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3 years ago

14

. Based on their position in the periodic table, predict which (of the two elements given) will be more reactive Circle your cho

ice.
a. V and Rb b. Ne and O c. FandO

1 answer:

Vladimir79 [104]3 years ago

8 0

Explanation:

your anawer to the question is A. V and Rb

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In Part III, the phenolphthalein indicator is used to monitor the equilibrium shifts of the ammonia/ammonium ion system. The phe

AnnZ [28]

The pink color in the solution fades. Some of the colored indicator ion converts to the colorless indicator molecule.

<h3>Explanation</h3>

What's the initial color of the solution?

$\text{NH}_4\text{Cl}$ is a salt soluble in water. $\text{NH}_4\text{Cl}$ dissociates into ions completely when dissolved.

$\text{NH}_4\text{Cl} \; (aq)\to {\text{NH}_4}^{+} \; (aq) +{\text{Cl}}^{-} \; (aq)$ .

The first test tube used to contain $\text{NH}_4\text{OH}$ . $\text{NH}_4\text{OH}$ is a weak base that dissociates partially in water.

$\text{NH}_4\text{OH} \; (aq) \rightleftharpoons {\text{NH}_4}^{+} \;(aq)+ {\text{OH}}^{-} \; (aq)$ .

There's also an equilibrium between $\text{OH}^{-}$ and ${\text{H}_3\text{O}}^{+}$ ions.

${\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l)$ .

$\text{OH}^{-}$ ions from $\text{NH}_4\text{OH}$ will shift the equilibrium between $\text{OH}^{-}$ and ${\text{H}_3\text{O}}^{+}$ to the right and reduce the amount of ${\text{H}_3\text{O}}^{+}$ in the solution.

The indicator equilibrium will shift to the right to produce more ${\text{H}_3\text{O}}^{+}$ ions along with the colored indicator ions. The solution will show a pink color.

What's the color of the solution after adding NH₄Cl?

Adding $\text{NH}_4\text{Cl}$ will add to the concentration of ${\text{NH}_4}^{+}$ ions in the solution. Some of the ${\text{NH}_4}^{+}$ ions will combine with $\text{OH}^{-}$ ions to produce $\text{NH}_4\text{OH}$ .

The equilibrium between $\text{OH}^{-}$ and ${\text{H}_3\text{O}}^{+}$ ions will shift to the left to produce more of both ions.

${\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l)$

The indicator equilibrium will shift to the left as the concentration of ${\text{H}_3\text{O}}^{+}$ increases. There will be less colored ions and more colorless molecules in the test tube. The pink color will fade.

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3 years ago

When solving a problem it is important to identify your given and needed units, but it is also important to understand the relat

Angelina_Jolie [31]

Answer:

The question has some details missing. here are the details ; Given the following ;

1. 43.2 g of tablet with 20 cm3 of space

2. 5 cm3 of tablets weighs 10.8 g

3. 5 g of balsa wood with density 0.16 g/cm3

4. 150 g of iron. With density 79g/cm 3

5. 32 cm3 sample of gold with density 19.3 g/cm3

6. 18 ml of cooking oil with density 0.92 g/ml

Explanation:

<u>Appropriate for calculating mass</u>

32 cm3 sample of gold with density 19.3 g/cm3

18 ml of cooking oil with density 0.92 g/ml

<u>Appropriate for calculating volume</u>

5 g of balsa wood with density 0.16 g/cm3

150 g of iron. With density 79g/cm 3

<u>Appropriate for calculating density</u>

43.2 g of tablet with 20 cm3 of space

5 cm3 of tablets weighs 10.8 g

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3 years ago

An electrolysis reaction is

Llana [10]

Answer:

the electrolysis reaction is a non- spontaneous reaction

Explanation:

Since electrons flow from it, the anode in an electrolytic cell is positive, while the cathode is negative when electrons flow into it. The device functions like a galvanic cell in that direction. In an electrolytic cell, an external voltage is applied and that is what causes a non spontaneous reaction

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3 years ago

HELP PLEASE WITH 1 QUESTION.

Nata [24]

The activity series goes top to bottom, most active to least active elements, going: Li, K, Ba, Sr, Ca, Na, Mg, Mn, Zn, Fe, Cd, Co, Ni, Sn, Pb, H, Cu, Ag, Hg, Au.

Thus, your list of metals would go from most reactive to least reactive: Li, K, Mg, Zn, Fe, Cu, Au

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3 years ago

Consider 0.01 m aqueous solutions of each of the following. a) NaI; b) CaCl2; c) K3PO4; and d) C6H12O6 (glucose) Arrange the sol

stealth61 [152]

Answer:

The solutions are ordered by this way (from lowest to highest freezing point): K₃PO₄ < CaCl₂ < NaI < glucose

Option d, b, a and c

Explanation:

Colligative property: Freezing point depression

The formula is: ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

We need to determine the i, which is the numbers of ions dissolved. It is also called the Van't Hoff factor.

Option d, which is glucose is non electrolyte so the i = 1

a. NaI → Na⁺ + I⁻ i =2

b. CaCl₂ → Ca²⁺ + 2Cl⁻ i =3

c. K₃PO₄ → 3K⁺ + PO₄⁻³ i=4

Potassium phosphate will have the lowest freezing point, then we have the calcium chloride, the sodium iodide and at the end, glucose.

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3 years ago