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denis23 [38]
2 years ago
9

An aqueous solution of calcium hydroxide is standardized by titration with a 0.112 M solution of hydrobromic acid. If 15.2 mL of

base are required to neutralize 12.4 mL of the acid, what is the molarity of the calcium hydroxide solution?
Chemistry
1 answer:
TiliK225 [7]2 years ago
7 0

Answer:

0.0457 M

Explanation:

The reaction that takes place is:

  • 2HBr + Ca(OH)₂ → CaBr₂ + 2H₂O

First we<u> calculate how many moles of acid reacted</u>, using the <em>HBr solution's concentration and volume</em>:

  • Molarity = Moles / Volume
  • Molarity * Volume = Moles
  • 0.112 M * 12.4 mL = 1.389 mmol HBr

Now we <u>convert HBr moles to Ca(OH)₂ moles</u>, using the stoichiometric ratio:

  • 1.389 mmol HBr * \frac{1mmolCa(OH)_{2}}{2mmolHBr} = 0.6944 mmol Ca(OH)₂

Finally we <u>calculate the molarity of the Ca(OH)₂ solution</u>, using the <em>given volume and calculated moles</em>:

  • 0.6944 mmol Ca(OH)₂ / 15.2 mL = 0.0457 M
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Suppose three neutrons are released when an atom in a sample of fissionable nuclei undergoes fission. Each of these neutrons has
Eddi Din [679]

Answer:

13 nuclei

Explanation:

From the question, the fission of one nucleus produces three neutrons which causes  more nuclei to undergo fission.

This implies that, after the first fission, three neutrons cause three nuclei to undergo fission. The three nuclei that underwent fission produces nine neutrons which causes the fission of nine nuclei.

All together we the number of nuclei that underwent fission as;

1 + 3 + 9 = 13 nuclei.

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2 years ago
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tatyana61 [14]

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Si compro un televisor a 30.000 pesos y lo pago en 12 cuotas con un recargo del 15% ¿Cual sera el valor de cada cuota?¿ Cual ser
pantera1 [17]
Each fee would be 2,875 pesos
The total amount of the TV would be 34,500 pesos

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4 0
3 years ago
The equilibrium of 2H 2 O(g) 2H 2 (g) + O 2 (g) at 2,000 K has a Keq value of 5.31 x 10-10. What is the Keq expression for this
maks197457 [2]
Answer is: Keq expression for this system is Keq = <span>[O</span>₂<span> ]</span> · [H₂<span>]</span>² ÷  [H₂O<span>]</span>².<span>
Chemical reaction: 2H</span>₂O(g) ⇄ O₂(g) + 2H₂(g).
The equilibrium constant<span> (Keq) is a ratio of the concentration of the products (in this reaction oxygen and hydrogen) to the concentration of the reactants (in this reaction water).</span>
7 0
3 years ago
Consider the reaction of peroxydisulfate ion (S2O2−8) with iodide ion (I−) in aqueous solution: S2O2−8(aq)+3I−(aq)→2SO2−4(aq)+I−
kolbaska11 [484]

Answer:

r = 3.61x10^{-6} M/s

Explanation:

The rate of disappearance (r) is given by the multiplication of the concentrations of the reagents, each one raised of the coefficient of the reaction.

r = k.[S2O2^{-8} ]^{x} x [I^{-} ]^{y}

K is the constant of the reaction, and doesn't depends on the concentrations. First, let's find the coefficients x and y. Let's use the first and the second experiments, and lets divide 1º by 2º :

\frac{r1}{r2} = \frac{0.018^{x} x0.036^{y} }{0.027^xx0.036^y}

\frac{2.6x10^{-6}}{3.9x10^{-6}} = (\frac{0.018}{0.027})^xx(\frac{0.036}{0.036})^y

0.67 = 0.67^x

x = 1

Now, to find the coefficient y let's do the same for the experiments 1 and 3:

\frac{r1}{r3} = \frac{0.018x0.036^y}{0.036x0.054^y}

\frac{2.6x10^{-6}}{7.8x10^{-6}} = (\frac{0.018}{0.036})x(\frac{0.036}{0.054})^y

0.33 = 0.5x 0.67^y

0.67 = 0.67^y

y = 1

Now, we need to calculate the constant k in whatever experiment. Using the first :

2.6x10^{-6} = kx0.018x0.036kx6.48x10^{-4} = 2.6x10^{-6}

k = 4.01x10^{-3} M^{-1}s^{-1}[/tex]

Using the data given,

r = 4.01x10^{-3}x1.8x10^{-2}x5.0x10^{-2}

r = 3.61x10^{-6} M/s

7 0
3 years ago
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