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1 year ago

Why is the center of the buffer region of a weak acid–strong base titration significant?

1 answer:

7 0

When a weak acid is titrated against a strong base, the pH of the solution rises, levels off through the buffer zone, and then rises quickly to reach the equivalence point.

<h3>What is titration?</h3>

Titration is a typical research center technique for quantitative investigation synthetic to decide the convergence of a recognized analyte. Titration is a crucial cycle in the creation of cheddar. For this situation, titration is significant while estimating the pH and acridity of the underlying milk being utilized to make the cheddar. Corrosive base titration is a quantitative investigation of convergence of an obscure corrosive or base arrangement. Titration, otherwise called titrimetry, is a typical lab strategy for quantitative substance investigation that is utilized to decide the convergence of a distinguished analyte. A reagent, called the titrant or titrator is ready as a standard arrangement. A known fixation and volume of titrant responds with an answer of analyte or titrand to decide focus. The volume of titrant responded is called titration volume.

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Which atmosphere layer is represented by F what are its characteristics

inysia [295]

That wouod be the ionosphere!

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3 years ago

The theoretical yield of a reaction is the amount of product obtained if the limiting reactant is completely converted to produc

Elis [28]

Answer: 9.9 grams

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

a) moles of $H_2$

$\text{Number of moles}=\frac{8.150g}{2g/mol}=4.08moles$

b) moles of $C_2H_4$

$\text{Number of moles}=\frac{9.330g}{28g/mol}=0.33moles$

$H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)$

According to stoichiometry :

1 mole of $C_2H_4$ combine with 1 mole of $H_2$

Thus 0.33 mole of $C_2H_4$ will combine with = $\frac{1}{1}\times 0.33=0.33$ mole of $H_2$

Thus $C_2H_4$ is the limiting reagent as it limits the formation of product.

As 1 mole of $C_2H_4$ give = 1 mole of $C_2H_6$

Thus 0.33 moles of $C_2H_4$ give = $\frac{1}{1}\times 0.33=0.33moles$ of $C_2H_6$

Mass of $C_2H_6=moles\times {\text {Molar mass}}=0.33moles\times 30g/mol=9.9g$

Thus theoretical yield (g) of $C_2H_6$ produced by the reaction is 9.9 grams

7 0

3 years ago

HELP ME NOW ASAP PLEASE, 10 POINTS HURRY

xxMikexx [17]

Which of those depend on mayfly nymphs. In other terms which animal eats mayfly nymphs?

I would say frogs

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2 years ago

How many grams are in 3.78 x 10^22 molecules of SO2?

Vesnalui [34]

X2 is the answer help me and I will help you

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3 years ago

CO(g)+2H2(g)⇌CH3OH(g) A reaction mixture in a 5.23-L flask at a certain temperature initially contains 27.2 g CO and 2.36 g H2.

irga5000 [103]

Answer: 5.70M

Explanation:

Molar mass of CO = 28.01 g/mol

Molar mass of H2 = 2.02 g/mol

Molar mass of CH3OH = 32.05 g/mol.

To determine the amount of each compound in the reaction mixture we use the formula.

Amount in mol = reacting mass/molar mass.

Inputing the given values we have,

26.6 g CO x (1 mol / 28.01 g ) = 0.9496608354 mol of CO.

To calculate the concentration of CO we use C=n/v, where n=amount and v= volume of CO.

Inputing the values in the formula

[CO] = 0.9496608354 mol CO / 5.23 L = 0.18158 M CO

Repeating thesame procedure for H

Amount of H=2.36 g H2 x ( 1 mol / 2.02 g ) = 1.168316832 mol of H2

Concentration of H2 in the mixture

[H2] = 1.168316832 mol H2 / 5.23 L = 0.223388 M of H2

Amount of CH3OH is determine similarly using rmass/molar mass

8.66 CH3OH x (1 mol / 32.05 g ) = 0.2702028081 mol of CH3OH

Concentration of

[CH3OH] = 0.2702028081 mol CH3OH/ 5.23 L = 0.051664 M CH3OH

Now equilibrium constant is determined by

Kc = [CH3OH] / [CO] [H2]^2

=0.051664/0.18158×0.223388×0.223388.

=5.70

7 0

3 years ago