It should be 8 O atoms. 3O atoms in Na2S2O3 and 5O atom in 5H2O. The reason there are 5 O atoms are because the 5 in front of H2O means you multiply each atom in the compound by that number (like the distributive property). The H2 molecule becomes 10 Hydrogen atoms (5*2) and the Oxygen becomes 5 Oxygen atoms (5*1). Then you add the 5O atoms to the 3O atoms which equals 8
Answer:
B. Na and Li
Both are group I elements.
Answer:
84.24 g
Explanation:
Given data:
Mass of oxygen = 75 g
Mass of Al required to react = ?
Solution:
Chemical equation:
4Al + 3O₂ → 2Al₂O₃
Number of moles of oxygen:
Number of moles = mass/ molar mass
Number of moles = 75 g/ 32 g/mol
Number of moles = 2.34 mol
Now we will compare the moles of oxygen with Al.
O₂ : Al
3 : 4
2.34 : 4/3×2.34 = 3.12 mol
Mass of Al required:
Mass = number of moles × molar mass
Mass = 3.12 mol × 27 g/mol
Mass = 84.24 g
Yes it could, but you'd have to set up the process very carefully.
I see two major challenges right away:
1). Displacement of water would not be a wise method, since rock salt
is soluble (dissolves) in water. So as soon as you start lowering it into
your graduated cylinder full of water, its volume would immediately start
to decrease. If you lowered it slowly enough, you might even measure
a volume close to zero, and when you pulled the string back out of the
water, there might be nothing left on the end of it.
So you would have to choose some other fluid besides water ... one in
which rock salt doesn't dissolve. I don't know right now what that could
be. You'd have to shop around and find one.
2). Whatever fluid you did choose, it would also have to be less dense
than rock salt. If it's more dense, then the rock salt just floats in it, and
never goes all the way under. If that happens, then you have a tough
time measuring the total volume of the lump.
So the displacement method could perhaps be used, in principle, but
it would not be easy.
<h3>
Answer:</h3>
5.55 mol C₂H₅OH
<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>
- Reading a Periodic Tables
- Moles
<u>Stoichiometry</u>
- Using Dimensional Analysis
- Analyzing Reactions RxN
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
[Given] 500. g C₆H₁₂O₆ (Glucose)
[Solve] moles C₂H₅OH (Ethanol)
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol C₆H₁₂O₆ → 2 mol C₂H₅OH
[PT] Molar mass of C - 12.01 g/mol
[PT] Molar Mass of H - 1.01 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol
<u>Step 3: Stoichiometry</u>
- [DA] Set up conversion:
- [DA} Multiply/Divide [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
5.55001 mol C₂H₅OH ≈ 5.55 mol C₂H₅OH
