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DochEvi [55]
3 years ago
9

Are the particles in a solid in constant motion, or are they "frozen in place?"

Chemistry
2 answers:
babymother [125]3 years ago
4 0
Particles that are in a solid are tightly locked in place so the answer to your question would be  "Frozen in place"
Tems11 [23]3 years ago
3 0
The molecules are frozen in place but still vibrate
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A sample of CO2 weighing 86.34g contains how many molecules?
irakobra [83]

Answer:

1.181 × 10²⁴ molecules CO₂

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

<u>Step 1: Define</u>

86.34 g CO₂

<u>Step 2: Identify Conversion</u>

Avogadro's Number

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol

<u>Step 3: Convert</u>

<u />86.34 \ g \ CO_2(\frac{1 \ mol \ CO_2}{44.01 \ g \ CO_2} )(\frac{6.022 \cdot 10^{23} \ molecules \ CO_2}{1 \ mol \ CO_2} ) = 1.18141 × 10²⁴ molecules CO₂

<u>Step 4: Check</u>

<em>We are given 4 sig figs. Follow sig fig rules and round.</em>

1.18141 × 10²⁴ molecules CO₂ ≈ 1.181 × 10²⁴ molecules CO₂

4 0
3 years ago
Who is Alfred Wegener? What did he discover?
Aliun [14]

Answer:

Alfred Wegener proposed the theory of continental drift – the idea that Earth's continents move. Despite publishing a large body of compelling fossil and rock evidence for his theory between 1912 and 1929, it was rejected by most other scientists.

Hope this helps!

3 0
3 years ago
In which of the following equilibrium systems will an increase in the volume cause the equilibrium to shift away from the reacta
Alenkinab [10]

Answer: 2NOBr(g) ⇌ 2NO(g) + Br2(g)

Explanation: For volume changes in equillibrium, the following are to be taken into consideration:

  • Volume changes have no effect on equillibrium system that contains solid or aqueous solutions.
  • An increase in volume of an equilibrium system will shift to favor the direction that produces more moles of gas.
  • A decrease in volume of an equilibrium system will shift to favor the direction that produces less moles of gas.
  • Volume changes will have no effect on the equillibrium system if there is an equal number of moles on both sides of the reaction.

2NOBr(g) ⇌ 2NO(g) + Br2(g) is the equillibrium system because there are more moles of products,therefore an increase in the volume of the reaction will shift to the right and produce more moles of products. Also both reactants and products exist in the gaseous state and does not have equal number of moles.

5 0
3 years ago
State the oxidation number assigned to each bold element in the formula: NH4+1 a 3 b -3 c -1 d 6
Leya [2.2K]
The fomula is NH4 (1+)


There are only two elements N and H.


As per oxidation state rules, the most electronegative element will have a negative oxidation state and the other element will have a positive oxidation state.


N is more electronative than H, so H will have a positive oxidation state and nitrogen will have a negative oxidation state.


You can also use the rule that states the hydrogen mostly has 1+ oxidation state,except when it is bonded to metals.


In conclusion the oxidation state of H in NH4 (1+) is 1+.


Now you must know that the sum of the oxidations states equals the charge of the ion, which in this case is 1+.


That implies that 4* (1+)  + x =   1+


=> x = (1+) - 4(+) = 3-


Answer:  the oxidation state of N is 3-, that is the option b.
8 0
4 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
4 years ago
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