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I am Lyosha [343]
3 years ago
13

If you reply with a link, I will report you​

Chemistry
1 answer:
tresset_1 [31]3 years ago
8 0

Answer:

1 2 2

Explanation:

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A 0.652-g sample of a pure strontium halide reacts with excess sulfuric acid. the solid strontium sulfate formed is separated, d
seraphim [82]

Answer:

The formula of the original halide is SrCl₂.

Explanation:

  • The balanced equation of this reaction is:

SrX₂ + H₂SO₄ → SrSO₄ + 2 HX, where X is the halide.

  • From the equation stichiometry, 1.0 mole of strontium halide will result in 1.0 mole of SrSO₄.
  • The number of moles of SrSO₄ <em>(n = mass/molar mass) </em>= (0.755 g) / (183.68 g/mole) = 4.11 x 10⁻³ mole.
  • The number of moles of SrX are  4.11 x 10⁻³ moles from the stichiometry of the balanced equation.
  • n = mass / molar mass, n =  4.11 x 10⁻³ moles and mass = 0.652 g.
  • The molar mass of SrX₂ = mass / n = (0.652) / (4.11 x 10⁻³ moles) = 158.62 g/mole.
  • The molar mass of SrX₂ (158.62 g/mole) = Atomic mass of Sr (87.62 g/mole) + (2 x Atomic mass of halide X).
  • The atomic mass of halide X = (158.62 g/mole) - (87.62 g/mole) / 2 = 71 / 2  g/mole = 35.5 g/mole.
  • This is the atomic mass of Cl.
  • <em>So, the formula of the original halide is SrCl₂</em>.
4 0
3 years ago
Determine the number of milliliters of 0.00300 M phosphoric acid required to neutralize 40.00 mL of 0.00150 M calcium hydroxide.
a_sh-v [17]

The volume of H₃PO₄ : 13.33 ml

<h3>Further explanation</h3>

Given

0.003 M  Phosphoric acid-H₃PO₄

40 ml of 0.00150 M Calcium hydroxide-Ca(OH)₂

Required

Volume of H₃PO₄

Solution

Acid-base titration formula  

Ma. Va. na = Mb. Vb. nb  

Ma, Mb = acid base concentration  

Va, Vb = acid base volume  

na, nb = acid base valence  (amount of H⁺/OH⁻)

H₃PO₄⇒3H⁺ + PO₄³⁻ ⇒ 3 H⁺ = valence = 3

Ca(OH)₂⇒Ca²⁺ + 2OH⁻⇒ 2 OH⁻ = valence = 2

Input the value :

a = H₃PO₄, b = Ca(OH)₂

0.003 x Va x 3 = 0.0015 x 40 x 2

Va = 13.33 ml

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2 years ago
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I think it’ll be Rr
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Define atomic mass unit?​
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