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devlian [24]
3 years ago
9

You have one test tube which contains a colorless solution that is either Cu+2 (aq) or Bi+3(aq). Select a reagent that will allo

w you to differentiate between the two chemical species. If the solution is Cu+2, what will happen when the reagent is added? If the solution is Bi+3, what will happen when the reagent is added?
Chemistry
1 answer:
Jlenok [28]3 years ago
8 0

Answer:

Explanation:

You can use KI and K4[Fe(CN)6].

When you add KI in a solution with only Cu2+ nothing will happen, but when you add K4[Fe(CN)6] it will form a solid of Cu2[Fe(CN)6] that will be reddish-brown.

When you add K4[Fe(CN)6] in a solution with only Bi3+ nothing will happen, but when you add KI it will form a solid of BiI3 that will be black.

So in a solution which contains both of then, you can add KI to determinate Bi3+, if you see a black solid. If nothing happens, you can affirme that it contains Cu2+.

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Topsoil consists of most weathered mineral and organic material. Biological agents are also responsible for the breakdown of complex organic matter which releases simple nutrients. This process of mineralisation make soil fertile.

Explanation:

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What scientists added negatively-charged particles to the model of the atom?
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J.J.Thomson discovered negatively charged particles by cathode ray tube experiment in the year 1897. The particles were name electrons.

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Which of earth's spheres is most affected by weathering and erosion
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Answer: the hydrosphere

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3 years ago
1. Pewien tlenek azotu o masie cząsteczkowej 108 u zawiera 74,07% tlenu. Wykonaj stosowne obliczenia i napisz wzór sumaryczny te
love history [14]

Answer:

1. Stąd empiryczny wzór substancji to N₂O₅

2. W związku z tym ilość w gramach chlorku sodu NaCl is 114,4 g.

Explanation:

1. Mamy tutaj;

Masa molowa tlenku azotu = 108u

Masa azotu = 14,0067u

Masa tlenu = 15,999 u

74,07% masy tlenku azotu to tlen

Dlatego masa obecnego tlenu = 108 × 74,07 / 100 = 79,9956 u

Masa obecnego azotu = 108 - 79,9956 = 28,0044u

Liczba moli tlenu = 79,9956 / 15,999 = 5,00003 ≈ 5

Liczba moli azotu = 28,0044 / 14,0067 = 1,99935 ≈ 2

Stąd empiryczny wzór substancji to N₂O₅.

2. Kiedy sód reaguje z chlorem, mamy;

2Na (s) + Cl₂ (g) → 2NaCl (s)

Dlatego 2 mole sodu Na reaguje z 1 molem chloru gazowego Cl₂, z wytworzeniem 2 moli chlorku sodu NaCl

W związku z tym 1 mol sodu Na reaguje z 1/2 molem chloru gazowego Cl₂ z wytworzeniem 1 mola chlorku sodu NaCl

Masa Na obecnego w reakcji = 45 g

Masa molowa sodu = 22,989769u

Liczbę moli sodu w 45 g sodu podano w następujący sposób;

Liczba \, \, moli \, \, Na= \frac{Mass \, of \, Na}{Molowy \, masa \, z \, Na} = \frac{45}{22.989769} = 1.96 \, mole

Z czego 1,96 moli sodu Na reaguje z 1/2 × 1,96 mola chloru gazowego Cl₂ z wytworzeniem 1,96 mola chlorku sodu NaCl

Masa molowa NaCl = 58,44 g / mol

Dlatego masa NaCl = liczba moli NaCl × masa molowa NaCl

Masa NaCl = 1,96 × 58,44 = 114,39001 g ≈ 114,4 g

W związku z tym ilość w gramach chlorku sodu NaCl = 114,4 g.

4 0
3 years ago
Two solutions are made by mixing sugar and enough water to make a 1 liter solution. One contains 10% sugar and the other 20% sug
zysi [14]

Two solutions are made by mixing sugar and enough water to make a 1 liter solution. 10% sugar has more solvent.

In chemistry, a solution is a homogeneous mixture of two or more compounds in their relative proportions that can be continually altered up to what is referred to as the limit of solubility. Although the word "solution" is frequently used to refer to the liquid state of matter, solutions of gases and solids are also possible. A solute is dissolved by a solvent (s), which is a substance (from the Latin solv, "loosen, untie, solve,") to produce a solution. The most common form of a solvent is a liquid, although other forms include solids, gases, and supercritical fluids. Depending on the temperature, a different amount of solute can dissolve in a given volume of solvent. Solvents are extensively utilized in paints, paint removers, inks, and dry cleaning.

Learn more about solution here :

brainly.com/question/7932885

#SPJ4

7 0
1 year ago
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