To calculate the mass of ethylene oxide needed to react with 10 g of water, you need to determine the molar ratio of water to ethylene oxide in the balanced chemical equation for the reaction. Once you have the molar ratio, you can use it to calculate the mass of ethylene oxide needed. The molar mass of ethylene oxide is 44.05 g/mol.
To determine the grams of ethylene needed to react with 0.0126 mole of water, you need to use the balanced chemical equation for the reaction between ethylene and water. Once you have the balanced equation, use the molar ratio between ethylene and water to convert moles of water to moles of ethylene. Then, use the molar mass of ethylene to convert moles of ethylene to grams of ethylene.
63 g of water are needed.
To react with 0.373 mol of ethylene (C2H4), you need an equal number of moles of water (H2O) based on the balanced chemical equation: C2H4 + H2O β C2H5OH This means you need 0.373 mol of water to react with 0.373 mol of ethylene. To convert moles to grams, you would multiply 0.373 mol by the molar mass of water (18.015 g/mol) to get the grams needed.
Ethene (ethylene) plus carbon dioxide can react to form ethylene carbonate through a chemical process called organic carbonation. This reaction is often used in the synthesis of organic carbonates, which have various industrial applications.
There is an energy problem.
To determine the grams of ethylene needed to react with 0.0126 mole of water, you need to use the balanced chemical equation for the reaction between ethylene and water. Once you have the balanced equation, use the molar ratio between ethylene and water to convert moles of water to moles of ethylene. Then, use the molar mass of ethylene to convert moles of ethylene to grams of ethylene.
63 g of water are needed.
To determine the amount of iron needed to react with 40 grams of iron(III) oxide, you should use the stoichiometry of the reaction. Calculate the molar mass of iron(III) oxide (Fe2O3) and determine the molar ratio between iron and iron(III) oxide in the balanced chemical equation. From there, you can calculate the amount of iron needed to fully react with 40 grams of iron(III) oxide.
To react with 0.373 mol of ethylene (C2H4), you need an equal number of moles of water (H2O) based on the balanced chemical equation: C2H4 + H2O β C2H5OH This means you need 0.373 mol of water to react with 0.373 mol of ethylene. To convert moles to grams, you would multiply 0.373 mol by the molar mass of water (18.015 g/mol) to get the grams needed.
Yes, the acetylide ion (C2H-) can react with ethylene oxide. The acetylide ion is a strong nucleophile and can attack the electrophilic carbon in ethylene oxide, resulting in the formation of a new carbon-carbon bond. This reaction is known as an alkylation reaction.
Ethylene oxide is flammable. It can react with oxygen and ignite, posing a fire hazard.
To form iron(III) oxide, the chemical equation is: 4 Fe(s) + 3 O2(g) β 2 Fe2O3(s) From the equation, it can be seen that 3 moles of oxygen (O2) are needed to react with 4 moles of iron (Fe) to produce 2 moles of iron(III) oxide (Fe2O3). The molar mass of oxygen (O2) is approximately 32 g/mol. Therefore, you would need 96 grams of oxygen to react with the iron needed to form iron(III) oxide.
Ethene (ethylene) plus carbon dioxide can react to form ethylene carbonate through a chemical process called organic carbonation. This reaction is often used in the synthesis of organic carbonates, which have various industrial applications.
To determine the amount of ethylene needed to react with 0.132 mol of H2O, we need to use the balanced chemical equation. The balanced equation for the reaction of ethylene (C2H4) with water (H2O) is: C2H4 + H2O β C2H5OH. From the balanced equation, we can see that 1 mol of ethylene (C2H4) reacts with 1 mol of water (H2O). Therefore, 0.132 mol of H2O would require 0.132 mol of ethylene (C2H4). To convert moles to grams, you would need to know the molar mass of ethylene (C2H4).
There is an energy problem.
The balanced equation for the reaction is: 4 Na + O2 -> 2 Na2O. From the equation, 4 moles of sodium will react to form 2 moles of sodium oxide. Calculate the molar mass of Na2O (sodium oxide) to find out how many grams will be formed.
For every 40 grams of calcium (Ca), 32 grams of oxygen (O) will be needed to react. This is based on the chemical formula for calcium oxide (CaO), where one calcium atom reacts with one oxygen atom to form one molecule of CaO.