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iVinArrow [24]
2 years ago
11

Which type of changes are melting, freezing, and boiling?​

Chemistry
2 answers:
sukhopar [10]2 years ago
8 0

Answer:

the changes are physical changes

tatiyna2 years ago
4 0

Answer:

Adding heat.

Decreasing and increasing temperatures.

Explanation:

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If 5.65 grams of zinc metal react with 21.6 grams of silver nitrate, how many grams of silver metal can be formed and how many g
Igoryamba
The balanced chemical reaction is:

Zn + 2AgNO3 =  Zn(NO3)2 + 2Ag

To determine the amount of the reactant left, we have to determine which is the limiting and the excess reactant. We do as follows:

5.65 g Zn ( 1 mol / 65.38 g) = 0.09 mol Zn
21.6 g AgNO3 (1 mol / 169.87 g) = 0.13 mol AgNO3

The limiting reactant would be silver nitrate since it is consumed completely in the reaction. The excess reactant would be zinc. 

Excess Zinc = 0.09 mol Zn - (0.13 / 2) mol Zn = 0.025 mol Zn left
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3 years ago
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What is the “Island of Stability”?
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Answer:

The island of stability is a term from nuclear physics that describes the possibility of elements with particularly stable "magic numbers" of protons and neutrons. This would allow certain isotopes of some transuranic elements to be far more stable than others, that is, decay much more slowly.

Explanation:

7 0
2 years ago
How long is a half life for carbon 14?
babunello [35]

Answer:

half-life of 5,700 ± 40 years

Explanation:

6 0
3 years ago
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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
How do differences in temperature effect air masses?
laiz [17]

Answer:

The difference in temperature causes differences in air pressure between the two spots. ... This causes the jet stream to dip farther south and the winds to blow stronger on either side of the jet stream in winter. The jet stream separates the very cold air mass at higher latitudes from the warmer air mass equatorward.

Explanation:

7 0
2 years ago
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