Answer:
A. amount of precipitation, average temperature
Explanation:
Precipitation and average temperature are factors that climate includes. These factors are determined by other factors such as location of an area(like how far a place is from large bodies of water like the sea), ocean currents, lattitude (distance from the equator), winds (prevailing winds) , topography (such as mountains) and the like.
Grams of oxygen are required to react with 13.0 grams of octane (C8H18) in the combustion of octane in gasoline is 45.5g
Octane is a hydrocarbon which burns in gasoline in presence of oxygen according to the given balanced chemical equation,
2C₈H₁₈ + 25O₂------> 16CO₂ + 18H₂0
Molar mass of octane = 114.23g/mol
Molar mass of Oxygen = 32g/mol
According to the stiochiometry of the balanced equation the mole ratio of Octane and Oxygen is 2:25
2 mole of octane needs 25 mole of oxygen
1 mole of octane needs 12.5 moleof oxygen
114.23g of octane needs 400g of oxygen
13g of octane needs 45.5g of oxygen
Mass of oxygen needed =45.5g
Hence, the Mass of oxygen needed is 45.5g for the combustion of octane in gasoline.
Learn more about Octane here, brainly.com/question/21268869
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Answer:
No
Explanation:
No, his mass remains the same no matter where he is in the universe.
But then again the moon has less gravitational pull, therefore your weight and mass will be smaller in space and on the moon than on earth
I hope this was helpful! ;)
Answer:
Spanish I dont know spanish.
Explanation:
Answer:
pH = 3.3
Explanation:
Buffer solutions minimize changes in pH when quantities of acid or base are added into the mix. The typical buffer composition is a weak electrolyte (wk acid or weak base) plus the salt of the weak electrolyte. On addition of acid or base to the buffer solution, the solution chemistry functions to remove the acid or base by reacting with the components of the buffer to shift the equilibrium of the weak electrolyte left or right to remove the excess hydronium ions or hydroxide ions is a way that results in very little change in pH of the system. One should note that buffer solutions do not prevent changes in pH but minimize changes in pH. If enough acid or base is added the buffer chemistry can be destroyed.
In this problem, the weak electrolyte is HNO₂(aq) and the salt is KNO₂(aq). In equation, the buffer solution is 0.55M HNO₂ ⇄ H⁺ + 0.75M KNO₂⁻ . The potassium ion is a spectator ion and does not enter into determination of the pH of the solution. The object is to determine the hydronium ion concentration (H⁺) and apply to the expression pH = -log[H⁺].
Solution using the I.C.E. table:
HNO₂ ⇄ H⁺ + KNO₂⁻
C(i) 0.55M 0M 0.75M
ΔC -x +x +x
C(eq) 0.55M - x x 0.75M + x b/c [HNO₂] / Ka > 100, the x can be
dropped giving ...
≅0.55M x ≅0.75M
Ka = [H⁺][NO₂⁻]/[HNO₂] => [H⁺] = Ka · [HNO₂]/[NO₂⁻]
=> [H⁺] = 6.80x010⁻⁴(0.55) / (0.75) = 4.99 x 10⁻⁴M
pH = -log[H⁺] = -log(4.99 x 10⁻⁴) -(-3.3) = 3.3
Solution using the Henderson-Hasselbalch Equation:
pH = pKa + log[Base]/[Acid] = -log(Ka) + log[Base]/[Acid]
= -log(6.8 x 10⁻⁴) + log[(0.75M)/(0.55M)]
= -(-3.17) + 0.14 = 3.17 + 0.14 = 3.31 ≅ 3.3
