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iogann1982 [59]
2 years ago
6

Pls Help! 100 Points!!!!!

Chemistry
2 answers:
Katen [24]2 years ago
6 0

Answer:

The answer is c,F has a smaller radius than F− because F− has an additional energy shell.

Explanation:

if correct mark brainliest

jarptica [38.1K]2 years ago
4 0
The answer is C.
Hope this helps:)
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A solid precipitate is formed from the reaction of lead nitrate and sodium chromate.
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I believe it's Filtration.
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Buck's Turf Care Company mowed 233 lawns over a 11 week period. What is the average weekly rate of mowing lawns?
masya89 [10]

Answer: 21 lawns per week

Explanation:

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= Lawns mowed / Weeks taken

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6 0
3 years ago
What is H20 and oxygen
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How many molecules of carbon dioxide are in 5.61 moles of carbon dioxide (CO2)?
gulaghasi [49]
<h3>Answer:</h3>

3.38 × 10²⁴ molecules CO₂

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

<u>Math</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>

  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

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

<u>Step 1: Define</u>

[Given] 5.61 moles CO₂

[Solve] molecules CO₂

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                      \displaystyle 5.61 \ mooles \ CO_2(\frac{6.022 \cdot 10^{23} \ molecules \ CO_2}{1 \ mol \ CO_2})
  2. [DA] Multiply/Divide [Cancel out units]:                                                          \displaystyle 3.37834 \cdot 10^{24} \ molecules \ CO_2

<u>Step 4: Check</u>

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

3.37834 × 10²⁴ molecules CO₂ ≈ 3.38 × 10²⁴ molecules CO₂

6 0
3 years ago
The leaves of the rhubarb plant contain high concentrations of diprotic oxalic acid (hooccooh) and must be removed before the st
Andrei [34K]
First we have to find Ka1 and Ka2
pKa1 = - log Ka1 so Ka1 = 0.059
pKa2 = - log Ka2  so Ka2 = 6.46 x 10⁻⁵
Looking at the values of equilibrium constants we can see that the first one is really big compared to second one. so, the pH will be affected mainly by the first ionization of the acid.
Oxalic acid is H₂C₂O₄
H₂C₂O₄      ⇄     H⁺   + HC₂O₄⁻
0.0356 M            0          0
0.0356 - x            x          x
Ka1 = \frac{[H^+][HC2O4^-]}{[H2C2O4]} = x² / 0.0356 - x
x = 0.025 M
pH = - log [H⁺] = - log (0.025) = 1.6
5 0
2 years ago
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