Answer:
MnSO₄.7H₂O
Explanation:
To solve this question, we need to convert the mass of the dehydrated MnSO₄. The difference between mass of the hydrate and dehydrated compound is the mass of water. With the mass we can find the moles of water and the formula of the hydrate:
<em>Moles MnSO₄ -Molar mass: 151g/mol-:</em>
17.51g * (1mol / 151g) = 0.116 moles
<em>Moles H₂O -Molar mass: 18g/mol-:</em>
32.14g-17.51g = 14.63g * (1mol / 18g) = 0.813 moles
The ratio of moles MnSO₄: Moles H₂O represent the amount of water molecules in the hydrate:
0.813mol / 0.116mol = 7 molecules of water.
The hydrate formula is:
<h3>MnSO₄.7H₂O</h3>
There are six atoms in the carbon
Answer:
LHS-RHS
1 C - 1 C
4 H-2 H
2 O- 3 O
So on right side 2 hydrogen are less and one oxygen is more..so
Balanced equation is
CH4+2O2==CO2 + 2H2O
Explanation:
<h3>
Answer:</h3>
91.2 g Mn
<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
- 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>
[Given] 1.00 × 10²⁴ atoms Mn
<u>Step 2: Identify Conversions</u>
Avogadro's Numer
[PT] Molar Mass of Mn - 54.94 g/mol
<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 3 sig figs.</em>
91.2321 g Mn ≈ 91.2 g Mn
Answer:
E - Be and O
A - Mg and N
E - Li and Br
F - Ba and Cl
B - Rb and O
Explanation:
Be and O
Be is a metal that loses 2 e⁻ to form Be²⁺ and O is a nonmetal that gains 2 e⁻ to form O²⁻. For the ionic compound to be neutral, it must have the form BeO (E-MX).
Mg and N
Mg is a metal that loses 2 e⁻ to form Mg²⁺ and N is a nonmetal that gains 3 e⁻ to form O³⁻. For the ionic compound to be neutral, it must have the form Mg₃N₂ (A-M₃X₂).
Li and Br
Li is a metal that loses 1 e⁻ to form Li⁺ and Br is a nonmetal that gains 1 e⁻ to form Br⁻. For the ionic compound to be neutral, it must have the form LiBr (E-MX).
Ba and Cl
Ba is a metal that loses 2 e⁻ to form Ba²⁺ and Cl is a nonmetal that gains 1 e⁻ to form Cl⁻. For the ionic compound to be neutral, it must have the form BaCl₂ (F-MX₂).
Rb and O
Rb is a metal that loses 1 e⁻ to form Rb⁺ and O is a nonmetal that gains 2 e⁻ to form O²⁻. For the ionic compound to be neutral, it must have the form Rb₂O (B-M₂X).
