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Anettt [7]
3 years ago
12

What alkene gives a mixture of acetone and propanoic acid on reaction with potassium permanganate in the presence of strong acid

and water?
Chemistry
1 answer:
navik [9.2K]3 years ago
4 0

Answer:

(CH3)2C=CHCH2CH3

2-methylpent-2-ene

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The ph of 0.015 m hno2 (nitrous acid) aqueous solution was measured to be 2.63. what is the value of pka of nitrous acid?
Licemer1 [7]
Nitrous acid<span> dissociates as follows:
</span>
HNO₂(s) ⇄ H⁺(aq) + NO₂⁻(aq) 
           
According to the equation, an acid constant has the following form:

Ka = [H⁺] × [NO₂⁻ ] / [HNO₂] 

From pH, we can calculate the concentration of H⁺ and NO₂⁻:

[H⁺] = 10^-pH = 10^-2.63 = 0.00234 M = [NO₂⁻]

Now, the acid constant can be calculated:

Ka = 0.00234 x 0.00234 / 0.015  = 3.66 x 10⁻⁴

And finally,

pKa = -log Ka = 3.44 


7 0
3 years ago
Making the simplistic assumption that the dissolved NaCl(s) does not affect the volume of the solvent water, determine the const
shtirl [24]

Answer:

  • m = 1,000/58.5
  • b = - 1,000 / 58.5

1) Variables

  • molarity: M
  • density of the solution: d
  • moles of NaCl: n₁
  • mass of NaCl: m₁
  • molar mass of NaCl: MM₁
  • total volume in liters: Vt
  • Volume of water in mililiters: V₂
  • mass of water: m₂

2) Density of the solution: mass in grams / volume in mililiters

  • d = [m₁ + m₂] / (1000Vt)

3) Mass of NaCl: m₁

    Number of moles = mass in grams / molar mass

    ⇒ mass in grams = number of moles × molar mass

        m₁ = n₁ × MM₁


4) Number of moles of NaCl: n₁

   Molarity = number of moles / Volume of solution in liters

   M = n₁ / Vt

   ⇒ n₁ = M × Vt


5) Substitue in the equation of m₁:

   m₁ = M × Vt × MM₁


6) Substitute in the equation of density:

    d = [M × Vt × MM₁ + m₂] / (1000Vt)


7) Simplify and solve for M

  • d = M × Vt × MM₁ / (1000Vt) + m₂/ (1000Vt)
  • d = M × MM₁ / (1000) + m₂/ (1000Vt)

Making the simplistic assumption that the dissolved NaCl(s) does not affect the volume of the solvent water means 1000Vt = V₂  

  • d = M × MM₁ / (1000) + m₂/ V₂

        m₂/ V₂ is the density of water: 1.00 g/mL

  • d = M × MM₁ / (1000) + 1.00 g/mL
  • M × MM₁ / (1000) = d - 1.00 g/mL
  • M = [1,000/MM₁] d - 1,000/ MM₁

8) Substituting MM₁ = 58.5 g/mol

  • M = [1,000/58.5] d - [1,000/ 58.5]

Comparing with the equation Molarity = m×density + b, you obtain:

  • m = 1,000/58.5
  • b = - 1,000/58.5
7 0
3 years ago
Read 2 more answers
What does the formula for water, H2O, tell you about what the molecule is made of ?
Sergeu [11.5K]

Answer:

To find out what water is made of, it helps to look at its chemical formula, which is H2O. This basically tells us that the water molecule is composed of two elements: hydrogen and oxygen or, more precisely, two hydrogen atoms (H2) and one oxygen atom (O). Hydrogen and oxygen are gases at room temperature.

Explanation:

4 0
2 years ago
1. Pewien tlenek azotu o masie cząsteczkowej 108 u zawiera 74,07% tlenu. Wykonaj stosowne obliczenia i napisz wzór sumaryczny te
love history [14]

Answer:

1. Stąd empiryczny wzór substancji to N₂O₅

2. W związku z tym ilość w gramach chlorku sodu NaCl is 114,4 g.

Explanation:

1. Mamy tutaj;

Masa molowa tlenku azotu = 108u

Masa azotu = 14,0067u

Masa tlenu = 15,999 u

74,07% masy tlenku azotu to tlen

Dlatego masa obecnego tlenu = 108 × 74,07 / 100 = 79,9956 u

Masa obecnego azotu = 108 - 79,9956 = 28,0044u

Liczba moli tlenu = 79,9956 / 15,999 = 5,00003 ≈ 5

Liczba moli azotu = 28,0044 / 14,0067 = 1,99935 ≈ 2

Stąd empiryczny wzór substancji to N₂O₅.

2. Kiedy sód reaguje z chlorem, mamy;

2Na (s) + Cl₂ (g) → 2NaCl (s)

Dlatego 2 mole sodu Na reaguje z 1 molem chloru gazowego Cl₂, z wytworzeniem 2 moli chlorku sodu NaCl

W związku z tym 1 mol sodu Na reaguje z 1/2 molem chloru gazowego Cl₂ z wytworzeniem 1 mola chlorku sodu NaCl

Masa Na obecnego w reakcji = 45 g

Masa molowa sodu = 22,989769u

Liczbę moli sodu w 45 g sodu podano w następujący sposób;

Liczba \, \, moli \, \, Na= \frac{Mass \, of \, Na}{Molowy \, masa \, z \, Na} = \frac{45}{22.989769} = 1.96 \, mole

Z czego 1,96 moli sodu Na reaguje z 1/2 × 1,96 mola chloru gazowego Cl₂ z wytworzeniem 1,96 mola chlorku sodu NaCl

Masa molowa NaCl = 58,44 g / mol

Dlatego masa NaCl = liczba moli NaCl × masa molowa NaCl

Masa NaCl = 1,96 × 58,44 = 114,39001 g ≈ 114,4 g

W związku z tym ilość w gramach chlorku sodu NaCl = 114,4 g.

4 0
3 years ago
Both HNO3(aq) and CH3COOH(aq) can be classified as *
lord [1]

Answer:

acids

Explanation:

HNO3 is a strong acid (Nitric Acid)

CH3COOH is a weak acid (Acetic Acid

6 0
3 years ago
Read 2 more answers
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