing another mechanism that has a lower activation energy, is true.
3k= 3 Potassium atoms
2S= 2 sulfur atoms
2O4= 8 oxygen atoms
The numbers in front of atoms are like coefficients(i.e. 2x or 3y). You should have 13 atoms in total.
Answer:
Mass of CO₂ produced = 5.72 g
Explanation:
Given data:
Mass of methane = 2.34 g
Mass of oxygen = 8.32 g
Mass of CO₂ produced = ?
Solution:
Chemical equation:
CH₄ + 2O₂ → CO₂ + 2H₂O
Number of moles of methane:
Number of moles = mass/molar mass
Number of moles = 2.34 g/ 16 g/mol
Number of moles = 0.146 mol
Number of moles of oxygen:
Number of moles = mass/molar mass
Number of moles = 8.32 g/ 32 g/mol
Number of moles = 0.26 mol
Now we will compare the moles of carbon dioxide with oxygen and methane.
CH₄ : CO₂
1 : 1
0.146 : 0.146
O₂ : CO₂
2 : 1
0.26 : 1/2×0.26 = 0.13 mol
Less number of moles of CO₂ are produced by oxygen thus oxygen will react as limiting reactant.
Mass of CO₂:
Mass = number of moles × molar mass
Mass = 0.13 mol × 44 g/mol
Mass = 5.72 g
Question: The question is incomplete. Below is the complete question and the answer;
While ethanol (CH3CH2OH is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene CH2CH2) with water vapor at elevated temperatures. A chemical engineer studying this reaction fills a 50.0 L tank at 22. °C with 24. mol of ethylene gas and 24. mol of water vapor. He then raises the temperature considerably, and when the mixture has come to equilibrium determines that it contains 15.4 mol of ethylene gas and 15.4 mol of water vapor The engineer then adds another 12. mol of water, and allows the mixture to come to equilibrium again. Calculate the moles of ethanol after equilibrium is reached the second time. Round your answer to 2 significant digits.
Answer:
Number of moles of ethanol = 11 mol
Explanation:
SEE THE ATTACHED FILE FOR THE CALCULATION