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HACTEHA [7]
3 years ago
15

Help soon plz giving brainliest

Chemistry
2 answers:
Sophie [7]3 years ago
8 0

Answer:

Carbon, nitrogen and water are recycled trough ecosystems many times. Of all these nitrogen is eventually broken down completely.

Final Answer: Nitrogen

Explanation:

All carbon, nitrogen and water are recycled by the biosphere.

But all of these nitrogen is eventually broken down because there are three bonds are present in nitrogen and these bonds are stronger so cannot be easily broken down.

yarga [219]3 years ago
6 0

Answer: The answer is nitrogen.

i just did the assignment

hope this helps you :D

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To determine the strength of potassium permanganate with a standard solution of oxalic acid.
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3 years ago
2C3H7OH + 9O2 --> 6CO2 + 8H2O
Ne4ueva [31]
<h3>Answer:</h3>

733 g CO₂

<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>

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<u>Stoichiometry</u>

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced]   2C₃H₇OH + 9O₂ → 6CO₂ + 8H₂O

[Given]   5.55 mol C₃H₇OH

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

[RxN] 2 mol C₃H₇OH → 6 CO₂

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: Stoichiometry</u>

  1. Set up conversion:                    \displaystyle 5.55 \ mol \ C_3H_7OH(\frac{6 \ mol \ CO_2}{2 \ mol \ C_3H_7OH})(\frac{44.01 \ g \ CO_2}{1 \ mol \ CO_2})
  2. Multiply/Divide:                                                                                               \displaystyle 732.767 \ g \ CO_2

<u>Step 4: Check</u>

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

732.767 g CO₂ ≈ 733 g CO₂

8 0
3 years ago
1. What pressure would be exerted by 46.0grams of hydrogen gas placed into a 3.00 L container at
dlinn [17]

Answer:

187.34 atm

Explanation:

From the question,

PV = nRT.................. Equation 1

Where P = Pressure, V = Volume, n = number of mole, R = molar gas constant, T = Temperature.

make P the subject of the equation

P = nRT/V.............. Equation 2

n = mass(m)/molar mass(m')

n = m/m'............... Equation 3

Substitute equation 3 into equation 2

P = (m/m')RT/V............ Equation 4

Given: m = 46 g, T = 25°C = (25+273) = 298 K, V = 3.00 L

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Substitute these values into equation 4

P = (46/2)(0.082×298)/3

P = (23×0.082×298)/3

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