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Goshia [24]
2 years ago
10

Gallium has two naturally occurring isotopes. 69Ga (68.9257 amu) is the more abundant isotope, at 60.400%. If the atomic mass of

gallium is 69.723 amu, what is the mass of the other isotope, 71Ga
Chemistry
1 answer:
LiRa [457]2 years ago
8 0

Answer:

70.9391 amu

Explanation:

The abundance of the second isotope is 100 - 60.4% = 39.6%.

We can solve this problem by keeping in mind the <em>following formula for atomic mass</em>:

  • Atomic mass = Mass of Isotope 1 * Abundance + Mass of Isotope 2 * Abundance

We<u> input the given data</u>:

  • 69.723 = 68.9257 * 60.4/100 + Mass of Isotope 2 * 39.6/100
  • Mass of Isotope 2 = 70.9391 amu
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Calculate the volume in mL of 0.279 M Ca(OH)2 needed to neutralize 24.5 mL of 0.390 M H3PO4 in a titration.
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The volume of the 0.279 M Ca(OH)₂ solution required to neutralize 24.5 mL of 0.390 M H₃PO₄ is 51.4 mL

<h3>Balanced equation </h3>

2H₃PO₄ + 3Ca(OH)₂ —> Ca₃(PO₄)₂ + 6H₂O

From the balanced equation above,

  • The mole ratio of the acid, H₃PO₄ (nA) = 2
  • The mole ratio of the base, Ca(OH)₂ (nB) = 3

<h3>How to determine the volume of Ca(OH)₂ </h3>
  • Molarity of acid, H₃PO₄ (Ma) = 0.390 M
  • Volume of acid, H₃PO₄ (Va) = 24.5 mL
  • Molarity of base, Ca(OH)₂ (Mb) = 0.279 M
  • Volume of base, Ca(OH)₂ (Vb) =?

MaVa / MbVb = nA / nB

(0.39 × 24.5) / (0.279 × Vb) = 2/3

9.555 / (0.279 × Vb) = 2/3

Cross multiply

2 × 0.279 × Vb = 9.555 × 3

0.558 × Vb = 28.665

Divide both side by 0.558

Vb = 28.665 / 0.558

Vb = 51.4 mL

Thus, the volume of the Ca(OH)₂ solution needed is 51.4 mL

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brainly.com/question/14356286

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