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aleksandrvk [35]
2 years ago
11

At the midpoint of a titration curve ________. At the midpoint of a titration curve ________. The concentration of a conjugate b

ase is twice that of the concentration of a conjugate acid The pH equals the pKa The ability of the solution to buffer is at its least effective The concentration of a conjugate base is 1/2 that of the concentration of a conjugate acid
Chemistry
1 answer:
VARVARA [1.3K]2 years ago
8 0

Answer:

The concentration of a conjugate base is twice that of the concentration of a conjugate acid

Explanation:

This is because at the midpoint, the number of moles of acid equals half the number of moles of base at the midpoint. This means that at the midpoint, half the analyte has been titrated.

Since the concentration of the conjugate acid is half that of the conjugate base at the midpoint, this implies that the concentration of the conjugate base is twice that of the conjugate acid.

<u>Thus, at the midpoint of a titration curve the concentration of a conjugate base is twice that of the concentration of a conjugate acid </u>

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Need help I don’t understand this at all need to show my work and strategy it’s stoichiometry gram to gram please help
laila [671]

Answer:

781×10¯² g of MgCl₂.

Explanation:

The balanced equation for the reaction is given below:

Mg + 2HCl —> MgCl₂ + H₂

Next, we shall determine the mass of HCl that reacted and the mass of MgCl₂ produced from the balanced equation. This is illustrated below:

Molar mass of HCl = 1 + 35.5

= 36.5 g/mol

Mass of HCl from the balanced equation = 2 × 36.5 = 73 g

Molar mass of MgCl₂ = 24 + (35.5×2)

= 24 + 71

= 95 g/mol

Mass of MgCl₂ from the balanced equation = 1 × 95 = 95 g

SUMMARY:

From the balanced equation above,

73 g of HCl reacted to produce 95 g of MgCl₂.

Finally, we shall determine the mass of MgCl₂ produced by the reaction of 6 g of HCl. This can be obtained as follow:

From the balanced equation above,

73 g of HCl reacted to produce 95 g of MgCl₂.

Therefore, 6 g of HCl will react to produce = (6 × 95)/73 = 781×10¯² g of MgCl₂.

Thus, 781×10¯² g of MgCl₂ were obtained from the reaction.

4 0
3 years ago
5) Determine the mass if lithium hydroxide that is produced when 12.87 g of lithium
netineya [11]

Answer: 26.54 grams

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of lithium nitride}=\frac{12.87g}{34.83g/mol}=0.369moles

Li_3N+3H_2O\rightarrow 3LiOH+NH_3  

Li_3N is the limiting reagent as it limits the formation of product and H_2O is the excess reagent

According to stoichiometry :

As 1 moles of Li_3N give = 3 moles of LiOH

Thus 0.369 moles of O_2 give =\frac{3}{1}\times 0.369=1.108moles  of LiOH

Mass of LiOH=moles\times {\text {Molar mass}}=1.108moles\times 23.95g/mol=26.54g

Thus  26.54 g of LiOH will be produced from the given mass.

4 0
3 years ago
18 grams<br> 4 kilograms<br> 8 kilograms
Rina8888 [55]

Answer:

1 18 grams

2 500g

4 960 grams

5 7 kilograms

5 0
3 years ago
Intermolecular forces dipole differences london dispersion
loris [4]
33233728793278237876548742787874578378572098-2932-=93788784787489
8 0
3 years ago
Calculate the reaction quotient Qp for the following redox reaction: 14H+ + Cr2O72- + 6Cl- ----&gt; 2Cr3+ + 3Cl2 + 7H2O The reac
stich3 [128]

Answer:

Value of Q_{p} for the given redox reaction is 1.0\times 10^{-8}

Explanation:

Redox reaction with states of species:

14H^{+}(aq.)+Cr_{2}O_{7}^{2-}(aq.)+6Cl^{-}(aq.)\rightarrow 2Cr^{3+}(aq.)+3Cl_{2}(g)+7H_{2}O(l)

Reaction quotient for this redox reaction:

Q_{p}=\frac{[Cr^{3+}]^{2}.P_{Cl_{2}}^{3}}{[H^{+}]^{14}.[Cr_{2}O_{7}^{2-}].[Cl^{-}]^{6}}

Species inside third braket represent concentration in molarity, P represent pressure in atm and concentration of H_{2}O is taken as 1 due to the fact that H_{2}O is a pure liquid.

pH=-log[H^{+}]

So, [H^{+}]=10^{-pH}

Plug in all the given values in the equation of Q_{p}:

Q_{p}=\frac{(0.10)^{2}\times (0.010)^{3}}{(10^{-0.0})^{14}\times (1.0)\times (1.0)^{6}}=1.0\times 10^{-8}

7 0
3 years ago
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