Fluoride is an anion of Fluorine
What this means is that the two have the same number of protons (9), but Fluoride has 10 electrons compared to Fluorine's 9.
So the answers are:
Protons - 9
Neutrons - 9
Electrons - 10
Atomic Number - 9
Atomic Mass - 19 g/mol
Answer:
80.27%
Explanation:
Let's consider the following balanced equation.
2 Fe³⁺(aq) + Sn²⁺(aq) ⇒ 2Fe²⁺(aq) + Sn⁴⁺(aq)
First, we have to calculate the moles of Sn²⁺ that react.
We also know the following relations:
- According to the balanced equation, 1 mole of Sn²⁺ reacts with 2 moles of Fe³⁺.
- 1 mole of Fe³⁺ is oxidized from 1 mole of Fe.
- The molar mass of Fe is 55.84 g/mol.
Then, for 1.348 × 10⁻3 moles of Sn²⁺:
If there are 0.1505 g of Fe in a 0.1875 g sample, the mass percentage of Fe is:
Answer:
The area of the given rectangular index card = <u>9677.4 mm²</u>
Explanation:
Area is defined as the space occupied by a two dimensional shape or object. The SI unit of area is square metre (m²).
<u>The area of a rectangle</u> (A) = length (l) × width (w)
Given dimensions of the rectangle: Length (l) = 5.0 inch, Width (w) = 3.0 inch
Since, 1 inch = 25.4 millimetres (mm)
Therefore, l = 5 × 25.4 = 127 mm, and w = 3 × 25.4 = 76.2 mm
Therefore, <u>the area of the given rectangular index card</u> = A= l × w = 127 mm × 76.2 mm = <u>9677.4 mm²</u>
<h3>Answer:</h3>
a) Moles of Caffeine = 1.0 × 10⁻⁴ mol
b) Moles of Ethanol = 4.5 × 10⁻³ mol
<h3>Solution:</h3>
Data Given:
Mass of Caffeine = 20 mg = 0.02 g
M.Mass of Caffeine = 194.19 g.mol⁻¹
Molecules of Ethanol = 2.72 × 10²¹
Calculate Moles of Caffeine as,
Moles = Mass ÷ M.Mass
Putting values,
Moles = 0.02 g ÷ 194.19 g.mol⁻¹
Moles = 1.0 × 10⁻⁴ mol
Calculate Moles of Ethanol as,
As we know one mole of any substance contains 6.022 × 10²³ particles (atoms, ions, molecules or formula units). This number is also called as Avogadro's Number.
The relation between Moles, Number of Particles and Avogadro's Number is given as,
Number of Moles = Number of Molecules ÷ 6.022 × 10²³
Putting values,
Number of Moles = 2.72 × 10²¹ Molecules ÷ 6.022 × 10²³
Number of Moles = 4.5 × 10⁻³ Moles
Answer:
11.4
Explanation:
Step 1: Given data
- Concentration of the base (Cb): 0.300 M
- Basic dissociation constant (Kb): 1.8 × 10⁻⁵
Step 2: Write the dissociation equation
NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq)
Step 3: Calculate the concentration of OH⁻
We will use the following expression.
![[OH^{-} ]=\sqrt{Kb \times Cb } = \sqrt{1.8 \times 10^{-5} \times 0.300 } = 2.3 \times 10^{-3} M](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%20%5D%3D%5Csqrt%7BKb%20%5Ctimes%20Cb%20%7D%20%3D%20%5Csqrt%7B1.8%20%20%5Ctimes%2010%5E%7B-5%7D%20%5Ctimes%200.300%20%7D%20%3D%202.3%20%5Ctimes%2010%5E%7B-3%7D%20M)
Step 4: Calculate the pOH
We will use the following expression.
![pOH =-log[OH^{-} ]= -log(2.3 \times 10^{-3} M) = 2.6](https://tex.z-dn.net/?f=pOH%20%3D-log%5BOH%5E%7B-%7D%20%5D%3D%20-log%282.3%20%5Ctimes%2010%5E%7B-3%7D%20M%29%20%3D%202.6)
Step 5: Calculate the pH
We will use the following expression.
