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Aneli [31]
3 years ago
7

In the important industrial process for producing ammonia (the Haber Process), the overall reaction is: N2(g) + 3H2(g) → 2NH3(g)

+ 100.4 kJ A yield of NH3 of approximately 98% can be obtained at 200°C and 1,000 atmospheres of pressure. This reaction is:
exothermic
endothermic
Chemistry
2 answers:
kakasveta [241]3 years ago
8 0
Endothermic i think? Using Le Chât's principle
Dahasolnce [82]3 years ago
4 0

Answer:

Exothermic

Explanation:

From the perspective of thermodynamics, chemical reactions can be broadly

classified as endothermic and exothermic reactions.

Endothermic reactions are accompanied by the absorption of heat and have a negative value for the reaction enthalpy i.e. ΔH is negative. In contrast, exothermic reactions are accompanied by the release of heat and have a positive value for the reaction enthalpy i.e. ΔH is positive.

The given reaction is:

N2(g) + 3H2(g)---- 2NH3(g) + 100.4 kJ

Based on the above reaction, production of NH3 is accompanied by the release of 100.4 kJ of heat, hence this is an exothermic reaction.

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How many grams of CO2 are produced from 6.7 L of O2 gas at STP?
Tasya [4]
<h3>Answer:</h3>

13 g CO₂

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

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Moles
  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

<u>Stoichiometry</u>

  • Using Dimensional Analysis

<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>
<h3>Explanation:</h3>

<u>Step 1: Define</u>

<em>Identify variables</em>

[Given] 6.7 L O₂

[Solve] g O₂

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

[STP] 22.4 L = 1 mol

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

[PT] Molar Mass of C: 12.01 g/mol

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

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 6.7 \ L \ O_2(\frac{1 \ mol \ O_2}{22.4 \ L \ O_2})(\frac{44.01 \ g \ O_2}{1 \ mol \ O_2})
  2. [DA] Divide/Multiply [Cancel out units]:                                                           \displaystyle 13.1637 \ g \ O_2

<u>Step 4: Check</u>

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

13.1637 g CO₂ ≈ 13 g CO₂

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a_sh-v [17]

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3 years ago
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alekssr [168]

Answer:

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Explanation:

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8 0
4 years ago
In UV region measurements, the üsed cell must be Oa. Plastic cell O b. KBr cell C. Fused silica cell d. Glass cell​
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