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11 days ago

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Chemistry

2 answers:

VMariaS [2.6K]11 days ago

8 0

Response:

a. 3 Br₂(l) + 6 OH⁻(aq) → 5 Br⁻(aq) + BrO₃⁻(aq) + 3 H₂O

b. Br₂

c. Br₂

Clarification:

Balancing a redox reactionis performed using the ion-electron method.

Step 1: Identify both half-reactions.

Reduction: Br₂(l) → Br⁻(aq)

Oxidation: Br₂(l) → BrO₃⁻(aq)

Step 2: Perform mass balance. This reaction occurs in basic conditions, thus we must add OH⁻ and H₂O as needed.

0.5 Br₂(l) → Br⁻(aq)

6 OH⁻(aq) + 0.5 Br₂(l) → BrO₃⁻(aq) + 3 H₂O

Step 3: Ensure electrical balance by incorporating electrons when necessary.

1 e⁻ + 0.5 Br₂(l) → Br⁻(aq)

6 OH⁻(aq) + 0.5 Br₂(l) → BrO₃⁻(aq) + 3 H₂O + 5 e⁻

Step 4: Scale both half-reactions to ensure the electron counts balance out.

5 × (1 e⁻ + 0.5 Br₂(l) → Br⁻(aq))

1 × (6 OH⁻(aq) + 0.5 Br₂(l) → BrO₃⁻(aq) + 3 H₂O + 5 e⁻)

Step 5: Combine both half-reactions and simplify as appropriate.

5 e⁻ + 3 Br₂(l) + 6 OH⁻(aq) → 5 Br⁻(aq) + BrO₃⁻(aq) + 3 H₂O + 5 e⁻

3 Br₂(l) + 6 OH⁻(aq) → 5 Br⁻(aq) + BrO₃⁻(aq) + 3 H₂O

The species that undergoes reduction is identified as the oxidizer. The species that undergoes oxidation is termed the reducer. In this situation, Br₂ qualifies as both.

Alekssandra [2.7K]11 days ago

6 0

Response:

the oxidizing agent is $Br_{2}$ while the reducing agent is $Br_{2}$

Clarification:

In reduction-oxidation reactions, electrons are transferred between compounds. During reduction, the oxidation number decreases, while it increases during oxidation. A reducing agent loses electrons, increasing its oxidation number, whereas an oxidizing agent gains electrons, reducing its oxidation number.

The unbalanced equation is:

$Br_{2(l)}$ → $BrO^{-} _{3(aq)}$ + $Br_{(aq)} ^{-}$

For oxidation reactions:

$Br_{2} + 6H_{2}O$ → $2BrO^{-} _{3(aq)}$ + 12 $H^{+}$ + $10e^{-}$

For reduction reactions:

$Br_{2(l)}$ + $2e^{-}$ → $2Br^{-}$

The next step is to equalize the electrons gained and lost:

$Br_{2} + 6H_{2}O$ → $2BrO^{-} _{3(aq)}$ + 12 $H^{+}$ + $10e^{-}$

+ $5Br_{2(l)}$ → $10e^{-}$ $10Br^{-}$

Finally, combine both equations:

⇌ $3Br_{2(aq)}+3H_{2}O_{(l)}$ + $2BrO^{-} _{3(aq)}$ + $5Br_{2(l)}$ $6H_{(aq)} ^{+}$

Thus, the oxidizing agent is $Br_{2}$ and the reducing agent is $Br_{2}$

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En un balneario necesitan calentarse 1 millón de litros de agua anuales, subiendo la temperatura desde 15 ºC a 50 ºC y para ello

KiRa [2711]

Answer:

a) $m_{CH_4}=2630kg$

b) 1657 €

Explanation:

Hola,

a) En esta cuestión analizaremos el millón de litros de agua anualmente, dado que este dato nos permite calcular el calor necesario para calentar dicha cantidad, considerando que la densidad del agua es de 1 kg/L:

$Q_{H_2O}=m_{H_2O}Cp(T_2-T_1)=1x10^6LH_2O*\frac{1kgH_2O}{1LH_2O}*4.18\frac{kJ}{kg\°C}(50-15) \°C\\Q_{H_2O}=146.3x10^6kJ$

A continuación, utilizamos la entalpía de combustión del metano para determinar la cantidad en kilogramos necesaria, ya que la energía calórica perdida por el metano es equivalente a la energía obtenida por el agua:

