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vichka [17]
3 years ago
12

For the Heisenburg uncertainty principle, where did the 4 pi come from in the equation, which is "x times p is greater than or e

qual to h divided from 4 pi"?
Chemistry
1 answer:
Setler [38]3 years ago
5 0
That factor appears when you actually do the quite complex maths that lead to the principle; it's quantum mechanics, and it's not something that can be shown easily.

What is important to note is that this factor \frac{1}{4\pi}is not important. What is important is that the uncertainty \Delta x\Delta p_x is approximately of the order of \hbar=\frac{h}{2\pi}, which is a characteristic value that appears in many (most) formulas dealing with quantum mechanics.
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How many particles of CuCr2O7 are present in a 64.5 gram sample?
kow [346]
<h3>Answer:</h3>

1.39 × 10²³ particles CuCr₂O₇

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

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

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 64.5 g CuCr₂O₇

[Solve] particles CuCr₂O₇

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

Avogadro's Number

[PT] Molar Mass of Cu - 63.55 g/mol

[PT] Molar Mass of Cr - 52.00 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of CuCr₂O₇ - 63.55 + 2(52.00) + 7(16.00) = 279.55 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                     \displaystyle 64.5 \ g \ CuCr_2O_7(\frac{1 \ mol \ CuCr_2O_7}{279.55 \ g \ CuCr_2O_7})(\frac{6.022 \cdot 10^{23} \ particles \ CuCr_2O_7}{1 \ mol \ CuCr_2O_7})
  2. [DA] Divide/Multiply [Cancel out units]:                                                         \displaystyle 1.38944 \cdot 10^{23} \ particles \ CuCr_2O_7

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1.38944 × 10²³ particles CuCr₂O₇ ≈ 1.39 × 10²³ particles CuCr₂O₇

8 0
3 years ago
How many bromine molecules are in 250 g of bromine? (R
Galina-37 [17]
Use the formula triangle for quantitive chemistry
5 0
4 years ago
Which element in period 2 has six electrons in its electron dot structure?
navik [9.2K]

Answer:

calcium

Explanation:

8 0
4 years ago
Harvey kept a balloon with a volume of 348 milliliters at 25.0˚C inside a freezer for a night. When he took it out, its new volu
ad-work [718]

Initial volume of the balloon = V_{1}= 348 mL

Initial temperature of the balloon T_{1} = 25.0^{0}C + 273 = 298 K

Final volume of the balloon V_{2} = 322 mL

Final temperature of the balloon = T_{2} = ?

According to Charles law, volume of an ideal gas is directly proportional to the temperature at constant pressure.

\frac{V_{1} }{T_{1} } =\frac{V_{2} }{T_{2} }

On plugging in the values,

\frac{348mL}{298 K} =\frac{322 mL}{T_{2} }

T_{2} =276 K

Therefore, the temperature of the freezer is 276 K

5 0
3 years ago
Materials expand when heated. Consider a metal rod of length L0 at temperature T0. If the temperature is changed by an amount ΔT
marysya [2.9K]

Answer:

(a) The length at temperature 180°C is 40.070 cm

(b) The length at temperature 90°C is 64.976 inches

(c) L(T, α) = 60·α·T - 9000·α + 60

Explanation:

(a) The given parameters are

The thermal expansion coefficient, α for steel = 1.24 × 10⁻⁵/°C

The initial length of the steel L₀ = 40 cm

The initial temperature, t₀ = 40°C

The length at temperature 180°C = L

Therefore, from the given relation, for change in length, ΔL, we have;

ΔL = α × L₀ × ΔT

The amount the temperature changed ΔT = 180°C - 40°C = 140°C

Therefore, the change in length, ΔL, is found as follows;

ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 40 × 140°C = 0.07 cm

Therefore, L =  L₀ + ΔL = 40 + 0.07 = 40.07 cm

The length at temperature 180°C = 40.07 cm

(b) Given that the length at T = 120°C is 65 in., we have;

The temperature at which the new length is sought = 90°C

The amount the temperature changed ΔT = 90°C - 120°C = -30°C

ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 65 × -30°C = -0.024375 inches

The length, L at 90°C is therefore, L = L₀ + ΔL = 65 - 0.024375 = 64.976 in.

The length at temperature 90°C = 64.976 inches

(c) L = L₀ + ΔL  = L₀ +  α × L₀ × ΔT = L₀ +  α × L₀ × (T - T₀)

Therefore;

L = 60 +  α × 60 × (T - 150°C)

L = 60 + α × 60 × T - 9000 × α

L(T, α) = 60·α·T - 9000·α + 60

7 0
4 years ago
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