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4 years ago

Hummingbirds eat approximately 9.0g of sugar water per day

Chemistry

1 answer:

Darya [45]4 years ago

4 0

Answer:

The answer to your question is A. 0.0066 moles or 6.6 x 10⁻³ moles

B. C₁₂H₂₂O₁₁ + 12 O₂ ⇒ 12 CO₂ + 11H₂O

Explanation:

Data

mass of sugar water = 9 g

concentration = 25%

moles of sugar (C₁₂H₂₂O₁₁)

Formula

Mass percent = mass of solute/mass of solution x 100

-Solve for mass of solute

Mass of solute = (mass percent x mass of solution) / 100

-Substitution

Mass of solute = (25 x 9)/100

-Result

Mass of solute = 225/100

= 2.25 g of sugar

-Calculate the molar mass of sugar

C₁₂H₂₂O₁₁ = (12 x 12) + (22 x 1) + (11 x 16)

= 144 + 22 + 176

= 342 g

-Calculate the moles of sugar

342 g ---------------------- 1 mol

2.25 g ------------------- x

x = (2.25 x 1) / 342

x = 2.25/342

x = 0.0066 moles or 6.6 x 10⁻³ moles

B. Balanced chemical reaction

C₁₂H₂₂O₁₁ + 12 O₂ ⇒ 12 CO₂ + 11H₂O

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(a) The average rate will be:

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

(b) The average rate will be:

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

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Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Thus, the rate of reaction will be:

$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$