I think C. Mutualism.
Hope this helps :)
The free energy change of the reaction; Fe (s) + Au3+ (aq) -> Fe3+ (aq) + Au (s) is calculated to be -443.83KJ/mol.
For the reaction shown in question 7, we can divide it into half equations as follows;
Oxidation half equation;
6 Al (s) -------> 6Al^3+(aq) + 18e
Reduction half equation;
3Cr2O7^2-(aq) + 42H^+(aq) + 18e -----> 6Cr^3+(aq) + 21H2O(l)
The balanced reaction equation is;
6Al(s) + 3Cr2O7^2-(aq) + 42H^+(aq) -----> 6Al^3+(aq) + 6Cr^3+(aq) + 21H2O(l)
The E° of this reaction is obtained from;
E° anode = -0.04 V
E°cathode = +1.50 V
E° cell = +1.50 V - (-0.04 V) = 1.54 V
Given that;
ΔG° = -nFE°cell
n = 3, F = 96500, E°cell = 1.54 V
ΔG° = -(3 × 96500 × 1.54)
ΔG° = -443.83KJ/mol
Learn more: brainly.com/question/967776
When a volcano erupts magma comes from the top when the magma cools down it hardens and leaves a hard surface
Answer is: <span>the coefficient of phosphoric acid is 12.
</span>Chemical reaction: P₄S₃ + NO₃⁻ + H⁺ → H₃PO₄ + SO₄⁻ + NO.
Reduction half reaction: NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O /·38
Oxidation half reaction: P₄S₃ + 28H₂O → 4H₃PO₄ + 3SO₄²⁻ + 44H⁺ + 38e⁻ /·3.
38NO₃⁻ + 152H⁺ + 3P₄S₃ + 84H₂O → 38NO + 76H₂O + 12H₃PO₄ + 9SO₄²⁻ + 132H⁺.
Balnced chemical reaction:
3P₄S₃ + 38NO₃⁻ + 20H⁺ + 8H₂O → 12H₃PO₄ + 9SO₄²⁻ + 38NO.
The complete question is;
Calculate the empirical formula for each of the following naturalflavors based on their elemental mass percent composition.
Q1)
methyl butyrate (component of apple taste andsmell): C -58.80 % H- 9.87 %
O -31.33.%Express your answer as a chemical formula.
Q2)
vanillin (responsible for the taste and smellof vanilla): C - 63.15% H- 5.30 %
O - 31.55%Express your answer as a chemical formula.
Q1)
empirical formula is the simplest ratio of whole number of elements in the compound. as the percentages have been given, lets calculate for 100 g of compound
C H O
mass 58.80 g 9.87 g 31.33
molar mass 12 g/mol 1 g/mol 16 g/mol
number of moles 58.80/12 9.87/1 31.33/16
= 4.9 =9.87 = 1.95
then divide number of moles by least number of moles - 1.95 in this case
4.9/1.95 = 2.51 9.87/1.95 = 5.06 1.95/1.95= 1
next multiply by 2 to get numbers that can be rounded off to whole numbers
2.51x2 = 5.02 5.06x2 = 10.12 1x2 = 2
when rounded off to the nearest whole number
C - 5
H - 10
O - 2
therefore empirical formula is C₅H₁₀O₂
Q2) for this too since elemental composition has been given in percentages lets calculate for 100 g of compound
C H O
mass 63.15 g 5.30 g 31.55 g
molar mass 12 g/mol 1 g/mol 16 g/mol
number of moles 63.15/12 5.30/1 31.55/16
=5.26 =5.30 =1.97
divide the number of moles by the least number of moles - 1.97
5.26/1.97 5.30/1.97 1.97/1.97
=2.67 = 2.69 = 1
multiply each by 3 to get numbers that can be rounded off to whole numbers
2.67x3 = 8.01 2.69x3 = 8.07 1x3 = 3
rounded off to the nearest whole numbers
C - 8
H - 8
O - 3
empirical formula = C₈H₈O₃
