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Alborosie
3 years ago
12

Can someone help me please??

Chemistry
1 answer:
Xelga [282]3 years ago
4 0

41) Answer is: the average atomic mass of chromium is 52.056.

Ar(⁵⁰Cr) = 50.

ω(⁵⁰Cr) = 4.34% ÷ 100%.

ω(⁵⁰Cr) = 0.0434; relative abundance of chromium-50.

Average mass can calculate by summary of percentege of each isotope.

Ar(Cr) = Ar(⁵⁰Cr) · ω(⁵⁰Cr) + Ar(⁵²Cr) · ω(⁵²Cr) + Ar(⁵³Cr) · ω(⁵³Cr) + Ar(⁵⁴Cr) · ω(⁵⁴Cr).

Ar(Cr)= 50 · 0.0434 + 52 · 0.8379 + 53 · 0.095 + 54 · 0.0237.

Ar(Cr) = 2.17 + 43.571 + 5.035 + 1.280

Ar(Cr) = 52.056.

42) Chlorine has atomic number 17, it means it has 17 protons and 17 electrons.

Electron configuration of chlorine atom: ₁₇Cl 1s² 2s² 2p⁶ 3s² 3p⁵.

Chloride is negative ion of chlorine. Chloride is formed when chlorine gain one lectron.

Chloride anion has 17 protons and 18 electrons (like argon-noble gas).

The electron configuration for the chloride ion: ₁₇Cl⁻ 1s²2s²2p⁶3s²3p⁶.

43) Anwser is: the percent by mass of hydrogen in aspirin is 4.48%.

% calculate by atomic mass of element multiply number of that element in molecule divided by molecular mass of molecule.

Ar(H) = 1.008; the relative atomic mass of hydrogen atom.

Mr(C₉H₈O₄) = 9 · Ar(C) + 8 · Ar(H) + 4 · Ar(O).

Mr(C₉H₈O₄) = 180.157.

ω(H)= 8 · Ar(H) ÷ Mr(C₉H₈O₄) · 100%.

ω(H) = 8 · 1.008 ÷ 180.157 · 100%.

ω(H) = 4.48 %.

44) Balanced chemical reaction:

Cd(NO₃)₂ + 2NH₄Cl → CdCl₂ + 2NH₄NO₃.

This chemical reaction is double displacement reaction - cations (Cd²⁺ and NH₄⁺) and anions (NO₃⁻ and Cl⁻) of the two reactants switch places and form two new compounds.

Cd(NO₃)₂ is cadmium nitrate.

NH₄Cl is amonium chloride.

CdCl₂ is cadmium chloride.

NH₄NO₃ is ammonium nitrate.

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Vika [28.1K]
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5 0
3 years ago
As the sun heats the surface of the oceans, two layers of water result. What separates the layers?
Basile [38]

Answer:

Thermocline separate these two layers.

Explanation:

Ocean consist of three major layers on the basis of temperature.

1. Upper layer

2. Deep layer

3. Thermocline

Upper layer:

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Deep layer

The deep layer is present below the thermocline. It is present in deep where sunlight can not approach to it and its temperature remain low.

Thermocline

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2 years ago
One should know properties of constituents of a mixture to separate the mixture .Why?​
klio [65]

Answer:

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8 0
3 years ago
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Please answer the other questions!!!
son4ous [18]

Answer:

Na + N ------- Na3N

Explanation:

It forms Sodium Nitride

8 0
2 years ago
Escriba en termino de moles, de moléculas y de masa las siguientes ecuaciones a. Fe +2HCl ________ FeCl2 + H2 b. CH4 + 2O2 _____
MA_775_DIABLO [31]

Answer:

a.

  • 1 mol de Hierro reacciona con 2 moles de acido clorhidrico para formar 1 mol de cloruro de hierro (II)
  • 6.02×10²³ moleculas de hierro reaccionan con 1.20×10²⁴ moleculas de acido clorhidrico para formar 6.02×10²³ moleculas de cloruro de hierro (II)
  • 55.85 g de hierro reaccionan con 72.9 gramos de acido clorhidrico para formar 126.75 g de cloruro férrico.

b.

