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victus00 [196]
3 years ago
13

A gas that effuses through a porous cylinder 1.87 times faster than chlorine gas. what is the molar mass and identity.

Chemistry
1 answer:
Rufina [12.5K]3 years ago
6 0
From the Graham's law of effusion;
R1/R2 = √MM2/√MM1
Molar mass of chlorine gas is 71
Therefore;
1.87= √ 71 /√mm1
= 1.87² = 71/mm1 
mm1 = 71/1.87²
         = 71/3.4969
         = 20.3
Thus, the molar mass of the other gas is 20.3 , and i think the gas is neon
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Answer:

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Explanation:

<u>1) Chemical reaction.</u>

a) Kind of reaction: neutralization

b) General form: acid + base → salt + water

c) Word equation:

  • sodium hydroxide + oxalic acid → sodium oxalate + water

d) Chemical equation:

  • NaOH + H₂C₂O₄ → Na₂C₂O₄ + H₂O

b) Balanced chemical equation:

  • 2NaOH + H₂C₂O₄ → Na₂C₂O₄ + 2H₂O

<u>2) Mole ratio</u>

  • 2mol Na OH : 1 mol H₂C₂O₄ :1 mol Na₂C₂O₄ : 2 mol H₂O

<u>3) Starting amount of oxalic acid</u>

  • mass = 28 mg = 0.028 g
  • molar mass = 90.03 g/mol
  • Convert mass in grams to number of moles, n:

        n = mass in grams / molar mass = 0.028 g / 90.03 g/mol =  0.000311 mol

<u>4) Titration</u>

  • Volume of base: 28.4 mL = 0.0248 liter
  • Concentration of base: x (unknwon)

  • Number of moles of acid: 2.52 mol (calculated above)
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\frac{2}{1} =\frac{x}{2.52}\\ \\ \\x=0.000311(2)=0.000622

That means that there are 0.000622 moles of NaOH (solute)

<u>5) Molarity of NaOH solution</u>

  • M = n / V (liter) = 0.000622 mol / 0.0284 liter = 0.0219 M

That is the correct number using <em>three signficant figures</em>, such as the starting data are reported.

5 0
3 years ago
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Explanation:

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Here element Co is having an oxidation state of +3 called as Co^{3+} cation and phosphprous forms P^{3-} anion with oxidation state of -3. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral CoP

The nomenclature of ionic compounds is given by:

1. Positive is written first followed by the oxidation state of metal in roman numerals in square brackets.

2. The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.

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