Answer:
6.022 x 10^23 particles
Explanation:
Chemists have chosen to count atoms and molecules using a unit called the mole (mol), from the Latin moles, meaning “pile” or “heap.”
One mole is 6.022 x 10^23 of the microscopic particles which make up the substance in question.
Hope this helped! :^)
The answer is C
Sbbsshhshsgssh
Answer : The characteristic properties of a substance always remains the same irrespective of the sample being observed is large or small.
Few examples of characteristic properties of any substances includes freezing or melting point, boiling or condensing point, density, viscosity and solubility. They are specific for specific substances and which makes them easily recognizable.
Whereas chemical properties are specific for one substance which chemically transformed into another substances.
Answer:
B) ) –1615.1 kJ mol^–1
Explanation:
since
SiO2(s) + 4 HF(aq) → SiF4(g) + 2 H2O(l) ∆Hºrxn = 4.6 kJ mol–1
the enhalpy of reaction will be
∆Hºrxn = ∑νp*∆Hºfp - ∑νr*∆Hºfr
where ∆Hºrxn= enthalpy of reaction , ∆Hºfp= standard enthalpy of formation of products , ∆Hºfr = standard enthalpy of formation of reactants , νp=stoichiometric coffficient of products, νr=stoichiometric coffficient of reactants
therefore
∆Hºrxn = ∑νp*∆Hºfp - ∑νr*∆Hºfr
4.6 kJ/mol = [1*∆HºfX + 2*(–285.8 kJ/mol)] - [1*(–910.9kJ/mol) + 4*(–320.1 kJ/mol)]
4.6 kJ/mol =∆HºfX -571.6 kJ/mol + 2191.3 kJ/mol
∆HºfX = 4.6 kJ/mol + 571.6 kJ/mol - 2191.3 kJ/mol = -1615.1 kJ/mol
therefore ∆HºfX (unknown standard enthalpy of formation = standard enthalpy of formation of SiF4(g) ) = -1615.1 kJ/mol
