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yaroslaw
2 months ago
10

A saturated solution of potassium iodide contains, in each 100 mL, 100 g of potassium iodide. The solubility of potassium iodide

is 1 g in 0.7 mL of water. Calculate the specific gravity of the saturated solution
Chemistry
1 answer:
KiRa [2.9K]2 months ago
5 0

Answer:

The specific gravity of the saturated solution is 2

Explanation:

Specific gravity represents the ratio of the density of a solution, in this case, a saturated potassium iodide (KI) solution, to the density of water. Assuming the density of water is 1:

Specific gravity  = Density

Density itself is defined as the mass divided by volume.

In 100mL of water, the mass of dissolve-able KI is:

100mL * (1g KI / 0.7mL) = 143g of KI

This indicates that all 100g of KI dissolves (Mass solute)

With 100mL of water corresponding to a mass of 100g (Mass solvent)

The overall mass of the solution computes to 100g + 100g = 200g

In a volume of 100mL, the solution's density is:

200g / 100mL = 2g/mL.

Specific gravity is a dimensionless quantity, thus the specific gravity of the saturated solution is 2

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What is the molar mass of citric acid (C6H8O7) and baking soda (NaHCO3)?
Alekssandra [3086]

Answer:

1. 192.0 g/mol.

2. 84.0 g/mol.

Explanation:

  • The molar mass refers to the weight of all atoms combined in a molecule measured in grams per mole.
  • To find a molecule's molar mass, we begin by looking up the atomic weights of the relevant elements from the periodic table. Next, we tally the atoms present and multiply that by their respective atomic weights.

1. Molar mass of citric acid (C₆H₈O₇):

Molar mass of C₆H₈O₇ = 6(atomic mass of C) + 8(atomic mass of H) + 7(atomic mass of O) = 6(12.0 g/mol) + 8(1.0 g/mol) + 7(16.0 g/mol) = 192.0 g/mol.

2. Molar mass of baking soda (NaHCO₃):

Molar mass of NaHCO₃ = (atomic mass of Na) + (atomic mass of H) + (atomic mass of C) + 3(atomic mass of O) = (23.0 g/mol) + (1.0 g/mol) + (12.0 g/mol) + 3(16.0 g/mol) = 84.0 g/mol.

4 0
3 months ago
The standard cell potential (E°cell) for the reaction below is +0.63 V. The cell potential for this reaction is ________ V when
castortr0y [3046]

Answer: The cell potential for the given reaction stands at 0.50 V

Explanation:

The provided cell reaction is:

Pb^{2+}(aq)+Zn(s)\rightarrow Zn^{2+}(aq)+Pb(s)

The half-reactions are:

Anode oxidation half reaction:  Zn\rightarrow Zn^{2+}+2e^-

Cathode reduction half reaction:  Pb^{2+}+2e^-\rightarrow Pb

Initially, we need to find the cell potential for this reaction.

Utilizing the Nernst equation:

E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}

where,

F = Faraday's constant = 96500 C

R = gas constant = 8.314 J/mol·K

T = room temperature = 25^oC=273+25=298K

n = electrons exchanged in oxidation-reduction = 2

E^o_{cell} = standard electrode potential for the cell = +0.63 V

E_{cell} = cell potential for the reaction =?

[Zn^{2+}] = concentration of Zn²⁺ = 3.5 M

[Pb^{2+}] = 2.0\times 10^{-4}M

Now substituting all known values into the equation, we arrive at:

E_{cell}=(+0.63)-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{3.5}{2.0\times 10^{-4}}

E_{cell}=0.50V

So, the resulting cell potential for this reaction is 0.50 V

5 0
2 months ago
Grignard reagents are air-and moisture-sensitive. List at least threereactants, solvents, and/or techniques that were utilized i
Anarel [2989]

Diethyl ether (DTH) and Tetrahydrofuran (THF).

Clarification:

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Therefore, solvents like anhydrous diethyl ether or tetrahydrofuran (THF), as well as poly(tetramethylene ether) glycol (PTMG), are used in experimental procedures to limit the exposure of Grignard reagents to air and moisture.
These solvents are chosen because the oxygen they contain stabilizes the magnesium reagent.
THF is a stable compound.
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lions [2927]

Answer:

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Explanation:

Within glycolysis, the phosphorylation of glucose to glucose-6-phosphate occurs first, facilitated by the hexokinase enzyme. This reaction is endergonic. This phosphorylation is a coupled reaction tied to ATP hydrolysis, where the free energy released by ATP hydrolysis drives glucose phosphorylation.

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A 3.81-gram sample of NaHCO3 was completely decomposed in an experiment. 2NaHCO3 → Na2CO3 + H2CO3 In this experiment, carbon dio
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Answer:

The yield percentage of H_2CO_3 is 24.44%

Explanation:

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