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Mnenie [13.5K]
2 years ago
6

Iron fillings sprinkled near a magnet arrange themselves into a pattern that illustrates the

Chemistry
1 answer:
Ipatiy [6.2K]2 years ago
5 0

the force of attraction of iron fillings towards magnet


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A. True

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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.
Vsevolod [243]

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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2 years ago
What is the mass of 8.0 x 10^26 UF6 molecules?
gulaghasi [49]
First, find the number of moles of UF6
Avagadro's number = 6.023 x 10^23

Number of moles = 8.0 x 10^26 / Avagadro's number = 8.0 x 10^26 / 6.023 x 10^23 = 1.328 x 10³ moles

Molecular weight of UF6 = Molecular weight of U (238.02891) + Molecular weight of F6 (6 x 18.9984032) = 238.02891 + 113.9904192 = 352.0193292 g/mol

Therefore mass of 8.0 x 10^26 UF6 molecules = 352.0193292 g/mol x 1.328 x 10³ moles = 467.481669 x 10³ grams




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2 years ago
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