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Sphinxa [80]
3 years ago
12

PLEASE HELP

Chemistry
1 answer:
Ede4ka [16]3 years ago
6 0

Answer:

Explanation:

E

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Hydrogen is a pure substance represented by the chemical symbol H. Which of the following best describes hydrogen?
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What mass of NH3 in grams must be used to produce 1.81 tons of HNO3 by the Ostwald process, assuming an 80.0 percent yield in ea
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The three reactions involved in the Ostwald process for the conversion of NH3 to HNO3 are:

4NH3(g) + 5O2(g) ==> 4NO(g) + 6H2O(l)   ------(1)

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3NO2(g) + H2O(l) ==> 2HNO3(aq) + NO(g) ---(3)

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Based on the stoichiometry of reaction (3):

Theoretical moles of NO2 = 1.5(Moles HNO2 ) = 1.5(26064) = 39096

Assuming 80% yield:

Actual moles of NO2 = 0.8*39096 = 31277 moles

Based on the stoichiometry in reaction (2):

Theoretical moles of NO = 31277 moles

Assuming 80% yield:

Actual moles of NO = 0.8*31277 = 25022 moles

Bases on stoichiometry in reaction(3):

Moles of NH3 = 25022 moles

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Actual moles of NH3 required = 0.8*25022 = 20018 moles

Mass of NH3 required = 20018 moles * 17 g/mole = 340306 g = 340.31 kg

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calculate the molarity of MgCl2 in the following solution: 5.34 g of MgCl2 is dissolved and diluted to 214 mL of solution.
anyanavicka [17]
<h3><u>Ⲁⲛ⳽ⲱⲉⲅ</u><u>:</u></h3>

\quad\hookrightarrow\quad \sf {0.262M }

<h3><u>Ⲋⲟⳑⳙⲧⳕⲟⲛ :</u></h3>

Molarity is used to measure the concentration of a solution , so it is also as molar concentration. It is denoted as M or Mol/L

<u>We </u><u>are </u><u>given </u><u>that </u><u>:</u>

  • Weight of \sf MgCl_{2} = 5.34g
  • Volume of solution = 214 ml , or 0.214 L

The molar mass of magnesium chloride ( \sf MgCl_{2} ) is 95.21 g / mol

We can calculate the molarity of the solution by dividing the number of moles of solute by volume of solvent in liter ,i.e:

\quad\longrightarrow\quad \sf  {M = \dfrac{n}{V} } ‎ㅤ‎ㅤ‎ㅤ⸻( 1 )

<em>Where,</em><em> </em>

  • M = molarity
  • n = number of moles
  • V = Volume

We can calculate the number of moles by dividing the actual mass by its molar mass ,i.e:

\quad\longrightarrow\quad \sf { n = \dfrac{w}{m}}‎ㅤ‎ㅤ‎ㅤ‎⸻ ( 2 )

<em>W</em><em>here,</em>

  • n = number of moles
  • m = molar mass
  • w = actual mass

<u>Therefore</u><u>,</u>

\implies\quad \tt {n =\dfrac{w}{m} }

\implies\quad \tt { n =\dfrac{5.35\: g}{95.21\: g /mol}}

\implies\quad{\pmb{ \tt {n = 0.056 mol}} }

<u>P</u><u>utting </u><u>the </u><u>values </u><u>in </u><u>equation </u><u>(</u><u> </u><u>1</u><u> </u><u>)</u><u>:</u>

\implies\quad \tt {M=\dfrac{n}{V} }

\implies\quad \tt { M =\dfrac{0.056\:mol}{0.214\:L}}

\implies\quad\underline{\pmb{ \tt { M = 0.262 \:M }}}

7 0
2 years ago
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