Answer:
340 grams Ca₃P₂ (2 sig. figs.)
Explanation:
3Ca + 2P => Ca₃P₂
5.6 mole + excess => ? grams
Convert the 'known' to a coefficient of 1 by dividing all coefficients by 3.
=> Ca + 2/3P => 1/3Ca₃P₂
From the above, 1 mole of Ca => 1/3 mole Ca₃P₂
∴ 5.6 mole Ca in an excess of P => 1/3(5.6 mole) Ca₃P₂
=> 1.8666 mol Ca₃P₂ (calculator answer) ≅ 1.9 mol Ca₃P₂
=> 1.9 mole x 182 g Ca₃P₂/mol Ca₃P₂ = 339.73333 grams Ca₃P₂
≅ 340 grams Ca₃P₂ (2 sig. figs.)
Answer:
Where is the results and what is the question or is there a picture
Explanation:
They do not last forever. And they’re not sufficient
Answer:
Group 8A
Explanation:
This is because the elements in Group 8A is stable, what I mean is this elements have the maximum number of electrons in there last orbit so they dont need to form any compund with any other element in the periodic table.
<em>HOPE</em><em> </em><em>THIS</em><em> </em><em>WILL</em><em> </em><em>HELP YOU</em><em /><em>❤️</em>
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