$Q_{H_2O}=-Q_{CH_4}=-146.3x10^5kJ=m_{CH_4}\Delta _cH_{CH_4}$

$m_{CH_4}= \frac{Q_{CH_4}}{\Delta _cH_{CH_4}} =\frac{-146.3x10^5kJ}{-890kJ/molCH_4} *\frac{16gCH_4}{1molCH_4} \\\\m_{CH_4}=2630112.36g=2630kg$

b) En este supuesto, tenemos que, bajo condiciones normales de 1 bar y 273 K, el precio de 1 metro cúbico de metano es 0,45 €, lo que nos permite calcular las moles de metano en esas condiciones:

$n_{CH_4}=\frac{PV}{RT}=\frac{1atm*1000L}{0.082\frac{atm*L}{mol*K}*273K} =44.67mol$

En consecuencia, los kilogramos de metano que se obtienen por 0,45 € son:

$44.67molCH_4*\frac{16gCH_4}{1molCH_4}*\frac{1kg}{1000g} =0.715kgCH_4$

Finalmente, usando regla de tres:

0.715 kg ⇒ 0.45 €

2630 kg ⇒ X

X = (2630 kg x 0.45 €) / 0.715 kg

X = 1657 €

Regards.

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29 days ago

A student puts 0.020 mol of methyl methanoate into an empty and rigid 1.0 L vessel at 450 K. The pressure is measured to be 0.74

KiRa [2711]

Explanation:

Initial moles of ethanoic acid = 0.020 mol

At equilibrium, half of the ethanoic acid molecules have reacted.

Thus, moles of ethanoic acid reacted = 0.020 mol * (50% / 100%)

= 0.010 mol

Moles of ethanoic acid remaining = 0.020 mol - 0.010 mol = 0.010 mol

The moles of product $(CH3COOH)^{2}$ gas formed are determined as follows:

0.010 mol CH3COOH * (1 mol $(CH3COOH)^{2}$ / 2 mol CH3COOH)

= 0.005 mol $(CH3COOH)^{2}$

Consequently, the total moles of gas present in the vessel at equilibrium are 0.010 mol CH3COOH and 0.005 mol $(CH3COOH)^{2}$

Total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol

Next, let’s determine the pressure:

0.020 mol of gas has a pressure of 0.74 atm; so under the same conditions, we find the pressure exerted by 0.015 mol of gas:

P1/n1 = P2/n2

P2 = P1*(n2 / n1)

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1 month ago

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lorasvet [2515]

Answer: The percentage abundance for $_{14}^{30}\textrm{Si}$ isotope is 3.09 %.

Explanation:

The average atomic mass of an element is calculated by taking the sum of the masses of all isotopes weighted by their respective natural fractional abundances.

To compute average atomic mass, the following formula is applied:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

From the information provided:

Let the fractional abundance for $_{14}^{28}\textrm{Si}$ isotope be 'x'

For $_{14}^{28}\textrm{Si}$ isotope:

Mass of $_{14}^{28}\textrm{Si}$ isotope = 27.9769 amu

Percentage abundance of $_{14}^{28}\textrm{Si}$ isotope = 92.22 %

Fractional abundance for $_{14}^{28}\textrm{Si}$ isotope = 0.9222

For $_{14}^{29}\textrm{Si}$ isotope:

Mass of $_{14}^{28}\textrm{Si}$ isotope = 28.9764 amu

Percentage abundance for $_{14}^{28}\textrm{Si}$ isotope = 4.68%

Fractional abundance of $_{14}^{28}\textrm{Si}$ isotope = 0.0468

For $_{14}^{30}\textrm{Si}$ isotope:

Mass of $_{14}^{30}\textrm{Si}$ isotope = 29.9737 amu

Fractional abundance for $_{14}^{30}\textrm{Si}$ isotope = x

The average atomic mass of silicon is 28.084 amu

By inserting these values into equation 1, we derive:

$28.084=[(27.9769\times 0.9222)+(28.9764\times 0.0468)+(29.9737\times x)]\\\\x=0.0309$

To convert this fractional abundance into a percentage, multiply by 100:

$\Rightarrow 0.0309\times 100=3.09\%$

This shows that the percentage abundance for $_{14}^{30}\textrm{Si}$ isotope is 3.09 %.

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1 month ago

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The inquiry is incomplete; here is the full question:

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Independent Variable-

Dependent Variable-

Constants

Control Group-

Answer:

A) The quantity of food given to the goldfish

B) The body fat of the goldfish

C) -Type of fish in the experiment (goldfish)

Time period for feeding the fish (six weeks)

Shape and size of the tanks

D) group of goldfish receiving the standard feeding amount

Explanation:

The objective of the experiment is to assess how the quantity of food affects the body fat of goldfish. Consequently, the amount of food serves as the independent variable while the body fat acts as the dependent variable.

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