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Nutka1998 [239]
3 years ago
12

Products of MnO + CL2

Chemistry
1 answer:
Alenkinab [10]3 years ago
3 0

If the equation is complete the products would be manganese chloride and oxygen gas would be given off.

MnCl2 + O2

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I need help with this question please asap I would really appreciate it
yanalaym [24]

Answer:

I think c

Explanation:

4 0
2 years ago
Which property determines an atom's ability to attract electrons shared in a chemical bond?
ASHA 777 [7]
It is electronegativity.
4 0
3 years ago
Why are salt and sugar both able to dissolve in water, even though the solutes have different types of chemical bonding? everyth
Bas_tet [7]

Sucrose and other simple sugars may dissolve in water because they are polar molecules with an unequal charge distribution. Water is also quite polar, capable of forming weak, temporary connections with other polar compounds.

Salt dissolves into ions, with Na being positively charged and CL being negatively charged. Because water is highly polar (parts of the molecule are negatively charged while others are positively charged), the sodium ions are surrounded by water molecules, with the negatively charged component of the water molecules surrounding the NA ion. The Cl ion experiences the inverse effect.

<h3>How does salt dissolve in water compared to sugar?</h3>

A solution's solute and solvent are two different types of substances that can dissolve one another. Different solvents have different levels of solubility for different solutes. For instance, sugar is far more soluble in water than salt. Even sugar, though, has a limit on how much may dissolve.

learn more about solubility refer:

brainly.com/question/23946616

#SPJ4

5 0
1 year ago
6
Butoxors [25]

Answer:

the 6 stands for the atomic number

the 12.01 stands for the atomic mass

6 0
2 years ago
Consider the reaction given below.
Drupady [299]

Answer:

  • <u>K =  0.167 s⁻¹</u>

Explanation:

<u>1) Rate law, at a given temperature:</u>

  • Since all the data are obtained at the same temperature, the equilibrium constant is the same.

  • Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:

        r = K [A]ᵃ [B]ᵇ

<u>2) Use the data from the table</u>

  • Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s

         Divide r₂ by r₁:     [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0

  • Use the first and second set of data to find the exponent a:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s

        Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]

                                  2ᵃ = 2 ⇒ a = 1

         

<u>3) Write the rate law</u>

  • r = K [A]¹ [B]⁰ = K[A]

This means, that the rate is independent of reactant B and is of first order respect reactant A.

<u>4) Use any set of data to find K</u>

With the first set of data

  • r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹

Result: the rate constant is K =  0.167 s⁻¹

6 0
3 years ago
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