Answer:
Explanation:
The expression can be solved mathematically as follows:
1) 
Given
2) 
Definition of sum
3) 
/Associative property
4) 
Distributive property
5) 
Multiplication
6) 
Subtraction.
7) 
Multiplication/Associative property
8) 
Associative property
9) /Result
This year course engages students in becoming skilled readers of prose written in a variety of periods, disciplines, and
rhetorical contexts and in becoming skilled writers who compose for a variety of purposes. More immediately, the course
prepares the students to perform satisfactorily on the A.P. Examination in Language and Composition given in the spring.
Both their writing and their reading should make students aware of the interactions among a writer’s purposes, audience
expectations, and subjects as well as the way generic conventions and the resources of language contribute to effectiveness
in writing. Students will learn and practice the expository, analytical, and argumentative writing that forms the basis of
academic and professional writing; they will learn to read complex texts with understanding and to write prose of
sufficient richness and complexity to communicate effectively with mature readers. Readings will be selected primarily,
but not exclusively, from American writers. Students who enroll in the class will take the AP examination.
Answer:
116.3 grCO2
Explanation:
1st - we balance the equation so that it finds the same amount of elements of the product side and of the reagent side
C6H6 +15/2 O2⟶ 6CO2 +3 H2O
2nd - we calculate the limiting reagent
39.2gr C6H6*(240grO2/78grC6H6)=120 grO2
we don't have that amount of oxygen so this is the excess reagent and oxygen the limiting reagent
3rd - we use the limiting reagent to calculate the amount of CO2 in grams
105.7grO2*(264grCO2/240grO2)=116.3 grCO2
Part (a) :
H₂(g) + I₂(s) → 2 HI(g)
From given table:
G HI = + 1.3 kJ/mol
G H₂ = 0
G I₂ = 0
ΔG = G(products) - G(reactants) = 2 (1.3) = 2.6 kJ/mol
Part (b):
MnO₂(s) + 2 CO(g) → Mn(s) + 2 CO₂(g)
G MnO₂ = - 465.2
G CO = -137.16
G CO₂ = - 394.39
G Mn = 0
ΔG = G(products) - G(reactants) = (1(0) + 2*-394.39) - (-465.2 + 2*-137.16) = - 49.3 kJ/mol
Part (c):
NH₄Cl(s) → NH₃(g) + HCl(g)
ΔG = ΔH - T ΔS
ΔG = (H(products) - H(reactants)) - 298 * (S(products) - S(reactants))
= (-92.31 - 45.94) - (-314.4) - (298 k) * (192.3 + 186.8 - 94.6) J/K
= 176.15 kJ - 84.78 kJ = 91.38 kJ
The answer is going to be mercury
