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Artist 52 [7]
3 years ago
7

Sodium thiosulfate, na2s2o3, is used as a "fixer" in black and white photography. identify the reducing agent in the reaction of

thiosulfate with iodine.2s2o32–(aq) + i2(aq) → s4o62–(aq) + 2i–(aq)
Chemistry
1 answer:
aev [14]3 years ago
8 0
Answer is: reducin agent in this reaction is thiosulfate (S₂O₃²⁻).
Balanced chemical reaction: 2S₂O₃²⁻(aq) + I₂(aq) → S₄O₆²⁻(aq) + 2I⁻<span>(aq).
Reducing agent is element or compound who loose electrons in chemical reaction. Sulfur in </span>thiosulfate change oxidation number from +2 to +5 tetrathionate anion (two<span> sulfur </span>atoms in the ion have oxidation state<span> 0 and two atoms have oxidation state +5).</span>
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artcher [175]
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I need the 3 questions !!! About Rubidium (Rb) <br><br> ASAP PLEASE DUE IN LIKE 20 min
hjlf

Answer:

  1. 419kJ/mol  
  2. 5,0,0,+12
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Explanation:

1. Topic: Chemistry

ElementFirst Ionization Energy (kJ/mol) Lithium520Sodium496Rubidium403Cesium376According to the above table, which is most likely to be the first ionization energy for potassium?  

  • 536kJ/mol  
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  • 419kJ/mol  
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2. Topic: Chemistry, Atom

The correct set of four quantum numbers for the valence electrons of the rubidium atom   (Z=37)  is:  

  • 5,1,1,+12  
  • 5,0,1,+12  
  • 5,0,0,+12
  • 5,1,0,+12

3. Rubidium and cesium are pyrophoric. Here the term pyrophoric means:

  • That hardly catches fire
  • That does not catch fire at all
  • That catches fire spontaneously
  • None of these
5 0
3 years ago
PLEASE ANSWER I GIVE BRAINLIEST AND MAX POINTS
d1i1m1o1n [39]

Answer:

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Explanation:

7 0
3 years ago
The crack shown in the lithosphere was created by the movement of two or more tectonic plates. What is this crack called
nikitadnepr [17]

Answer:

The lithosphere is made up of pieces of tectonic plates. These plates are constantly changing and move towards the mantle. Non-stop movement of tectonic plates causes stress on the earth's outermost layer i.e,the crust. When these stresses extends it leads to cause cracks called faults.

Explanation:

4 0
3 years ago
A 50.0 mL sample of a 1.00 M solution of CuSO4 is mixed with 50.0 mL of 2.00 M KOH in a calorimeter. The temperature of both sol
Reika [66]

Answer : The enthalpy change for the process is 52.5 kJ/mole.

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the solution

q=[q_1+q_2]

q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]

where,

q = heat released by the reaction

q_1 = heat absorbed by the calorimeter

q_2 = heat absorbed by the solution

c_1 = specific heat of calorimeter = 12.1J/^oC

c_2 = specific heat of water = 4.18J/g^oC

m_2 = mass of water or solution = Density\times Volume=1/mL\times 100.0mL=100.0g

\Delta T = change in temperature = T_2-T_1=(26.3-20.2)^oC=6.1^oC

Now put all the given values in the above formula, we get:

q=[(12.1J/^oC\times 6.1^oC)+(100.0g\times 4.18J/g^oC\times 6.1^oC)]

q=2623.61J

Now we have to calculate the enthalpy change for the process.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat released = 2626.61 J

n = number of moles of copper sulfate used = Concentration\times Volume=1M\times 0.050L=0.050mole

\Delta H=\frac{2623.61J}{0.050mole}=52472.2J/mole=52.5kJ/mole

Therefore, the enthalpy change for the process is 52.5 kJ/mole.

8 0
3 years ago
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