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marissa
1 month ago
8

The pKs of succinic acid are 4.21 and 5.64. How many grams of monosodium succinate (FW = 140 g/mol) and disodium succinate (FW =

162 g/mol) must be added to 1 L of water to produce a solution with a pH 5.28 and a total solute concentration of 100 mM? (Assume the total volume remains 1 liter, answer in grams monosodium succinate, grams disodium succinate, respectively.)
Chemistry
1 answer:
alisha [2.9K]1 month ago
5 0
9.744g of monosodium succinate. 4.925g of disodium succinate. To determine the pH of the buffer created from the mixture of monosodium succinate and disodium succinate, we utilize the Henderson-Hasselbalch equation: pH = pKa + log ([Na₂Suc] / [NaHSuc]). Aiming for a pH of 5.28 with a pKa of 5.64 gives us the equation 5.28 = 5.64 + log ([Na₂Suc] / [NaHSuc]), which simplifies to -0.36 = log ([Na₂Suc] / [NaHSuc]). This results in 0.4365 = ([Na₂Suc] / [NaHSuc]). The total buffer concentration is 100mM, equating to 0.100M, hence the relation 0.100M = [Na₂Suc] + [NaHSuc]. Plugging this into the earlier equation gives us 0.4365 = (0.100M - [NaHSuc]) / [NaHSuc]. Solving this leads to [NaHSuc] = 0.0696M and subsequently [Na₂Suc] = 0.0304M. Since the buffer volume is 1L, this implies 0.0696 moles of NaHSuc and 0.0304 moles of Na₂Suc. Using the molar mass, we calculate the mass of monosodium succinate: 0.0696moles * (140g / 1mol) = 9.744g of monosodium succinate. For disodium succinate: 0.0304moles * (162g / 1mol) = 4.925g of disodium succinate.
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This indicates that the enzyme is a type of protein.

Explanation:

It is important to remember that proteins are composed of vast numbers of amino acids. Because these amino acids are tiny units, they cannot function as a catalyst on their own.

However, when they form a polymer, the protein enzyme will possess varying shapes, sizes, and both physical and chemical attributes differing from a single monomer.

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A bottle containing 1,665 g of sulfuric acid (H2SO4, 98.08 g/mol) was spilled in a laboratory. The emergency spill kit contained
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Answer: Yes, there is sufficient sodium carbonate available.

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In this scenario, according to the specified reaction:

Using stoichiometry, one can figure out the grams of sodium carbonate required to neutralize 1,665 g of sulfuric acid as outlined below:

H_2SO_4(aq) + Na_2CO_3(s) \rightarrow Na_2SO_4(aq) + CO_2(g) + H_2O(l)

Hence, the amount on hand is 2.0 kg, which leaves 0.2 kg as surplus, therefore:

A. Yes, there is sufficient sodium carbonate available.

1,665gH_2SO_4*\frac{1molH_2SO_4}{98.08gH_2SO_4}*\frac{1molNa_2CO_3}{1molH_2SO_4} *\frac{105.99gNa_2CO_3}{1molNa_2CO_3}*\frac{1kgNa_2CO_3}{1000gNa_2CO_3}\\m_{Na_2CO_3}=1.80gNa_2CO_3Best regards.

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18 days ago
What mass of water in grams will fill a tank 100 cm long, 50 cm wide, and 30 cm high? Knowing that the density of water is 1 g/m
castortr0y [3046]

The mass is 150,000 grams. Multiply 100 by 50 by 30 to determine the container's volume, which equals 150,000 cm^3. Since a milliliter is equivalent to one cubic centimeter, and given that the density of water is one gram per milliliter, it follows that the mass of water is 150,000 grams.

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15 days ago
A 25.0 g sample of an alloy was heated to 100.0 oC and dropped into a beaker containing 90 grams of water at 25.32 oC. The tempe
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Response:

The specific heat of the alloy C_{a} = 0.37 \frac{KJ}{Kg K}

Clarification:

Weight of the alloy m_{a} = 25 gm

Initial temperature T_{a} = 100°c = 373 K

Weight of the water m_{w} = 90 gm

Initial temperature of water T_{w} = 25.32 °c = 298.32 K

Final temperature T_{f} = 27.18 °c = 300.18 K

Using the energy balance equation,

Heat released by the alloy = Heat absorbed by the water

m_{a} C_{a} [[T_{a} - T_{f}] = m_{w} C_w (T_{f} -T_{w} )

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C_{a} = 0.37 \frac{KJ}{Kg K}

This gives us the specific heat of the alloy.

4 0
1 month ago
Freon-11, CCl3F has been commonly used in air conditioners. It has a molar mass of 137.35 g/mol and its enthalpy of vaporization
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Answer:

180.56 kilojoules of heat energy is extracted when 1.00 kg of freon-11 evaporates.

Explanation:

The molar mass of freon-11 is 137.35 g/mol

The enthalpy of vaporization for freon-11 is 24.8 kJ/mol at its normal boiling point of 24°C. Given that,\Delta H_{vap}=24.8 kJ/mol

Mass of freon-11 evaporated = 1.00 kg = 1000 g

Moles of freon-11 evaporated can be calculated as

\frac{1000 g}{137.35 g/mol}=7.2807 mol

The energy removed in the form of heat when 1.00 kg of freon-11 vaporizes is:

7.28067 mol\times \Delta H_{vap}=7.2807 mol\times 24.8 kJ/mol

=180.56 kJ

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9 days ago
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