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forsale [732]
3 years ago
14

The higher the number of living organisms in an area, the ___ fertile the soil. Less, more.

Chemistry
1 answer:
goldenfox [79]3 years ago
4 0
I think it is the more fertile the soil
You might be interested in
n a gas-phase equilibrium mixture of SbCl5, SbCl3, and Cl2 at 500 K, pSbCl5 = 0.17 bar and pSbCl3 = 0.22 bar. Calculate the equi
Lorico [155]

Answer:

2.7 × 10⁻⁴ bar

Explanation:

Let's consider the following reaction at equilibrium.

SbCl₅(g) ⇄ SbCl₃(g) + Cl₂(g)

The pressure equilibrium constant (Kp) is 3.5 × 10⁻⁴. We can use these data and the partial pressures at equilibrium of SbCl₅ and SbCl₃, to find the partial pressure at equilibrium of Cl₂.

Kp = pSbCl₃ × pCl₂ / pSbCl₅

pCl₂ = Kp × pSbCl₅ / pSbCl₃

pCl₂ = 3.5 × 10⁻⁴ × 0.17 / 0.22

pCl₂ = 2.7 × 10⁻⁴ bar

7 0
3 years ago
How many molecules are contained in 103.4g of sulfuric acid?
Ierofanga [76]
<h3>Answer:</h3>

1.827 × 10²⁴ molecules H₂S

<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>Compounds</u>

  • Writing Compounds
  • Acids/Bases

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>Explanation:</h3>

<u>Step 1: Define</u>

103.4 g H₂S (Sulfuric Acid)

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

Avogadro's Number

Molar Mass of H - 1.01 g/mol

Molar Mass of S - 32.07 g/mol

Molar Mass of H₂S - 2(1.01) + 32.07 = 34.09 g/mol

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 103.4 \ g \ H_2S(\frac{1 \ mol \ H_2S}{34.09 \ g \ H_2S})(\frac{6.022 \cdot 10^{23} \ molecules \ H_2S}{1 \ mol \ H_2S})
  2. Multiply:                                                                                                            \displaystyle 1.82656 \cdot 10^{24} \ molecules \ H_2S

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 4 sig figs.</em>

1.82656 × 10²⁴ molecules H₂S ≈ 1.827 × 10²⁴ molecules H₂S

4 0
3 years ago
What is the mass of potassium chloride when 6.75 g of potassium reacts with an excess of chlorine gas? the balanced chemical equ
lorasvet [3.4K]
The balanced equation for the above reaction is;
2K + Cl₂ ---> 2KCl
Stoichiomtery of K to KCl is 2:2
Potassium is the limiting reactant which is fully consumed in the reaction. The amount of product formed depends on amount of limits reactant present.
Number of moles of K reacted - 6.75 g/ 39 g/mol = 0.17 mol
Therefore number of KCl moles formed - 0.17 mol
Mass of KCl formed - 0.17 mol x 74.5 g/mol = 12.67 g 
6 0
3 years ago
Please help I will reward brainly
cupoosta [38]
The answer is glycolsis, I'm pretty sure.
4 0
3 years ago
Read 2 more answers
HELLPPPP PLZZZ
Charra [1.4K]
C. all the other choices are positive impacts to the environment unlike C
3 0
3 years ago
Read 2 more answers
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