Answer:
The answer to your question is 88.7 ml
Explanation:
Data
Volume = ?
Concentration of NaOH = 0.142 M
Volume of H₂C₄H₄O₆ = 21.4 ml
Concentration of H₂C₄H₄O₆ = 0.294 M
Balanced chemical reaction
2 NaOH + H₂C₄H₄O₆ ⇒ Na₂C₄H₄O₆ + 2H₂O
1.- Calculate the moles of H₂C₄H₄O₆
Molarity = moles/volume
Solve for moles
moles = Molarity x volume
Substitution
moles = 0.294 x 21.4/1000
Result
moles = 0.0063
2.- Use proportions to calculate the moles of NaOH
2 moles of NaOH ------------------ 1 moles of H₂C₄H₄O₆
x ------------------ 0.0063 moles
x = (0.0063 x 2) / 1
x = 0.0126 moles of NaOH
3.- Calculate the volume of NaOH
Molarity = moles / volume
Solve for volume
Volume = moles/Molarity
Substitution
Volume = 0.0126/0.142
Result
Volume = 0.088 L or 88.7 ml
<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
A beta particle is an electron and it has a -1 charge and zero mass.
Beta decay by
emitting an electron is called as β⁻
decay. When this happens, a neutron of the element converts into a proton by
emitting an electron. Hence, the mass of daughter nucleus is same as parent
atom but atomic number/number of protons is higher by 1 than atomic number of
parent atom.
In a β⁻ decay, the symbol is used as ₋₁⁰β or ₋₁⁰e.
-1 is for charge
<span> 0 is for the mass of the particle
</span>
Answer: A physical property that will be the same regardless of the amount of matter. Extensive Properties: A physical property that will change if the amount of matter changes.
