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grigory [225]
3 years ago
6

Identify the oxidizing agent and the reducing agent in the following:

Chemistry
1 answer:
Sergeeva-Olga [200]3 years ago
8 0

Explanation:

Oxidizing agent is the reactant in the chemical reaction which oxidizes others and reduces itself.

Reducing agent is the reactant in the chemical reaction which reduces others and oxidizes itself.

(a) In the given reaction :-

8H^+_{(aq)} + [Cr_2O_7]^{2-}_{(aq)} + 3SO_3^{2-}_{(aq)}\rightarrow 2Cr^{3+}_{(aq)} + 3SO_4^{2-}_{(aq)} + 4H_2O_{(l)}

Cr exists in +6 oxidation state in [Cr_2O_7]^{2-} and forms Cr^{3+} in which it has +3 oxidation state.

Thus, it is reduced in the chemical reaction. Thus, [Cr_2O_7]^{2-} is oxidizing agent and SO_3^{2-} is the reducing agent.

(b) In the given reaction :-

NO_3^-_{(aq)} + 4Zn_{(s)} + 7OH^-_{(aq)} + 6H_2O_{(l)}\rightarrow 4Zn(OH)_4^{2-}_{(aq)} + NH_3_{(aq)}

Zn exists in 0 oxidation state in Zn and forms Zn(OH)_4^{2-} in which it has +2 oxidation state.

Thus, it is oxidized in the chemical reaction. Thus, Zn is reducing agent and NO_3^- is the oxidizing agent.

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Answer:

D

Explanation:

6 0
3 years ago
Sodium electron configuration in longhand notation
Elden [556K]

Answer:

Na₁₁ = 1s² 2s² 2p⁶ 3s¹

Explanation:

Sodium is present in group 1.

It is alkali metal.

It has one valence electron.

The atomic number of sodium is 11.

Its atomic mass is 23 amu.

The longhand notation of electronic configuration of sodium can be written as,

Na₁₁ = 1s² 2s² 2p⁶ 3s¹

The electronic configuration in shorthand notation( noble gas) would be written as,

Na₁₁ = [Ne] 3s¹

Sodium loses its one valence electron to complete the octet and get stable thus form +1 cation.

It react with halogen and form salt. Such as sodium chloride.

2Na + Cl₂  →  2NaCl

5 0
3 years ago
If the mass is 10g and the volume<br> is 2cm3 what is the density?
SSSSS [86.1K]

Answer:

The density is 5 g/cm3

Explanation:

The density (δ) is the ratio between the mass and the volume of a compound:

δ=m/v= 10 g/2 cm3= 5 g/cm3

3 0
3 years ago
In an acid-base titration, a student uses 21.35 mL of 0.150 M NaOH to neutralize 25.00 mL of H2SO4. How many moles of acid are i
GalinKa [24]

Answer: There are 0.006 moles of acid in the flask.

Explanation:

Given: V_{1} = 21.35 mL,        M_{1} = 0.150 M

V_{2} = 25.0 mL,           M_{2} = ?

Formula used to calculate molarity of H_{2}SO_{4} is as follows.

M_{1}V_{1} = M_{2}V_{2}

Substitute the values into above formula as follows.

M_{1}V_{1} = M_{2}V_{2}\\0.15 M \times 21.35 mL = M_{2} \times 25.0 mL\\M_{2} = 0.1281 M

As molarity is the number of moles of a substance present in a liter of solution.

Total volume of solution = V_{1} + V_{2}

= 21.35 mL + 25.0 mL

= 46.36 mL  (1 mL = 0.001 L)

= 0.04636 L

Therefore, moles of acid required are calculated as follows.

Molarity = \frac{no. of moles}{Volume (in L)}\\0.1281 M = \frac{no. of moles}{0.04635 L}\\no. of moles = 0.006 mol

Thus, we can conclude that there are 0.006 moles of acid in the flask.

3 0
3 years ago
Drag each label to the correct location on the equation. Each label may be used more than once.
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Hope this helps! If so, brainliest would be appreciated!
8 0
2 years ago
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