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AysviL [449]
3 years ago
10

Use the ruler to determine the length of this object. Record your answer to the nearest tenth. The object is ____ long.

Chemistry
1 answer:
VLD [36.1K]3 years ago
7 0
The object is 2.7 cm long
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3 years ago
A 30.7 g sample of Strontium nitrate, Sr(NO3)2•nH2O, is heated to a constant mass of 22.9 g. Calculate the hydration number.
Elodia [21]

Answer:

  • <em>Hydration number:</em> 4

Explanation:

<u>1) Mass of water in the hydrated compound</u>

Mass of water = Mass of the hydrated sample - mass of the dehydrated compound

Mass of water = 30.7 g - 22.9 g = 7.8 g

<u>2) Number of moles of water</u>

  • Number of moles = mass in grams / molar mass

  • molar mass of H₂O = 2×1.008 g/mol + 15.999 g*mol = 18.015 g/mol

  • Number of moles of H₂O = 7.9 g / 18.015 g/mol = 0.439 mol

<u>3) Number of moles of Strontium nitrate dehydrated, Sr (NO₃)₂</u>

  • The mass of strontium nitrate dehydrated is the constant mass obtained after heating = 22.9 g

  • Molar mass of Sr (NO₃)₂ :  211.63 g/mol (you can obtain it from a internet or calculate using the atomic masses of each element from a periodic table).

  • Number of moles of Sr (NO₃)₂ = 22.9 g / 211.63 g/mol =  0.108 mol

<u>4) Ratio</u>

  • 0.439 mol H₂O / 0.108 mol Sr(NO₃)₂ ≈  4 mol H₂O : 1 mol Sr (NO₃)₂

Which means that the hydration number is 4.

4 0
4 years ago
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