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never [62]
2 years ago
14

The volume of an ideal gas is held constant. Determine the ratio P2/P1 of the final pressure to the initial pressure when the te

mperature of the gas rises (a) from 46 to 92 K and (b) from 35.4 to 69.0 oC.
Chemistry
1 answer:
cricket20 [7]2 years ago
5 0

Answer:

A. P₂ / P₁ = 2

B. P₂ / P₁ = 1.1

Explanation:

A. Determination of the ratio P₂/P₁

Volume = constant

Initial temperature (T₁) = 46 K

Final temperature (T₂) = 92 K

Final pressure /Initial pressure (P₂/P₁) =?

P₁/T₁ = P₂/T₂

P₁/46 = P₂/92

Cross multiply

46 × P₂ = P₁ × 92

Divide both side by P₁

46 × P₂ / P₁ = 92

Divide both side by 46

P₂ / P₁ = 92 / 46

P₂ / P₁ = 2

B. Determination of the ratio P₂/P₁

Volume = constant

Initial temperature (T₁) = 35.4 °C = 35.4 + 273 = 308.4 K

Final temperature (T₂) = 69.0 °C = 69 + 273 = 342 K

Final pressure /Initial pressure (P₂/P₁) =?

P₁/T₁ = P₂/T₂

P₁/308.4 = P₂/342

Cross multiply

308.4 × P₂ = P₁ × 342

Divide both side by P₁

308.4 × P₂ / P₁ = 342

Divide both side by 308.4

P₂ / P₁ = 342 / 308.4

P₂ / P₁ = 1.1

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Answer:

³⁸₂₀Ca.

Explanation:

³⁸₁₉K –> __ + ⁰₋₁β

Let ʸₓA represent the unknown.

Thus the equation above can be written as:

³⁸₁₉K –> ʸₓA + ⁰₋₁β

Thus, we can obtain the value of y an x as follow:

38 = y + 0

y = 38

19 = x + (–1)

19 = x – 1

Collect like terms

19 + 1 = x

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Thus,

ʸₓA => ³⁸₂₀A => ³⁸₂₀Ca

Therefore, the equation is:

³⁸₁₉K –> ³⁸₂₀Ca + ⁰₋₁β

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3 years ago
How many moles are in 1.51x10^26 atoms of xenon (Xe)? Please and thank you :)!!
RoseWind [281]
<h3>Answer:</h3>

251 mol Xe

<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
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<u>Chemistry</u>

<u>Atomic Structure</u>

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<u>Stoichiometry</u>

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 1.51 × 10²⁶ atoms Xe

[Solve] moles Xe

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                     \displaystyle 1.51 \cdot 10^{26} \ atoms \ Xe(\frac{1 \ mol \ Xe}{6.022 \cdot 10^{23} \ atoms \ Xe})
  2. [DA] Multiply/Divide [Cancel out units]:                                                         \displaystyle 250.747 \ mol \ Xe

<u>Step 4: Check</u>

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

250.747 mol Xe ≈ 251 mol Xe

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