Explanation:
We have to find the number of molecules that are present in 0.54 g of Ca(NO₃)₂.
First we have to convert the mass of our sample into moles of Ca(NO₃)₂. We will use the molar mass of Ca(NO₃)₂ to do that.
molar mass of Ca = 40.08 g/mol
molar mass of N = 14.01 g/mol
molar mass of O = 16.00 g/mol
molar mass of Ca(NO₃)₂ = 1 * 40.08 g/mol + 2 * 14.01 g/mol + 6 * 16.00 g/mol
molar mass of Ca(NO₃)₂ = 164.10 g/mol
mass of Ca(NO₃)₂ = 0.54 g
moles of Ca(NO₃)₂ = 0.54 g * 1 mol/(164.10 g)
moles of Ca(NO₃)₂ = 0.00329 moles
According to Avogadro's number there are 6.022 *10^23 molecules in 1 mol of molecules. We can use that relationship to find the number of molecules that are present in our sample.
6.022 *10^23 molecules = 1 mol
molecules of Ca(NO₃)₂ = 0.00329 moles * 6.022 *10^23 molecules/(1 mol)
molecules of Ca(NO₃)₂ = 2.0 * 10^21 molecules
Answer: there are 2.0 * 10^21 molecules of Ca(NO₃)₂ in 0.54 g of it.
C. Neon is the element which is not likely to form bonds. This is because neon is a noble gas, which means that it already has the maximum of 8 valence electrons, and doesn't need more, whereas gold, oxygen, and mercury can bond more electrons.
Answer:
7.28 moles Ag°
Explanation:
Cu° + 2 AgNO₃ => Cu(NO₃)₂ + 2Ag°
Given 7.28 moles 7.28 moles
To determine limiting reactant, divide the mole values by the respective coefficient of balanced equation. The resulting smallest value is the limiting reactant. Note: this is a short cut method for determining limiting reactant only. Once the limiting reactant is determined one must use the given mole values of the limiting reactant to solve problem. That is ...
Limiting reactant determination:
Cu° + 2 AgNO₃ => Cu(NO₃)₂ + 2Ag°
Cu: 7.28 / 1 = 7.28
AgNO₃ : 7.28 / 2 = 3.64 => Limiting Reactant is AgNO₃
Solving Problem depends on AgNO₃; Cu will be in excess.
Since coefficients of AgNO₃ & Ag° are equal, then the moles AgNO₃ used equals moles Ag° produced and is therefore 7.28 moles Ag°.
Gas is the most compressible at room temperature.
