Answer:
C11H25SO4
Explanation:
The total mass of the compound is 253.4 g, so, the mass of each element will be:
C: 52.14% of 253.4 = 0.5214x253.4 = 132.12 g
H: 9.946% of 253.4 = 0.09946x253.4 = 25.20 g
S: 12.66% of 253.4 = 0.1266x253.4 = 32.08 g
O: 25.26% of 253.4 = 0.2526x253.4 = 64.00 g
The molar mass are: C = 12 g/mol, H 1 g/mol, S = 32 g/mol, and O = 16 g/mol
So, to know how much moles will be, just divide the mass calculated above for the molar mass:
C: 132.12/12 = 11 moles
H: 25.20/ 1 = 25 moles
S: 32.08/32 = 1 mol
O: 64.00/16 = 4 moles
So the molecular formula is C11H25SO4
4...........................
<h3>
Answer:</h3>
1.83 × 10⁻⁷ mol Au
<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 Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
3.60 × 10⁻⁵ g Au (Gold)
<u>Step 2: Identify Conversions</u>
Molar Mass of Au - 196.97 g/mol
<u>Step 3: Convert</u>
- Set up:
- Multiply:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
1.82769 × 10⁻⁷ mol Au ≈ 1.83 × 10⁻⁷ mol Au
Answer:
3Mg(s) +2P(s) -------> Mg3P2(s) + energy
Keq= [Mg3P2]/[Mg]^3 [P]^2
Explanation:
The equation for the formation of magnesium phosphide from its elements is;
3Mg(s) +2P(s) -------> Mg3P2(s) + energy
Hence we can see that three moles of magnesium atoms combines with two moles of phosphorus atoms to yield one mole of magnesium phosphide. The equation written above is the balanced chemical reaction equation for the formation of the magnesium phosphide.
The equilibrium expression for the reaction K(eq) will be given by;
Keq= [Mg3P2]/[Mg]^3 [P]^2
