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Ksivusya [100]
3 years ago
10

What is the total number of valence electrons for cesium?​?

Chemistry
1 answer:
GuDViN [60]3 years ago
5 0
It's an alkali metal so one, they sit on the very first of the column in the periodic table :DDDD
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How does the equilibrium change to counter the removal of A in this reaction?
sattari [20]

Answer : The correct option for blank 1 is, Shifts left.

The correct option for blank 2 is, Reverse.

Explanation :

According to the Le Chatelier's Principle, when the addition of the reactant in reaction system then the equilibrium will shift to the right (forward) direction of the reaction.

Or, if we remove the reactants from the reaction system then the equilibrium will be shifted to the left (backward) direction of the reaction. And simultaneously, there will be increase in the reverse reaction for the attainment of the equilibrium.


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3 years ago
What is the periodic table and why is it neccessary?
wlad13 [49]

Answer:

The periodic table is a table displaying each element and information about the elements, for example atomic number and chemical properties. This is necessary because without it it would be very hard to find information on the elements.

6 0
3 years ago
PLZ HELP ME ANSWER THIS!!!
Vedmedyk [2.9K]

Answer:

Water

Explanation:

5 0
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How many grams of H2O would be made from 7.9 moles of H2? (Round to one decimal place)
Paraphin [41]

Answer:

2.01588 grams

Explanation:

4 0
3 years ago
A sample of CO2 weighing 86.34g contains how many molecules?
irakobra [83]

Answer:

1.181 × 10²⁴ molecules CO₂

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

<u>Step 1: Define</u>

86.34 g CO₂

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

Avogadro's Number

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: Convert</u>

<u />86.34 \ g \ CO_2(\frac{1 \ mol \ CO_2}{44.01 \ g \ CO_2} )(\frac{6.022 \cdot 10^{23} \ molecules \ CO_2}{1 \ mol \ CO_2} ) = 1.18141 × 10²⁴ molecules CO₂

<u>Step 4: Check</u>

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

1.18141 × 10²⁴ molecules CO₂ ≈ 1.181 × 10²⁴ molecules CO₂

4 0
3 years ago
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