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2 years ago

13

Write an equation to show how HC2O4− can act as a base with HS− acting as an acid.

1 answer:

Artyom0805 [142]2 years ago

7 0

An acid donates $H^{+}$ ion in aqueous solution. A base accepts $H^{+}$ ion in aqueous solution.

The equation representing the acid base reaction of $HC_{2}O_{4}^{-}$ and $HS^{-}$ :

$HC_{2}O_{4}^{-}(aq) + HS^{-}(aq) ----> H_{2}C_{2}O_{4}(aq)+S^{2-}(aq)$

In the above reaction, as $HC_{2}O_{4}^{-}$ acts as a base it is accepting the hydrogen ion from $HS^{-}$ . Similarly, $HS^{-}$ donates its hydrogen ion to $HC_{2}O_{4}^{-}$ acting as an acid.

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What is the millimolar solubility of oxygen gas, o2, in water at 16 ∘c, if the pressure of oxygen is 1.00 atm?

Artyom0805 [142]

Partial pressure is the amount of pressure or force that is exerted by the atoms into the outer environment. it is dependent on the temperature and pressure of the present surroundings. in this case, we are asked in this problem to determine the partial pressure of oxygen at 16oC and 1 atm. We have to look into a solubility data table commonly found in handbooks and determined via experiments and correlations. According to literature, the value of the partial pressure is equal to 0.617 mM.This is under the assumption that the salinity of the water in which oxygen is dissolved is equal to zero.

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6. The half life for uranium-235 is 7.0x108 years.

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2 years ago

What would the final freezing point of water be if 3 mol of sugar were added to 1 kg of water (Kf = 1.86C/(mol/kg) for water and

Fudgin [204]

is this for a test or are you genuinely interested? molality = mols sugar/kg solvent

Solve for molality

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Suggest why sodium and hydrogen ions do not diffuse at the same rate

Troyanec [42]

Answer:

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8 0

1 year ago

Carbonic acid, H2CO3, has two acidic hydrogens. A solution containing an unknown concentration of carbonic acid is titrated with

stepan [7]

Answer:

1) Net ionic equation :

$2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)$

2) 0.765 M is the molarity of the carbonic acid solution.

Explanation:

1) In aqueous carbonic acid , carbonate ions and hydrogen ion is present.:

$H_2CO_3(aq)\rightarrow 2H^+(aq)+CO_3^{2-}(aq)$ ..[1]

In aqueous potassium hydroxide , potassium ions and hydroxide ion is present.:

$KOH(aq)\rightarrow K^+(aq)+OH^{-}(aq)$ ..[2]

In aqueous potassium carbonate , potassium ions and carbonate ion is present.:

$K_2CO_3(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq)$ ..[3]

$H_2CO_3(aq)+2KOH(aq)\rightarrow K_2CO_3(aq)+2H_2O(l)$

From one:[1] ,[2] and [3]:

$2H^+(aq)+CO_3^{2-}(aq)+2K^+(aq)+2OH^{-}(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq)+H_2O(l)$

Cancelling common ions on both sides to get net ionic equation :

$2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)$

2)

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2CO_3$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is KOH.

We are given:

$n_1=2\\M_1=?\\V_1=50.0 mL\\n_2=1\\M_2=3.840 M\\V_2=20.0 mL$

Putting values in above equation, we get:

$M_1=\frac{1\times 3.840 M\times 20.0 mL}{2\times 50.2 mL}=0.765 M$

0.765 M is the molarity of the carbonic acid solution.

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2 years ago