<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
The molar mass of NH4NO3 in g/mol is 80g/mol.
HOW TO CALCULATE MOLAR MASS:
The molar mass of a compound can be calculated by summing the atomic masses of its constituent elements.
In ammonium nitrate (NH4NO3), there are nitrogen, hydrogen, and oxygen elements.
- Atomic mass of nitrogen = 14
- Atomic mass of oxygen = 16
- Atomic mass of hydrogen = 1
Molar mass of NH4NO3 = 14 + 1(4) + 14 + 16(3)
Molar mass of NH4NO3 = 80g/mol
- Therefore, the molar mass of NH4NO3 in g/mol is 80g/mol.
Learn more about molar mass at: brainly.com/question/8101390?referrer=searchResults
Answer:
C.) 2
Explanation:
The pH equation is:
pH = -log[H⁺]
In this equation, [H⁺] is the molarity of the acid. In this case, the acid is HCl. Molarity can be found using the equation:
Molarity (M) = moles / volume (L)
Since you were given moles and volume, you can find the molarity of HCl.
Molarity = moles / volume
Molarity = 0.01 moles / 1.00 L
Molarity = 0.01 M
Now, you can plug the molarity of the acid into the pH equation.
pH = -log[H⁺]
pH = -log[0.01]
pH = 2
Answer:
A) [H3PO4] will increase, [KH2PO4] will decrease, and pH will slightly decrease.
Explanation:
A buffer is a solution which resists changes to its pH when a small amount of acid or base is added to it.
Buffers consist of a weak acid (HA) and its conjugate base (A–) or a weak base and its conjugate acid. Weak acids and bases do not completely dissociate in water, and instead exist in solution as an equilibrium of dissociated and undissociated species. When a small quantity of a strong acid is added to a buffer solution, the conjugate base, A-, reacts with the hydrogen ions from the added acid to form the weak acid and a salt thereby removing the extra hydrogen ions from the solution and keeping the pH of the solution fairly constant. On the other hand, if a small quantity of a strong base is added to the buffer solution, the weak acid dissociates further to release hydrogen ions which then react with the hydroxide ions of the added base to form water and the conjugate base.
For example, if a small amount of strong acid is added to a buffer solution that is 0.700 M H3PO4 and 0.700 M KH2PO4, the following reaction is obtained:
KH₂PO₄ + H+ ----> K+ + H₃PO₄
Therefore, [H₃PO₄] will increase, [KH₂PO₄] will decrease, and pH will slightly decrease.:
