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What is the mole of 3.5 g AL(NO3)3?​

Sagot :

MrChrisidn

[tex]\underline{\underline{\large{\green{\cal{✒GIVEN:}}}}}[/tex]

[tex]\bullet \: \rm{3.5 g AL(NO3)3}[/tex]

[tex]\underline{\underline{\large{\green{\cal{REQUIRED:}}}}}[/tex]

Find the mole of 3.5 g AL(NO3)3.

[tex]\underline{\underline{\large{\green{\cal{SOLUTION:}}}}}[/tex]

The formula for calculating the [tex]\tt{\purple{mole \: of \: the \: compound}}[/tex] is expressed as:

[tex]\large{\boxed{\bm{\red{ mole= \dfrac{Mass}{Molar \: mass}}}}}[/tex]

Determine the molar mass of Al(NO3)3:

[tex]\tt{molar \: mass=27+3(14)+3(3×16)}[/tex]

[tex]\tt{molar \: mass=27+42+3(48)}[/tex]

[tex]\tt{molar \: mass=69+144}[/tex]

[tex]\tt{molar \: mass=213gmol^{−1}}[/tex]

Determine the required moles of Al(NO3)3, using the formula:

[tex]\tt{mole = \dfrac{3.5g}{213 {gmol}^{ - 1} } }[/tex]

[tex]\large{\tt{\purple{mole \approx 0.015 \: mol}}}[/tex]

Final Answer:

Hence, the mole of 3.5 g AL(NO3)3 [tex]\large{\rm{\purple{0.015 \: mol}}}[/tex]

NoahBadillaidn

MOLAR MASS

[tex]\normalsize{\blue{\overline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}}}[/tex]

What is the mole of 3.5 g AL(NO3)3?​

[tex]\\[/tex]

[tex]\rm{\underline{SOLUTION:}}[/tex]

To calculate how many moles are in 3.5 g of AL(NO3)3 (aluminum nitrate), first find the atomic masses of what aluminum nitrate is composed of. According to the periodic table:

  • Aluminum ≈ 27 g/mol
  • Nitrogen ≈ 14 g/mol
  • Oxygen ≈ 16 g/mol

[tex]\\[/tex]

In [tex]\sf{AL(NO_{3})_{3}}[/tex], there are:

  • 1 atom of aluminum
  • 3 atoms of nitrogen
  • 9 atoms of oxygen

[tex]\\[/tex]

Calculate the total molar mass of all three:

  • Aluminum: 1 atoms × 27 g/mol = 27 g/mol
  • Nitrogen: 3 atoms × 14 g/mol = 42 g/mol
  • Oxygen: 9 atoms × 16 g/mol = 144 g/mol

[tex]\\[/tex]

Calculate the number of moles using the formula:

Moles = mass/molar mass

  • [tex]\rm{Moles = \dfrac{mass}{molar \: mass} }[/tex]

  • [tex]\rm{Moles = \dfrac{3.5 \: g}{27 \: g/mol+42 \: g/mol+144 \: g/mol} }[/tex]

  • [tex]\rm{Moles = \dfrac{3.5 \: g}{213 \: g/mol} }[/tex]

  • [tex]\rm{Moles = \green{\underline{0.01643}}}[/tex]

[tex]\\[/tex]

Hence, there are 0.0164 moles in 3.5 g of [tex]\sf{AL(NO_{3})_{3}}[/tex].

[tex]\normalsize{\blue{\overline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}}}[/tex]

[tex]\large{\boxed{\tt{\blue{06/20/2024}}}}[/tex]

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