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In-s [12.5K]
3 years ago
10

Chemistry help! Thanks to anyone who helps!

Chemistry
1 answer:
zmey [24]3 years ago
5 0
A: 2-methyl-2-propanom
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What thype of bond requires the give and take of electrons ?
DedPeter [7]
Ionic bond involves electrostatic attraction between oppositely charged ions.
The ions are atoms that have gained 1 or more electrons and atoms that have lost 1 or more electrons.
<span>Answer: The type of bond that requires the give and take of electrons is Ionic bond</span>
6 0
4 years ago
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What particle limitation to the widespread of this reusable water bottles? Pls help someone
Sauron [17]

Answer: Plastic water bottles

Explanation:

If you use disposable water bottles, here are some important concerns you should know about how they’re made as well as the problems they cause for the planet, your health, and your wallet.

3 0
3 years ago
If you start with 0.30 m mn2 , at what ph will the free mn2 concentration be equal to 4.6 x 10-11 m?
aksik [14]

If you start with 0.30 m Mn₂ , at 12.5 pH, free Mn₂ concentration be equal to 4.6 x 10⁻¹¹ m

Initial molarity of Mn₂ = 0.30 M

Final molarity of Mn₂ = 4.6 x 10⁻¹¹

pH = ?

Ksp [Mn(OH)₂] = 4.6 x 10⁻¹⁴ (standard value)

Write the ionic equation

    Mn(OH)₂   →    Mn⁺² + 2OH⁻

    [Mn⁺²] = 4.6 x 10⁻¹¹

We will calculate the concentration of OH⁻ by using Ksp expression

    Ksp = [Mn⁺²][OH-]²

    [Mn⁺²][OH⁻]² = 4.6 x 10⁻¹⁴

    [OH⁻]² = 4.6 x 10⁻¹⁴ / 4.6 x 10⁻¹¹

    [OH⁻]² = 10⁻³

    [OH⁻] = (10⁻³)¹⁽²

    [OH⁻] = 0.0316 M

Calculate the pOH

    pOH = -log [OH⁻]

    pOH =  -log [0.0316]

    pOH = 1.5

Now calculate pH

   pH = 14 - pOH

   pH = 14 - 1.5

   pH = 12.5

You can also learn about molarity from the following question:

brainly.com/question/14782315

#SPJ4

7 0
1 year ago
Calculate how many grams of the first reactant are necessary to completely react with 17.3 g of the second reactant. the reactio
soldier1979 [14.2K]

Taking into account the reaction stoichiometry, 16.611 grams of Na₂CO₃ are necessary to completely react with 17.3 g of CuCl₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Na₂CO₃ + CuCl₂  → CuCO₃ + 2 NaCl

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Na₂CO₃: 1 mole
  • CuCl₂: 1 mole
  • CuCO₃: 1 mole
  • NaCl: 2 moles

The molar mass of the compounds is:

  • Na₂CO₃: 129 g/mole
  • CuCl₂: 134.45 g/mole
  • CuCO₃: 123.55 g/mole
  • NaCl: 58.45 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Na₂CO₃: 1 mole ×129 g/mole= 129 grams
  • CuCl₂: 1 mole ×134.45 g/mole= 134.45 grams
  • CuCO₃: 1 mole ×123.55 g/mole= 123.55 grams
  • NaCl: 2 mole ×58.45 g/mole=116.9 grams

<h3>Mass of CuCl₂ required</h3>

The following rule of three can be applied: If by reaction stoichiometry 134.35 grams of CuCl₂ react with 129 grams of Na₂CO₃, 17.3 grams of CuCl₂ react with how much mass of Na₂CO₃?

mass of Na₂CO₃= (17.3 grams of CuCl₂× 129 grams of Na₂CO₃)÷ 134.35 grams of CuCl₂

<u><em>mass of Na₂CO₃= 16.611 grams</em></u>

Finally, 16.611 grams of Na₂CO₃ is required.

Learn more about the reaction stoichiometry:

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6 0
1 year ago
0.0136 g + 2.70 × 10-4 g - 4.21 × 10-3 g = ?
melisa1 [442]

Answer choice is , 2

3 0
3 years ago
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