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Anni [7]
3 years ago
5

What is the conjugate acid of HCO3- ?OH-H2CO3CO32-H2OH3O+

Chemistry
1 answer:
adell [148]3 years ago
5 0

<u>Answer:</u> The conjugate acid of HCO_3^- is H_2CO_3

<u>Explanation:</u>

According to the Bronsted-Lowry conjugate acid-base theory:

  • An acid is defined as a substance which looses donates protons and thus forming conjugate base.
  • A base is defined as a substance which accepts protons and thus forming conjugate acid.

To form a conjugate acid of HCO_3^-, this compound will accept one proton to form H_2CO_3

The chemical equation for the formation of conjugate acid follows:

HCO_3^-+H^+\rightarrow H_2CO_3

The conjugate acid formed is named as carbonic acid.

Hence, the conjugate acid of HCO_3^- is H_2CO_3

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B. 13 electrons

Explanation:

Bcoz protons are the same with electrons and the neutrons too.

8 0
3 years ago
(Help would be greatly appreciated) What is the molarity of a solution which contains 22.41 grams of NaCl in 50.0 mL of solution
luda_lava [24]

Answer:

7.67

Explanation:

4 0
3 years ago
You wish to make a 0.299 M hydroiodic acid solution from a stock solution of 6.00 M hydroiodic acid. How much concentrated acid
Art [367]

Answer:

V_1=2.49mL

Explanation:

Hello,

In this case, considering that the moles of hydrioiodic acid remain unchanged during the dilution process:

n_{HI}=n_{HI}

One could apply the following equality in terms of molarity:

V_1M_1=V_2M_2

Whereas the subscript 1 accounts for the solution before the dilution and 2 after the dilution, therefore, the required volume of 6.00 M acid is:

V_1=\frac{V_2M_2}{M_1} =\frac{50.0mL*0.299M}{6.00M}=2.49mL

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5 0
3 years ago
What are particles that are close together and locked in place??
Grace [21]
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3 0
3 years ago
A gas used to extinguish fires is composed of 75 % CO2 and 25 % N2. It is stored in a 5 m3 tank at 300 kPa and 25 °C. What is th
tatyana61 [14]

Answer : The partial pressure of the CO_2 in the tank in psia is, 32.6 psia.

Explanation :

As we are given 75 % CO_2 and 25 % N_2 in terms of volume.

First we have to calculate the moles of CO_2 and N_2.

\text{Moles of }CO_2=\frac{\text{Volume of }CO_2}{\text{Volume at STP}}=\frac{75}{22.4}=3.35mole

\text{Moles of }N_2=\frac{\text{Volume of }N_2}{\text{Volume at STP}}=\frac{25}{22.4}=1.12mole

Now we have to calculate the mole fraction of CO_2.

\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CO_2+\text{Moles of }N_2}

\text{Mole fraction of }CO_2=\frac{3.35}{3.35+1.12}=0.75

Now we have to calculate the partial pressure of the CO_2 gas.

\text{Partial pressure of }CO_2=\text{Mole fraction of }CO_2\times \text{Total pressure of gas}

\text{Partial pressure of }CO_2=0.75mole\times 300Kpa=225Kpa=225Kpa\times \frac{0.145\text{ psia}}{1Kpa}=32.625\text{ psia}

conversion used : (1 Kpa = 0.145 psia)

Therefore, the partial pressure of the CO_2 in the tank in psia is, 32.6 psia.

3 0
3 years ago
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