Answer:
Rubidium is used in vacuum tubes as a getter, a material that combines with and removes trace gases from vacuum tubes. It is also used in the manufacture of photocells and in special glasses. Since it is easily ionized, it might be used as a propellant in ion engines on spacecraft.
Symbol: Rb (37)
Atomic Weight: 85.4678
Atomic Number: 37
Number of Stable Isotopes: 1 (View all isotope .
Step 1: write the equation:
P₄(s) + 6F₂(g) → 4PF₃(g)
Step 2: Molar mass of P₄ = 30.97 g/mol × 4 = 123.88 g/mol
Step 3: Number of moles of phosphorus
n = m/M
n = 8.5 g/123.88g/mol
n = 0.07 moles
Step 4: 0.07 × 12 = 0.84 moles of fluorine.
Fluorine is diatomic gas so we multiplied the number of moles by 12.
Step 5: To find the mass of fluorine we multiply the number of moles with the molar mass.
Mass of fluorine = 0.84 × 228
= 191.52 grams.
117 mL of 0.210 M K₂S solution
Explanation:
The question asks about the volume of 0.210 M K₂S (potassium sulfide) solution required to completely react with 175 mL of 0.140 M Co(NO₃)₂ (cobalt(II) nitrate).
We have the chemical reaction:
K₂S + Co(NO₃)₂ → CoS + 2 KNO₃
molar concentration = number of moles / volume (L)
number of moles = molar concentration × volume
number of moles of Co(NO₃)₂ = 0.140 × 175 = 24.5 mmoles
We see from the chemical reaction that 1 mmole of Co(NO₃)₂ is reacting with 1 mmole of K₂S, so 24.5 mmoles of Co(NO₃)₂ are reacting with 24.5 mmoles of K₂S.
volume = number of moles / molar concentration
volume of K₂S solution = 24.5 / 0.210 = 117 mL
Learn more about:
molar concentration
brainly.com/question/2767031
brainly.com/question/8858482
#learnwithBrainly
<h3>
Answer:</h3>
3.3 × 10²³ molecules Cu(NO₃)₂
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<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>
0.55 mol Cu(NO₃)₂
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- [DA] Set up:

- [DA] Multiply/Divide [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
3.3121 × 10²³ molecules Cu(NO₃)₂ ≈ 3.3 × 10²³ molecules Cu(NO₃)₂
Answer:
troposphere
stratosphere
mesosphere
thermospjere
exosphere
Explanation:
this is from the lowest to highest