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tester [92]
3 years ago
7

Write a balanced equation for the single-replacement oxidation-reduction reaction described, using the smallest possible integer

coefficients. The reaction that takes place when chlorine gas combines with aqueous potassium bromide. (Use the lowest possible coefficients. Omit states of matter.)
Chemistry
1 answer:
jeyben [28]3 years ago
7 0

Answer:

\rm Cl_2 + 2\; KBr \to Br_2 + 2\; KCl.

One chlorine molecule reacts with two formula units of (aqueous) potassium bromide to produce one bromine molecule and two formula units of (aqueous) potassium chloride.

Explanation:

<h3>Formula for each of the species</h3>

Start by finding the formula for each of the compound.

  • Both chlorine \rm Cl and bromine \rm Br are group 17 elements (halogens.) Each
  • On the other hand, potassium \rm K is a group 1 element (alkaline metal.) Each

Therefore, the ratio between \rm K atoms and \rm Br atoms in potassium bromide is supposed to be one-to-one. That corresponds to the empirical formula \rm KBr. Similarly, the ratio between

The formula for chlorine gas is \rm Cl_2, while the formula for bromine gas is \rm Br_2.

<h3>Balanced equation for the reaction</h3>

Write down the equation using these chemical formulas.

\rm ?\; Cl_2 + ?\; KBr \to ?\;Br_2 + ?\; KCl.

Start by assuming that the coefficient of compound with the largest number of elements is one. In this particular equation, both \rm KBr and \rm KCl features two elements each.

Assume that the coefficient of \rm KCl is one. Hence:

\rm ?\; Cl_2 + 1 \; KBr \to ?\;Br_2 + ?\; KCl.

Note that \rm KBr is the only source of \rm K and \rm Br atoms among the reactants of this reaction.

There would thus be one \rm K atom and one \rm Br atom on the reactant side of the equation.

Because atoms are conserved in a chemical equation, there should be the same number of \rm K and \rm Br atoms on the product side of the equation.

In this reaction, \rm Br_2 is the only product with \rm Br atoms.

One \rm Br atom would correspond to 0.5 units of \rm Br_2.

Similarly, in this reaction, \rm KCl is the only product with \rm K atoms.

One \rm K atom would correspond to one formula unit of \rm KCl.

Hence:

\displaystyle \rm ?\; Cl_2 + 1 \; KBr \to \frac{1}{2}\;Br_2 + 1\; KCl.

Similarly, there should be exactly one \rm Cl atom on either side of this equation. The coefficient of \rm Cl_2 should thus be 0.5. Hence:

\displaystyle \rm \frac{1}{2}\; Cl_2 + 1 \; KBr \to \frac{1}{2}\;Br_2 + 1\; KCl.

That does not meet the requirements, because two of these coefficients are not integers. Multiply all these coefficients by two (the least common multiple- LCM- of these two denominators) to obtain:

\displaystyle \rm 1\; Cl_2 + 2 \; KBr \to 1\;Br_2 + 2\; KCl.

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3 years ago
23.495 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 72.311 g of wate
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Answer:

1.86% NH₃

Explanation:

The reaction that takes place is:

  • HCl(aq) + NH₃(aq) → NH₄Cl(aq)

We <u>calculate the moles of HCl that reacted</u>, using the volume used and the concentration:

  • 32.27 mL ⇒ 32.27/1000 = 0.03227 L
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The moles of HCl are equal to the moles of NH₃, so now we <u>calculate the mass of NH₃ that was titrated</u>, using its molecular weight:

  • 3.4852x10⁻³ mol NH₃ * 17 g/mol = 0.0592 g NH₃

The weight percent NH₃ in the aliquot (and thus in the diluted sample) is:

  • 0.0592 / 12.949 * 100% = 0.4575%

Now we <u>calculate the total mass of NH₃ in the diluted sample</u>:

Diluted sample total mass = Aqueous waste Mass + Water mass = 23.495 + 72.311 = 95.806 g

  • 0.4575% * 95.806 g = 0.4383 g NH₃

Finally we calculate the weight percent NH₃ in the original sample of aqueous waste:

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Answer:

487.33 K.

Explanation:

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where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant.

T is the temperature of the gas in K.

  • If n is constant, and have two different values of (P, V and T):

<em>P₁V₁T₂ = P₂V₂T₁</em>

<em></em>

P₁ = 5.4 atm, V₁ = 1.0 L, T₁ = 33°C + 273 = 306 K.

P₂ = 4.3 atm, V₂ = 2.0 L, T₂ =??? K.

<em>∴ T₂ = P₂V₂T₁/P₁V₁</em> = (4.3 atm)(2.0 L)(306 K)/(5.4 atm)(1.0 L) = <em>487.33 K.</em>

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