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What is the enthalpy of cadmium carbonate and cadmium oxide?
Standard enthalpy of formation (kJ/mol) I could not find cadmium carbonate Cadmium oxide: -258.4 From the CRC Handbook of Chemistry and Physics
A metal forms two oxidesThe higher oxide contains 80 percent metal 0.72g of lower oxide gave 0.8g of the higher oxide when oxidised Show that the data illustrates the law of multiple proportions?
In higher oxide Metal = 80% Oxygen = (100-80)% = 20% Therefore, we can say that 4 parts of metal combines with 1 part of oxygen. Now, 0.72 g of lower oxide on oxidation gives 0.8 g of higher oxide. It can be assumed that the mass percent of metal in 0.8 g is same as that of 0.72 g of lower oxide. So, Mass of metal in higher oxide = (80/100) x 0.8 g = 0.64 g If in 0.8 g of higher oxide 0.64 g is metal then mass of oxygen present in higher oxide will be (0.8 - 0.64) g = 0.16 g Since, lower oxide contains the same mass of metal as that of higher oxide, we need to calculate the mass of oxygen in lower oxide. Mass of oxygen in lower oxide = (0.72-0.64) g = 0.08 g According to Law to multiple proportions if two elements combine with each other to form two different compounds then the ratio of masses of that element which combines with the other element whose mass is fixed in both the compounds, will be in small whole numbers. Now, in the given problem the mass of metal in both the oxide is fixed so, for the data to illustrate the law of multiple proportion the ratio of mass of oxygen in both the oxides should be in whole numbers. Now, mass of oxygen in higher oxide : mass of oxygen in lower oxide = 0.16 : 0.08 = 2 : 1. Therefore, it can be said that the given data depicts law of multiple proportion.
What compound has a lower percent of mass by nitrogen- NO or N2O3?
The oxide N2O3 has a lower percent of mass nitrogen.
How much energy is involved in the formation of 5 grams of rust?
The energy involved in the formation of 5 grams of rust can be calculated using the enthalpy of formation of iron(III) oxide (rust), which is -824 kJ/mol. First, convert the mass of rust to moles, then use the molar enthalpy of formation to calculate the energy involved.
Does aluminum melt at a higher temperature than aluminum oxide?
Aluminum oxide has a higher melting point than aluminum. Aluminum oxide melts at around 2072°C, while aluminum melts at a lower temperature of around 660°C.
Lattice enthalpy of magnesium oxide and sodium chloride?
The lattice enthalpy of sodium chloride is 789 kJ/mol.
What is the enthalpy of cadmium carbonate and cadmium oxide?
Standard enthalpy of formation (kJ/mol) I could not find cadmium carbonate Cadmium oxide: -258.4 From the CRC Handbook of Chemistry and Physics
What is the synthesis of phenyl acetaldehyde from styrene oxide please give detail process?
Heat (300º?) styrene oxide in a closed system, you will obtain a mixture of styrene oxide and phenyl acetaldehyde. IIRC you can heat at reflux S.O. and some kind of silica catalyst with a solvent and obtain that same product.
How do you convert acetic acid to acetaldehyde?
Acetic acid can be converted to acetaldehyde using an oxidizing agent such as silver oxide or chromic acid. The reaction involves breaking the carbon-carbon bond in acetic acid to form acetaldehyde as a primary product. This reaction is commonly known as dehydrogenation of acetic acid.
What are 10 gases?
Argon, Ether, Oxygen, Hydrogen, Methane, Nitrogen, Helium, carbon dioxide, Nitrous Oxide, carbon monoxide, xenon, Neon, Radon, Krypton, Chlorine, Ozone, Fluorine, Acetaldehyde,
A metal forms two oxidesThe higher oxide contains 80 percent metal 0.72g of lower oxide gave 0.8g of the higher oxide when oxidised Show that the data illustrates the law of multiple proportions?
In higher oxide Metal = 80% Oxygen = (100-80)% = 20% Therefore, we can say that 4 parts of metal combines with 1 part of oxygen. Now, 0.72 g of lower oxide on oxidation gives 0.8 g of higher oxide. It can be assumed that the mass percent of metal in 0.8 g is same as that of 0.72 g of lower oxide. So, Mass of metal in higher oxide = (80/100) x 0.8 g = 0.64 g If in 0.8 g of higher oxide 0.64 g is metal then mass of oxygen present in higher oxide will be (0.8 - 0.64) g = 0.16 g Since, lower oxide contains the same mass of metal as that of higher oxide, we need to calculate the mass of oxygen in lower oxide. Mass of oxygen in lower oxide = (0.72-0.64) g = 0.08 g According to Law to multiple proportions if two elements combine with each other to form two different compounds then the ratio of masses of that element which combines with the other element whose mass is fixed in both the compounds, will be in small whole numbers. Now, in the given problem the mass of metal in both the oxide is fixed so, for the data to illustrate the law of multiple proportion the ratio of mass of oxygen in both the oxides should be in whole numbers. Now, mass of oxygen in higher oxide : mass of oxygen in lower oxide = 0.16 : 0.08 = 2 : 1. Therefore, it can be said that the given data depicts law of multiple proportion.
What compound has a lower percent of mass by nitrogen- NO or N2O3?
The oxide N2O3 has a lower percent of mass nitrogen.
Why does calcium oxide react with silicon oxide to form slag?
Calcium oxide reacts with silicon oxide to form slag because the calcium oxide is a strong base that can react with acidic silicon oxide to form calcium silicate, which is a compound that has a lower melting point than the individual oxides. This helps to lower the overall melting point of the mixture and promotes the separation of impurities from the desired metal during the smelting process.
What is the bond length of nitrous oxide?
This might not be of much help, but it is longer than Nitrogen monoxide, as the bond enthalpy is higher in a N-O bond. There is not much on this issue on the Internet unfortunately.
What is the difference between nitrous oxide and halothane in relation to blood solubility?
nitrous oxide : lower solubility halothane : higher solubility
What is the balanced symbol equation for converting limestone into calcium oxide?
CaCO3(s) --> CaO(s) +CO2(g) enthalpy is +178 kJ molˉ¹
How much energy is involved in the formation of 5 grams of rust?
The energy involved in the formation of 5 grams of rust can be calculated using the enthalpy of formation of iron(III) oxide (rust), which is -824 kJ/mol. First, convert the mass of rust to moles, then use the molar enthalpy of formation to calculate the energy involved.
