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For the experiments from 1-3 with the same temperature change, what other parameters are the same at the initial reaction conditions?a. mmol HCl total mL solution
b. mmol NaOH Initial concentration of NaOH
c. mL H2O Initial concentration of HCl

Answers

Answer 1
Answer:

Answer:  

The following parameters:  

  1. total mL solution
  2. mmol NaOH
  3. initial concentration of NaOH

Explanation:  

The options have been incorrectly arranged in the question statement given here. The three conditions mentioned above are what would be observed to remain the same when this experiment is carried out.  

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Which is an example of plasmas in nature? solar cells plasma balls auroras clouds

Answers

The best example of natural plasma is the ionosphere of the atmosphere including solar corona and lightning including auroras cloud.

What is natural plasma?

the heated or hot  matter that is so hot that the electrons are thrown away from the atoms and forms the ionized gas. Comprises over 99% of the visible universe.

Lightning strikes create plasma by doing a stricking of electricity. Mostly the Sun, and some stars, are in a plasma state. Certain regions of Earth's atmosphere contain some plasma created primarily by ultraviolet radiation from the Sun.

One reason plasma is not so common is because this needs high temperatures required to keep gas in the plasma state. At average temperatures on Earth, there just isn't enough energy for atoms to remain ionized.

However, at thousands to millions of degrees Kelvin, these energies are available, and plasmas dominate

Therefore, aurora clouds  are natural plasma.

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Answer:

aurorus

Explanation:

One gram of a compound requires the following quantities of solvent to dissolve: 47 mL of water, 8.1 mL of chloroform, 370 mL of diethyl ether, or 86 mL of benzene. Calculate the solubility of the compound in these four solvents (as g/100 mL). Estimate the partition coefficient of the compound between chloroform and water, ethyl ether and water, and benzene and water. Which solvent would you choose to extract the compound from an aqueous solution

Answers

Answer:

Chloroform.

Explanation:

Given,

Solvent requires 1g of compound per 100 mL

For water,

= 1g/47ml

= 2.1

For Chloroform,

= 1 g/8.1 mL

= 12.345679

For Diethyl ether,

= 1 g/370 mL

= 0.27

For Benzene,

=  1 g/86 mL

= 1.2

Partition coefficients:

Water = -

chloroform = 5.9

Diethyl = .13

Benzene  = .57

The solvent chloroform would be chosen for drawing out the compound out of an aqueous solution as it has the maximum solubility.

Final answer:

The solubility of a compound in different solvents will determine its concentration in each solvent. The partition coefficient represents the relative solubility of a compound in two immiscible solvents. Chloroform would be the best choice to extract the compound from an aqueous solution.

Explanation:

The solubility of a compound is usually expressed as grams of solute per 100 mL of solvent. To calculate the solubility, you can use the following formula:

Solubility (g/100 mL) = (mass of solute / volume of solvent) * 100

Using this formula, the solubility of the compound in water is 47 g/100 mL, in chloroform is 97.53 g/100 mL, in diethyl ether is 2.70 g/100 mL, and in benzene is 1.16 g/100 mL.

The partition coefficient is a measure of the compound's solubility in two immiscible solvents. To calculate it, divide the solubility of the compound in one solvent by its solubility in another solvent. For example, the partition coefficient between chloroform and water would be:

Partition coefficient = Solubility in chloroform / Solubility in water = 97.53 g/100 mL / 47 g/100 mL = 2.07

The larger the partition coefficient, the more soluble the compound is in the first solvent compared to the second solvent. Based on the partition coefficients, chloroform would be the best choice to extract the compound from an aqueous solution.

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Write balanced net ionic equations and the corresponding equilibrium equations for the stepwise dissociation of the triprotic acid H3PO4.

Answers

Answer:

H_3PO_4(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+H_2PO_4^(-)(aq); \ Ka_1

H_2PO_4^(-)(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+HPO_4^(-2)(aq); \ Ka_2

HPO_4^(-2)(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+PO_4^(-3)(aq); \ Ka_3

Explanation:

Hello!

In this case, since the phosphoric acid is a triprotic acid, we infer it has three stepwise ionization reactions in which one hydrogen ion is released at each step, considering they are undergone due to the presence of water, thus, we proceed as follows:

H_3PO_4(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+H_2PO_4^(-)(aq); \ Ka_1

H_2PO_4^(-)(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+HPO_4^(-2)(aq); \ Ka_2

HPO_4^(-2)(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+PO_4^(-3)(aq); \ Ka_3

Moreover, notice each step has a different acid dissociation constant, which are quantified in the following order:

Ka1 > Ka2 > Ka3

Best regards!

Help me assiment will close after an hour ​

Answers

Answer :

(a) The molecular equation will be,

CaCO_3(aq)+2HCl(aq)\rightarrow CaCl_2(aq)+H_2O(l)+CO_2(g)

(b) The complete ionic equation in separated aqueous solution will be,

Ca^(2+)(aq)+CO_3^(2-)(aq)+2H^(+)(aq)+2Cl^(-)(aq)\rightarrow Ca^(2+)(aq)+2Cl^-(aq)+H_2O(l)+CO_2(g)

(c) The net ionic equation will be,

CO_3^(2-)(aq)+2H^(+)(aq)\rightarrow H_2O(l)+CO_2(g)

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

(a) The molecular equation will be,

CaCO_3(aq)+2HCl(aq)\rightarrow CaCl_2(aq)+H_2O(l)+CO_2(g)

(b) The complete ionic equation in separated aqueous solution will be,

Ca^(2+)(aq)+CO_3^(2-)(aq)+2H^(+)(aq)+2Cl^(-)(aq)\rightarrow Ca^(2+)(aq)+2Cl^-(aq)+H_2O(l)+CO_2(g)

In this equation, Ca^(2+)\text{ and }Cl^- are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

(c) The net ionic equation will be,

CO_3^(2-)(aq)+2H^(+)(aq)\rightarrow H_2O(l)+CO_2(g)

Round each answer to the number of significant figures in the parentheses.

Answers

Answer:

87.100. 4.4000. 0.012000. 9000. 6,040,850. 630.00. 100.1000. 0.00024000

calculate the mol fraction of ethanol and water in a sample of rectified spirit which contains 95% of ethanol by mass​

Answers

Answer:

We are given that there is 95% ethanol by mass in rectified spirit

so, we can say that in a 100g sample, we have 95 grams of ethanol and 5 grams of water

we will find the number of moles of ethanol and water in 100g solution of rectified spirit and use that to calculate the mole fraction

Moles of Ethanol:

Molar mass of ethanol = 46 grams / mol

Number of moles = Given mass / molar mass

Number of moles = 95 / 46

Moles of Ethanol = 2 moles (approx)

Moles of Water:

Molar mass of water = 18 grams per mol

Number of moles = Given mass / molar mass

Moles of water = 5 / 18

Moles of water  =  0.28 moles   (approx)

Mole Fractions:

Mole fraction of a specific compound is the number of moles of that compound divided by the total number of moles in the solution

Mole fraction of Ethanol:

Moles of ethanol / (moles of ethanol + moles of water)

2 / (2 + 0.28)

2 / (2.28) = 0.9 (approx)

Mole fraction of Water:

Moles of water / (Moles of ethanol + moles of water)

0.28 / (2 + 0.28)

0.28 / (2.28) = 0.1 (approx)
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