The number of Ml of C₅H₈ that can be made from 366 ml C₅H₁₂ is 314.7 ml of C₅H₈
<u><em>calculation</em></u>
step 1: write the equation for formation of C₅H₈
C₅H₁₂ → C₅H₈ + 2 H₂
Step 2: find the mass of C₅H₁₂
mass = density × volume
= 0.620 g/ml × 366 ml =226.92 g
Step 3: find moles Of C₅H₁₂
moles = mass÷ molar mass
from periodic table the molar mass of C₅H₁₂ = (12 x5) +( 1 x12) = 72 g/mol
moles = 226.92 g÷ 72 g/mol =3.152 moles
Step 4: use the mole ratio to determine the moles of C₅H₈
C₅H₁₂:C₅H₈ is 1:1 from equation above
Therefore the moles of C₅H₈ is also = 3.152 moles
Step 5: find the mass of C₅H₈
mass = moles x molar mass
from periodic table the molar mass of C₅H₈ = (12 x5) +( 1 x8) = 68 g/mol
= 3.152 moles x 68 g/mol = 214.34 g
Step 6: find Ml of C₅H₈
=mass / density
= 214.34 g/0.681 g/ml = 314.7 ml
Answer:
49.2 g/mol
Explanation:
Let's first take account of what we have and convert them into the correct units.
Volume= 236 mL x (
) = .236 L
Pressure= 740 mm Hg x (
)= 0.97 atm
Temperature= 22C + 273= 295 K
mass= 0.443 g
Molar mass is in grams per mole, or MM= 
or MM= 
. They're all the same.
We have mass (0.443 g) we just need moles. We can find moles with the ideal gas constant PV=nRT. We want to solve for n, so we'll rearrange it to be
n=
, where R (constant)= 0.082 L atm mol-1 K-1
Let's plug in what we know.
n=
n= 0.009 mol
Let's look back at MM= and plug in what we know.
MM= 
MM= 49.2 g/mol
Explanation:
2.04 % hydrogen
32.65% sulphur
65.31% is oxygen
atomic ratio
hydrogen =2.04÷1=2.04
sulphur =32.65÷32=1.02
oxygen =65.31÷16=4.08
simplest ratio
hydrogen = 2.04÷1.02=2
sulphur =1.02÷1.02=1
oxygen =4.08÷1.02=4
empirical formula is H2SO4
<h3>
Answer:</h3>
3.67 mol Al
<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>
2.21 × 10²⁴ atoms Al
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- Set up:
- Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
3.66988 mol Al ≈ 3.67 mol Al
Yes, free electrons appear in balanced redox reaction equations. However, this is only true for half-reactions. This is because redox reactions primarily involve the transfer of electrons, which are better visualized if explicitly shown in the balanced reactions. In reduction reactions, electrons are placed on the left side of the equation. Oxidation reactions show electrons on the right side of the equation.
Explanation:
A half reaction is either the chemical reaction or reduction reaction part of an oxidoreduction reaction. A half reaction is obtained by considering the amendment in chemical reaction states of individual substances concerned within the oxidoreduction reaction. Half-reactions are usually used as a way of leveling oxidoreduction reactions.The half-reaction on the anode, wherever chemical reaction happens, is Zn(s) = Zn2+ (aq) + (2e-).
The metal loses 2 electrons to create Zn2+. The half-reaction on the cathode wherever reduction happens is Cu2+ (aq) + 2e- = Cu(s).
Here, the copper ions gain electrons and become solid copper.
