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densk [106]
3 years ago
15

How many particles can be stored in a 4.0 L container at room temperature (23 C) and standard atmospheric pressure

Chemistry
1 answer:
velikii [3]3 years ago
5 0

Answer: 9.91×10²³ particles

Explanation:

To find the amount of particles, you will need to use the Ideal Gas Law with what we are given.

Ideal Gas Law: PV=nRT

After we find moles, we can use Avogadro's number to convert to particles.

n=\frac{PV}{RT}

P=101.3kPa=1.00 atm

V=4.0 L

T=23°C+273.15=296.15 K

R=0.08206 Latm/Kmol

n=\frac{(1.00 atm)(4.0 L)}{(0.08206Latm/Kmol)(296.15K)}

n=0.164595 mol

Now that we have moles, we can convert to particles.

0.164595mol*\frac{6.022*10^2^3 particles}{mol} =9.91*10^2^2 particles

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Given 1kg=2.20 lbs, how many kilograms are in 125 lbs?<br> (Metric conversion)
Nadya [2.5K]

Answer:

56.82 Kg

Explanation:

Given data:

1 Kg = 2.20 lbs

Number of kilogram in 125 lbs = ?

Solution:

lbs is used for pound. lb is abbreviation of libra. It is Latin word meaning balance.

Both kilogram and pounds are units of mass. Pound is smaller unit than kilogram.

one Kg = 2.20 lbs

Number of kg in 125 lbs:

125 lbs × 1 Kg/2.20 lbs

125 lbs.Kg/2.20 lbs

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3 years ago
CH4 + 202 → CO2 + 2H2O<br> How many grams of O2 needed to produce 36 grams of H2O?
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<h3>Answer:</h3>

64 g O₂

<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>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

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

<u>Step 1: Define</u>

[RxN - Balanced]   CH₄ + 2O₂ → CO₂ + 2H₂O

[Given]   36 g H₂O

[Solve]   x g O₂

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

[RxN] 2 mol O₂ → 2 mol H₂O

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

[PT] Molar Mas of H - 1.01 g/mol

Molar Mass of O₂ - 2(16.00) = 32.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up conversion:                     \displaystyle 36 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})(\frac{2 \ mol \ O_2}{2 \ mol \  H_2O})(\frac{32.00 \ g \ O_2}{1 \ mol \ O_2})
  2. Divide/Multiply [Cancel Units]:                                                                       \displaystyle 63.929 \ g \ O_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

63.929 g O₂ ≈ 64 g O₂

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