  • 6.02×10²³ moleculas de metano reaccionan con 1.20×10²⁴ moleculas de oxigeno para formar 6.02×10²³ moleculas de dioxido de carbono y 1.20×10²⁴ moleculas de agua.
  • 1 mol de metano reacciona con dos moles de oxigeno para generar 1 mol de dióxido de carbono y dos moles de agua en estado de vapor.
  • 16 gramos de metano reaccionan con 64 g de oxigeno para formar 44 gramos de dioxido de carbono y 36 gramos de agua.

c.

  • 3 moles de plata sólida reaccionan con 4 moles de acido nitrico para formar 3 moles de nitrato de plata, 1 mol de monoxido de nitrogeno y 2 moles de agua.
  • 1.80×10²⁴ moleculas de plata reaccionan con 2.41×10²⁴ moleculas de acido nitrico para formar 1.80×10²⁴ moleculas de nitrato de plata, 6.02×10²³ moleculas de monoxido de nitrogeno y 1.20×10²⁴ moleculas de agua.
  • 323.58 g de plata reaccionan con 252 g de acido nitrico para formar 509.58 g de nitrato de plata, 30 g de monoxido de nitrogeno y 36 g de agua.

Explanation:

a. Fe (s)  +2 HCl (aq) → FeCl₂ (aq)

1 mol de Hierro reacciona con 2 moles de acido clorhidrico para formar 1 mol de cloruro de hierro (II)

Calculamos cuanto son dos moles de moleculas sabiendo que:

6.02×10²³ moleculas / 1 mol  . 2 mol = 1.20×10²⁴ moleculas. Entonces

6.02×10²³ moleculas de hierro reaccionan con 1.20×10²⁴ moleculas de acido clorhidrico para formar 6.02×10²³ moleculas de cloruro de hierro (II)

Calculamos las masas molares de cada reactivo y producto

Fe = 55.85 g

HCl = 36.45 g

FeCl₂ = 126.75 g

55.85 g de hierro reaccionan con 72.9 gramos de acido clorhidrico para formar 126.75 g de cloruro férrico.

b. CH₄(g) + 2O₂ (g) → CO₂ (g) + 2H₂O (g)

1 mol de metano reacciona con dos moles de oxigeno para generar 1 mol de dióxido de carbono y dos moles de agua en estado de vapor.

6.02×10²³ moleculas de metano reaccionan con 1.20×10²⁴ moleculas de oxigeno para formar 6.02×10²³ moleculas de dioxido de carbono y 1.20×10²⁴ moleculas de agua.

Calculamos las masas molares:

CH₄ = 16 g

O₂ = 32 g

CO₂ = 44 g

H₂O g = 18 g

16 gramos de metano reaccionan con 64 g de oxigeno para formar 44 gramos de dioxido de carbono y 36 gramos de agua.

c. 3 Ag (s) + 4HNO3 (aq) → 3 AgNO3 (aq) + NO (g) + 2H₂O (aq)

Calculamos cuantos moleculas contienen 3 y 4 moles:

6.02×10²³  . 3 = 1.80×10²⁴ moleculas

6.02×10²³  . 4 = 2.41×10²⁴ moleculas

3 moles de plata sólida reaccionan con 4 moles de acido nitrico para formar 3 moles de nitrato de plata, 1 mol de monoxido de nitrogeno y 2 moles de agua.

1.80×10²⁴ moleculas de plata reaccionan con 2.41×10²⁴ moleculas de acido nitrico para formar 1.80×10²⁴ moleculas de nitrato de plata, 6.02×10²³ moleculas de monoxido de nitrogeno y 1.20×10²⁴ moleculas de agua.

Calcualmos las masas molares:

Ag = 107.86 g

HNO₃ = 63 g

AgNO₃ = 169.86 g

NO = 30 g

H₂O = 18 g

323.58 g de plata reaccionan con 252 g de acido nitrico para formar 509.58 g de nitrato de plata, 30 g de monoxido de nitrogeno y 36 g de agua.

6 0
3 years ago
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