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slamgirl [31]
3 years ago
8

Leave the answer blank if no precipitate will form. (Express your answer as a chemical formula.) Formula of precipitate ZnCl2(aq

) + KOH(aq)→ (s) K2CO3(aq) + AgNO3(aq)→ (s) (NH4)3PO4(aq) + Ni(NO3)2(aq)→ (s) NaCl(aq) + KNO3(aq)→ (s) Na2CO3(aq) + NH4Cl(aq)→
Chemistry
2 answers:
Rasek [7]3 years ago
7 0

Answer:

1. Zn(OH)₂ (s)

2. Ag₂CO₃ (s)

3. Ni₃(PO₄)₂(s)

4. No reaction

5. (NH₄)₂CO₃(s)

Explanation:

Let's state the equations and we analyse some solubility and precipitation information:

ZnCl₂(aq) + 2KOH(aq) → Zn(OH)₂ (s)  +  2KCl (aq)

All the salts from the halogens with group 1, are soluble.

The OH⁻ reacts to Zn cation in order to produce a precipitate. This is ok, but if the base is in excess, the Zn(OH)₂ will be soluble

K₂CO₃(aq) + 2AgNO₃(aq) → Ag₂CO₃ (s) ↓ + 2KNO₃(aq)

All salts from nitrate are soluble

All salts from carbonates are insoluble

2(NH₄)₃PO₄(aq) + 3Ni(NO₃)₂(aq) → Ni₃(PO₄)₂(s) ↓ + 6NH₄NO₃(aq)

Salts from phosphates are insoluble

All salts from nitrate are soluble

NaCl(aq) + KNO3(aq) → NO REACTION

All salts from nitrate are soluble

All the salts from the halogens with group 1, are soluble

Na₂CO₃(aq) + 2NH₄Cl(aq) → 2NaCl(aq) + (NH₄)₂CO₃(s) ↓

All salts from carbonates are insoluble

All the salts from the halogens with group 1, are soluble

Ivan3 years ago
4 0

Answer:

1) <u>Zn(OH)2(s)</u>

<u>2)Ag2CO3(s)</u>

<u>3)Ni3(PO4)2(s)</u>

<u>4)/</u>

<u>5)(NH4)2CO3(s)</u>

Explanation:

ZnCl2(aq) + 2KOH(aq)→ Zn^2+ +2Cl- + 2K+ + 2OH-

ZnCl2(aq) + 2KOH(aq)→ <u>Zn(OH)2(s)</u> + 2KCl(aq)

K2CO3(aq) + 2AgNO3(aq)→ 2K+ + CO3^2- + 2Ag+ + 2NO3-

K2CO3(aq) + 2AgNO3(aq)→ 2KNO3(aq) + <u>Ag2CO3(s)</u>

2(NH4)3PO4(aq) + 3Ni(NO3)2(aq) → 6NH4+ + 2PO4^3- + 3NI^2+ + 6NO3-

2(NH4)3PO4(aq) + 3Ni(NO3)2(aq) → <u>Ni3(PO4)2(s)</u> + 6NH4NO3(aq)

NaCl(aq) + KNO3(aq)→ Na+ + Cl- + K+ + NO3-

NaCl(aq) + KNO3(aq)→ NaNO3(aq) + KCl(aq)

Na2CO3(aq) + 2NH4Cl(aq)→ 2Na+ + CO3^2- + 2NH4+ + 2Cl-

Na2CO3(aq) + 2NH4Cl(aq)→ 2NaCl(aq) +( NH4)2CO3(s)

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Answer:

a) v₀ = 4.41 × 10¹⁴ s⁻¹

b) W₀ = 176 KJ/mol of ejected electrons

c) From the graph, light of frequency less than v₀ will not cause electrons to break free from the surface of the metal. Electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.

d) When frequency of the light exceeds v₀, there is an increase of electron kinetic energy from zero steadily upwards with a constant slope. This is because, once light frequency exceeds, v₀, its energy too exceeds the work function of the metal and the electrons instantaneously gain the energy of incident light and convert this energy to kinetic energy by breaking free and going into motion. The energy keeps increasing as the energy and frequency of incident light increases and electrons gain more speed.

e) The slope of the line segment gives the Planck's constant. Explanation is in the section below.

Explanation:

The plot for this question which is attached to this solution has Electron kinetic energy on the y-axis and frequency of incident light on the x-axis.

a) Wavelength, λ = 680 nm = 680 × 10⁻⁹ m

Speed of light = 3 × 10⁸ m/s

The frequency of the light, v₀ = ?

Frequency = speed of light/wavelength

v₀ = (3 × 10⁸)/(680 × 10⁻⁹) = 4.41 × 10¹⁴ s⁻¹

b) Work function, W₀ = energy of the light photons with the wavelength of v₀ = E = hv₀

h = Planck's constant = 6.63 × 10⁻³⁴ J.s

E = 6.63 × 10⁻³⁴ × 4.41 × 10¹⁴ = 2.92 × 10⁻¹⁹J

E in J/mol of ejected electrons

Ecalculated × Avogadros constant

= 2.92 × 10⁻¹⁹ × 6.023 × 10²³

= 1.76 × 10⁵ J/mol of ejected electrons = 176 KJ/mol of ejected electrons

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As evident from the graph, electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.

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e) The slope of the line segment gives the Planck's constant. From the mathematical relationship, E = hv₀,

And the slope of the line segment is Energy of ejected electrons/frequency of incident light, E/v₀, which adequately matches the Planck's constant, h = 6.63 × 10⁻³⁴ J.s

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Answer:

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Explanation:

Chemical reaction

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                                       14.4            ?

Process

1.- Calculate the molecular mass of bromine and potassium bromide

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Potassium bromide = 2(79.9 + 39.1) = 238 g

2.- Solve it using proportions

              159.8 g of Bromine ------------ 238 g of potassium bromide                    

                14.4 g of Bromine  ------------  x

                        x = (14.4 x 238) / 159.8

                        x = 3427.2 / 159.8

                        x = 21.45g of KBr

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