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LiRa [457]
3 years ago
10

Explain how you would use the IR spectra to characterize ferrocene, acetyl ferrocene and diacetyl ferrocene.

Chemistry
1 answer:
motikmotik3 years ago
6 0

Answer:

..........................

Explanation:

........................

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Answer:

When atoms lose or gain electrons, they become positively or negatively charged ions. In an ionic compound, the ions are arranged in a three-dimensional structure called a crystal. ... When an atoms gains or loses electrons, it gains a charge, thus becoming an ion.

Explanation:

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2. What happens to the pH when you add more H+ ions to a solution that has no buffers?
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It will go over the amount it needs to.
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Green (Gg) Is crossed with a yellow pea plant (Gg )
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Calculate the solubility of CaF2 in g/L (Ksp = 4.0 x 10-8). 2. What is the pH of a solution containing a hydrogen ion concentrat
Pepsi [2]

Answer:

\large \boxed{1. \text{ 0.17 g/L; 2. 3.52; 3. Cl; 4. (a) +3; (b) +4; (c) +6}}

Explanation:

1. Solubility of CaF_2

(a) Molar solubility

CaF₂ ⇌ Ca²⁺ + 2F⁻

K_{\text{sp }} = \text{[Ca$^{2+}$]}\text{[F$^{-}$]}^{2}= 4.0 \times 10^{-8}\\s(2s)^{2}=4.0 \times 10^{-8}\\4s^{3} = 4.0 \times 10^{-8}\\s^{3} = 1.0 \times 10^{-8}\\s =2.2 \times 10^{-3}\text{ mol/L}

(b) Mass solubility

\text{Solubility} = 2.2 \times 10^{-3} \text{ mol/L} \times \dfrac{\text{78.07 g}}{\text{1 L }} = \text{0.17 g/L}\\\\\text{The solubility of CaF$_{2}$ is $\large \boxed{\textbf{0.17 g/L}}$}

2. pH

pH = -log [H⁺] = -log(3.0 × 10⁻⁴) = 3.52

3. Oxidizing and reducing agents

Zn + Cl₂ ⟶ ZnCl₂

\rm \stackrel{\hbox{0}}{\hbox{Zn}} + \stackrel{\hbox{0}}{\hbox{ Cl}_{2} }\longrightarrow \stackrel{\hbox{+2}}{\hbox{Zn}}\stackrel{\hbox{-1}}{\hbox{Cl}_{2}}

The oxidation number of Cl has decreased from 0 to -1.

Cl has been reduced, so Cl is the oxidizing agent.

4. Oxidation numbers

(a) Al₂O₃

\stackrel{\hbox{$\mathbf{+3}$}}{\hbox{Al}_{2}}\stackrel{\hbox{-2}}{\hbox{O}_{3}}

1O = -2; 3O = -6; 2Al  = +6; 1Al = +3

(b) XeF₄

\stackrel{\hbox{$\mathbf{+4}$}}{\hbox{Xe}}\stackrel{\hbox{-1}}{\hbox{F}_{4}}

1F = -1; 4F = -4; 1 Xe = +4

(c) K₂Cr₂O₇

\stackrel{\hbox{${+1}$}}{\hbox{K}_{2}}\stackrel{\hbox{$\mathbf{+6}$}}{\hbox{Cr}_{2}}\stackrel{\hbox{-2}}{\hbox{O}_{7}}

1K = +1; 2K = +2; 1O = -2; 7O = -14

+2 - 14 = -12

2Cr = + 12; 1 Cr = +6

8 0
4 years ago
I need help please this is my final
aliina [53]

Answer:

hmm fart

Explanation:

yep my final

8 0
2 years ago
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