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gayaneshka [121]
4 years ago
8

What happens to a line when the y-intercept is changed? Check all that apply.

Chemistry
1 answer:
tamaranim1 [39]4 years ago
5 0

Answer:

As the y-intercept increases, the graph of the line shifts up;

As the y-intercept decreases, the graph of the line shifts down

Explanation:

There are two ways to think about this problem. The first way would be the graphical approach:

  • if we only change the y-intercept, this means we keep the same slope;
  • y-axis is the vertical axis;
  • if we change the point at which the line crosses the y-axis, we either shift it upward for a higher y-intercept or downward for a lower y-intercept.

Now, thinking algebraically, a line has the following equation in a general form:

y = mx + b

The y-intercept is essentially obtained when x = 0, then:

y = b:

  • if we increase b value, the y value increases, so the graph shifts upward;
  • if we decrease b value, the y value decreases, so the graph shifts downward.
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P4 (s) + 5O2 (g) ⟶ P4 O10(s) ΔG° = −2697.0 kJ/mol 2H2 (g) + O2 (g) ⟶ 2H2 O(g) ΔG° = −457.18 kJ/mol 6H2 O(g) + P4 O10(s) ⟶ 4H3 PO
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Answer:

the standard free energy of formation of phosphoric acid H3 PO4 is -1010 kJ/mol

Explanation:

Knowing that

1) P4 (s) + 5O2 (g) ⟶ P4 O10(s) ΔG° = −2697.0 kJ/mol

2) 2H2 (g) + O2 (g) ⟶ 2H2 O(g) ΔG° = −457.18 kJ/mol

3) 6H2 O(g) + P4 O10(s) ⟶ 4H3 PO4 (l) ΔG° = −428.66 kJ/mol

since

ΔG° reaction = ν * ΔGf °  products - v *ΔGf °  reactives

for reaction 1

ΔG° = ∑ν * ΔGf °  products - ∑v *ΔGf °  reactives

ΔG° = 1 *ΔGf ° P4 O10 - ( 5 *ΔGf ° O2 + 1 *ΔGf ° P4)

ΔG° = ΔGf ° P4 O10 - (5*0 +1*0)

ΔGf ° P4 O10 = ΔG° = −2697.0 kJ/mol

for reaction 2

ΔG° = ∑ν * ΔGf °  products - ∑v *ΔGf °  reactives

ΔG° = 2 *ΔGf ° H20 - ( 1*ΔGf ° O2 + 2 *ΔGf ° H2)

ΔG° = 2* ΔGf ° H20 - (1*0 +2*0)

ΔGf ° H20 = ΔG° /2  = -457.18 kJ/mol/2 = -228.59 kJ/mol

for reaction 3

ΔG° = ∑ν * ΔGf °  products - ∑v *ΔGf °  reactives

ΔG° = 4 *ΔGf ° H3 PO4 - ( 6*ΔGf ° H2O + 1*ΔGf ° P4O10)

−428.66 kJ/mol = 4 *ΔGf ° H3 PO4 - [ 6*(-228.59 kJ/mol) + 1*(−2697.0 kJ/mol)]

−428.66 kJ/mol = 4 *ΔGf ° H3 PO4 + 3611.36 kJ/mol

ΔGf ° H3 PO4 = (−428.66 kJ/mol - 3611.36 kJ/mol)/4 = -1010 kJ/mol

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