Answer:
A solid material whose constituents (such as atoms, molecules, or ions) are arranged in a highly ordered microscopic structure, forming a crystal lattice that extends in all directions.
Explanation:
First let's find out the oxidation number of Fe in K₄[Fe(CN)₆] compound.
The oxidation number of cation, K is +1. Hence, the total charge of the anion, [Fe(CN)₆] is -4. CN has charge has -1. There are 6 CN in anion. Let's assume the oxidation number of Fe is 'a'.
Sum of the oxidation numbers of each element = Charge of the compound
a + 6 x (-1) = -4
a -6 = -4
a = +2
Hence, oxidation number of Fe in [Fe(CN)₆]⁴⁻ is +2.
Now Fe has the atomic number as 26. Hence, number of electrons in Fe at ground state is 26.
Electron configuration = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² = [Ar] 3d⁶ 4s²
When making Fe²⁺, Fe releases 2 electrons. Hence, the number of electrons in Fe²⁺ is 26 - 2 = 24.
Hence, the electron configuration of Fe²⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶
= [Ar] 3d⁶
Hence, the number of 3d electrons of Fe in K₄[Fe(CN)₆] compound is 6.
<span>The symbol on a line that indicates a cold front on a map is a triangle. The point of the triangle will indicate the direction of movement of the cold front.</span>
Answer:
0.5077 moles
Explanation:
Data Given:
Moles = n = <u>???</u>
Temperature = T = 300 K
Pressure = P = 380 mmHg = 0.50 atm
Volume = V = 25 L
Formula Used:
Let's assume that the hydrogen gas in balloon is acting as an Ideal gas, the according to Ideal Gas Equation,
P V = n R T
where; R = Universal Gas Constant = 0.082057 atm.L.mol⁻¹.K⁻¹
Solving Equation for n,
n = P V / RT
Putting Values,
n = (0.50 atm × 25 L) / (0.082057 atm.L.mol⁻¹.K⁻¹ × 300 K)
n = 0.5077 moles
Answer:
0.3229 M HBr(aq)
0.08436M H₂SO₄(aq)
Explanation:
<em>Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown concentration HBr(aq) required 18.45 mL of 0.3500 M NaOH(aq) to neutralize it, to the point where thymol blue indicator changed from pale yellow to very pale blue. Calculate the concentration (molarity) of Stu's HBr(aq) sample.</em>
<em />
Let's consider the balanced equation for the reaction between HBr(aq) and NaOH(aq).
NaOH(aq) + HBr(aq) ⇄ NaBr(aq) + H₂O(l)
When the neutralization is complete, all the HBr present reacts with NaOH in a 1:1 molar ratio.
<em>Kemmi Major also does a titration. She measures 25.00 mL of unknown concentration H₂SO₄(aq) and titrates it with 0.1000 M NaOH(aq). When she has added 42.18 mL of the base, her phenolphthalein indicator turns light pink. What is the concentration (molarity) of Kemmi's H₂SO₄(aq) sample?</em>
<em />
Let's consider the balanced equation for the reaction between H₂SO₄(aq) and NaOH(aq).
2 NaOH(aq) + H₂SO₄(aq) ⇄ Na₂SO₄(aq) + 2 H₂O(l)
When the neutralization is complete, all the H₂SO₄ present reacts with NaOH in a 1:2 molar ratio.
