A would have a higher pitch
-3 is the answer your looking for
I don't have a graph but here's what I think. The relationship is that the volume will change depending on the temperature. So think of water for an example. The volume of it will stay the same at room temperature, but if you put a glass of it in the freezer for a few hours, take it out, measure the volume, the volume would have changed greatly. Or heating and evaporating the water will do the same.
BF3 .... BP = −100.3 °C
<span>RbCl ..... solid </span>
<span>CH3SCH3 ..... BP = 35-41 °C </span>
<span>SbH3 .... BP = −17 °C </span>
<span>SiS2 ..... solid </span>
<span>Ethanol solid --> ethanol melts --> ethanol liquid </span>
<span>-135C ---------------> -114C --------------> -50C </span>
<span>............ ΔT = 21C ....... ....... ΔT = 64C </span>
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
