Answer:
Pure substance B) Consists of a single element or type of compound.
Homogeneous A) Mixture that has its different components mixed evenly within the substance.
Heterogeneous D) Mixture that has its different components mixed unevenly within the substance.
Solution C) Liquid homogeneous mixture in which the solute is distributed evenly within the solvent.
Explanation:
Pure substances are a form of matter with definite constant composition and distinct properties. They consist of a single element or type of compound, as can be seen in its formula. Na, O₂, NaCl and H₂O are examples of pure substances.
When 2 or more pure substances are mixed together they form a mixture. If the mixture has its different components mixed <u>evenly</u> within the substance it is a homogeneous mixture. Whereas if the mixture has its different components mixed <u>unevenly</u> within the substance it is a heterogeneous mixture. The different parts observable in a heterogeneous mixture are known as phases.
In liquid homogeneous mixtures, we can recognize one or more substances that are in lower proportions (solutes) and one substance that is in greater proportion (solvent). This kind of mixture is known as a solution.
Answer:
There are 1.8021 ⋅ 1024 molecules of CH4 in 48 grams of CH4. To answer this question, you must understand how to convert grams of a molecule into the number of molecules. To do this, you have to utilize the concepts of moles and molar mass. A mole is just a unit of measurement. Avogadro's number is equal to 6.022 ⋅1023 molecules/mole. i think please dont complain to me if its wrong im sorry
Explanation:
B .pure water it’s made up of oxygen and hydrogen “H2O”
The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.