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3 years ago

Which chemical element was discovered first microsoft rewards.

Chemistry

1 answer:

djverab [1.8K]3 years ago

5 0

Answer:

phosphorous

Explanation:

A necessary prerequisite to the construction of the periodic table was the discovery of the individual elements. Although elements such as gold, silver, tin, copper, lead and mercury have been known since antiquity, the first scientific discovery of an element occurred in 1649 when Hennig Brand discovered phosphorous.

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The reaction type is Synthesis

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1. Based on this graph, how does metal A differ from metal B?

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B.Metal B has higher density than metal A

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3 years ago

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

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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

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