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3 years ago

The equivalence point of a weak acid: a. is at 7 b. Is above pH 7 c. Is below pH 7

Chemistry

1 answer:

almond37 [142]3 years ago

8 0

Answer:

b) is above pH 7

Explanation:

In the case of titration of a weak acid with a strong base, equivalence point occurs at pH> 7; there is complete salt formation, then the pH is calculated through the salt.

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3 years ago

How many mL of 0.715 M HCl is required to neutralize 1.25 grams of sodium carbonate? (producing carbonic acid)

jonny [76]

Answer:

34 mL

Explanation:

We'll begin by calculating the number of mole in 1.25 g of sodium carbonate, Na₂CO₃. This can be obtained as follow:

Mass of Na₂CO₃ = 1.25 g

Molar mass of Na₂CO₃ = (23×2) + 12 + (16×3)

= 46 + 12 + 48

= 106 g/mol

Mole of Na₂CO₃ =?

Mole = mass /molar mass

Mole of Na₂CO₃ = 1.25 / 106

Mole of Na₂CO₃ = 0.012 mole

Next, we shall determine the number of mole HCl needed to react with 0.012 mole of Na₂CO₃.

The equation for the reaction is given below:

Na₂CO₃ + 2HCl —> H₂CO₃ + 2NaCl

From the balanced equation above,

1 mole of Na₂CO₃ reacted with 2 moles of HCl.

Therefore, 0.012 mole of Na₂CO₃ will react with = 0.012 × 2 = 0.024 mole of HCl.

Next, we shall determine the volume of HCl required for the reaction. This is illustrated:

Mole of HCl = 0.024 mole

Molarity of HCl = 0.715 M

Volume of HCl =?

Molarity = mole /Volume

0.715 = 0.024 / volume of HCl

Cross multiply

0.715 × volume of HCl = 0.024

Divide both side by 0.715

Volume of HCl = 0.024 / 0.715

Volume of HCl = 0.034 L

Finally, we shall convert 0.034 L to mL

This can be obtained as follow:

1 L = 1000 mL

Therefore,

0.034 L = 0.034 L × 1000 mL / 1 L

0.034 L = 34 mL

Therefore, 34 mL of HCl is needed for the reaction.

6 0

3 years ago

In a galvanic cell,a oxidation occurs at the (name of electrode)b.the cathode is the (sign) electrodec.cations flow in solution

Tatiana [17]

Answer: a) anode

b) positive

c) cathode

d) anode to cathode

Explanation:

Galvanic cell is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy.

In the galvanic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode. Thus the electrons are produced at anode and travel towards cathode.

Due to loss of electrons at anode, positive charge in concentrated in anode and negative charge is concentrated in cathode. To balance these charges , positive ions or cations from salt bridge move towards cathode and negative ions or anions from salt bridge move towards anode.

The balanced two-half reactions will be:

Oxidation half reaction : $M(s)\rightarrow M^{n+}(aq)+ne^-$

Reduction half reaction : $X^{n+}(aq)+ne^-\rightarrow X(s)$

Thus the overall reaction will be,

$M(s)+X^{n+}(aq)\rightarrow M^{n+}(aq)+X(s)$

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