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____ [38]
2 years ago
11

A reagent bottle on the shelf labeled 0.5m nacl was used in place of the 0.5 m cacl2. assuming c2o4^2- to be in excess, what wou

ld be observed as a result of using this wrong reagent in this test? explain
Chemistry
1 answer:
Alchen [17]2 years ago
8 0
Let's see the reaction for each situation.

2 NaCl + C₂O₄²⁻ → Na₂C₂O₄ + Cl₂

CaCl₂+ C₂O₄²⁻ → CaC₂O₄ + Cl₂

Thus, it would differ whether the main product is sodium oxalate or calcium oxalate. In their solid forms, both are white. However, calcium oxalate is insoluble in water, while sodium oxalate is slightly soluble. So, if you used NaCl instead of CaCl₂, you wouldn't observe any precipitate in your solution. If there is, it would only be minute.
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Hydrazine (N2H4) and dinitrogen tetroxide (N2O4) form a self-igniting mixture that has been used as a rocket propellant. The rea
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Answer:

Explanation:

An oxidizing accepts an electron and becomes reduced while a reducing agent loses an electron and become oxidized.

Chemical equation:

1) 2 N₂H₄ + N₂O₄ → 3 N₂ + 4 H₂O

2) Hydrazine ( N₂H₄)  is being oxidized

Dinitrogen tetroxide N₂O₄ is being reduced

3) The reducing agent is Hydrazine ( N₂H₄) and the oxidizing agent is dinitrogen tetroxide (N₂O₄)

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2 years ago
**PLATO QUESTION, PLEASE ANSWER CORRECTLY, THANK YOU**
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4 0
2 years ago
Exactly 2.00 g of an ester A containing only C, H, and O was saponified with 15.00 mL of a 1.00 M NaOH solution. Following the s
Anna11 [10]

Answer:

Explanation:

check the attachment for the propose neutral  structure for each compound that is  consistent with the data.

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2 years ago
What is the concentration of x2??? in a 0.150 m solution of the diprotic acid h2x? for h2x, ka1=4.5??10???6 and ka2=1.2??10???11
lutik1710 [3]
The first dissociation for H2X:
                        H2X +H2O ↔ HX + H3O
initial                0.15                     0      0
change             -X                     +X      +X
at equlibrium 0.15-X                  X        X
because Ka1 is small we can assume neglect x in H2X concentration
     Ka1      = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15) 
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
        HX + H2O↔ X^2 + H3O
    8.2x10^-4          Y         8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2       = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m


4 0
2 years ago
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