What is the freezing point in °C) of a 0.743 m
aqueous solution of KCI?
Enter your rounded answer with
3 decimal places.

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b. a positive cation and a negative anion

✏ Anions are negative in nature while cations are positive in nature. Together they come together by an attractive electrostatic force to form an ionic bond.

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Answer:

Beta emission

Explanation:

In beta emission, a neutron is converted into a proton thereby emitting an electron and a neutrino. A neutrino is a particle that serves to balance the spins.

When a nucleus undergoes beta emission, the mass number of the parent and daughter nuclei remain the same while the atomic number of the daughter nucleus is greater than that of its parent by one unit.

Hence, in beta emission, the daughter nucleus is found one pace to the right of the parent in the periodic table.

The heat of reaction for the reaction 2H₂O₂ → 2H₂O +O₂ is -196 KJ

From the question, we are to calculate the heat of reaction for the reaction

2H₂O₂ → 2H₂O +O₂               ∆H=?

Using Hess's law

Hess's Law of constant heat summation states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes

From the given equations,

2H₂ +O₂ → 2H₂O                   ∆H= -572 KJ ---------- (1)

2H₂O₂ → 2H₂ + 2O₂              ∆H= 376 KJ ----------- (2)

Adding equations (1) and (2), we get

2H₂O₂ → 2H₂O +O₂               ∆H= -572 KJ + 376 KJ

2H₂O₂ → 2H₂O +O₂               ∆H= -196 KJ

Hence, the heat of reaction for the reaction 2H₂O₂ → 2H₂O +O₂ is -196 KJ.

Learn more on Calculating heat of reaction using Hess's law here: https://brainly.com/question/26491956

Solution :

Compound                      Ksp

[tex]$PbF_2$[/tex]                               [tex]$3.3 \times 10^{-8}$[/tex]

[tex]$Ni(CN)_2$[/tex]                        [tex]$3 \times 10^{-23}$[/tex]

FeS                                [tex]$8 \times 10^{-19}$[/tex]

[tex]$CaSO_4$[/tex]                           [tex]$4.93 \times 10^{-5}$[/tex]

[tex]$Mg(OH)_2$[/tex]                      [tex]$5.61 \times 10^{-12}$[/tex]

Ksp of [tex]$Ni(CN)_2 << Ksp \text{ of}\ \ Mg(OH)_2$[/tex] and both compounds dissociate the same way. Hence [tex]$Mg(OH)_2$[/tex] is more soluble than [tex]$(B). \ Ni(CN)_2$[/tex]

[tex]$Mg(OH)_2$[/tex]  is less soluble than [tex]$(A). \ \ PbF_2 \ ()Ksp \ PbF_2 > Ksp \ \text{ of } \ Mg(OH)_2$[/tex]

It is not possible to determine CD - [tex]$FeS \text{ or} \ CaSO_4$[/tex]  is more or less soluble than [tex]$Mg(OH)_2$[/tex]  as though they have a different Ksp values their molecular dissociation is also different and they may have a close solubility values.

[tex]$Ni(OH)_2$[/tex]  can be directly compared with PbS, [tex]$AlPO_4, MnS$[/tex]

[tex]$\text{For } \ Ni(OH)_2$[/tex]

[tex]$AB_2(s) \rightarrow A^{2+} + 2B^{-}$[/tex]

[tex]$Ni(OH)_2(s) \rightarrow Ni^{2+} + 2OH^-$[/tex]

100

1-s s 2s

Ksp = [tex][A2+][B-]^2 = s \times (2s)^2 = 4s^3[/tex]

Hence they can be directly compared by Ksp values, smaller the Ksp, smaller the solubility.

For Silver Chloride

[tex]$AB(s) \rightarrow A^{x+}+B^{x-}$[/tex]

[tex]$AgCl(s) \rightarrow Ag^+ + Cl^-$[/tex]

1 0 0

1 - s s s

Ksp [tex]$=[A^{x+}][B^{x-}]=s \times s = s^2$[/tex]

Hence, they can be directly compared by Ksp values, smaller the Ksp, smaller the solubility.

Answer:

mass of chlorine = 10.4 grams

Explanation:

Mass of the solution = 5.20 × 10⁶ g

ppm chlorine by mass = 2.00

Recall that:

[tex]ppm = \dfrac{mass \ of \ chorine \ gas (solute)}{ mass of the solution }\times 10^6[/tex]

[tex]2 = \dfrac{mass \ of \ chorine \ gas (solute)}{ 5.20\times 10^6}\times 10^6[/tex]

mass of chlorine = 5.20 × 2.0

mass of chlorine = 10.4 grams

Answer: The pressure 12.2 psi is expressed in millimeters of mercury (mm Hg) as 630.9221 mm Hg.

Explanation:

Given: Pressure = 12.2 psi

According to the standard conversion of units, 14.7 psi = 1 atm = 760mmHg.

Therefore, 1 psi equal to how many mm Hg is calculated as follows.

[tex]14.7 psi = 760 mm Hg\\1 psi = \frac{760}{14.7} mm Hg\\= 51.7149 mm Hg[/tex]

Hence, 12.2 psi will be converted into mm Hg as follows.

[tex]1 psi = 51.7149 mm Hg\\12.2 psi = 12.2 psi \times \frac{51.7149 mm Hg}{1 psi}\\= 630.9221 mm Hg[/tex]

Thus, we can conclude that the pressure 12.2 psi is expressed in millimeters of mercury (mm Hg) as 630.9221 mm Hg.

Answer:

%m/m = 0.9975%

Xₐ = 0.0039

Explanation:

In order to do this, we need various data. First to all, we need tje molecular mass of the ethanol. this can be obtained in handbooks, or simply taking the atomic weights of carbon (12 g/mol), Hydrogen (1 g/mol) and oxygen (16 g/mol) and summing those values:

MM C₂H₆O = (2*12) + (6*1) + (16*1) = 46 g/mol

Now, there is an expression that is commonly used to determine the molarity of a solution given the mass percent and density, and assuming of course, 1 liter of solution:

M = d * %m/m * 1000 / MM * 100     (1)

From here, we can solve for %m/m:

%m/m = M * MM * 100 / d * 1000

As the problem is not saying the volume of solution, we can easily assume we have 1 liter of solution. Therefore, the %m/m replacing the given data is:

%m/m = 0.216 *46 * 100 / 0.996 * 1000

To get the mole fraction, we first need to get the volume of solvent. From the density, we can get the mass of solution:

m = V * d

m = 1000 * 0.996 = 996 g of solution.

The mass of solute is:

m = 0.216 mol/L * 46 g/mol

m = 9.936 g/L, or simply we have 9.936 g of ethanol in 1 L of solution.

The mass of solvent:

solvent = 996 - 9.936 = 986.064 g

The molecular mass of water, so we can get the moles is 18 g/mol so:

moles water = 986.064 / 18 = 54.78

Finally the mole fraction:

Xₐ = 0.216 / (0.216 + 54.78)

Hope this helps

Answer:

2.13 M

Explanation:

To solve this problem we need to keep in mind the definition of molarity:

As the problem gives us both the number of moles and the volume of solution, we can proceed to calculate the molarity:

The answer is 2.13 M.

Answer:

The usage is explained below

Explanation:

Amines are identified from the use of the hinsberg reaction whereby the amine is mixed well with Hinsberg reagent and done in the presence of an aqueous alkali. In this reaction, the amine will attack the electrophile which will result in the chloride being displaced and the amides being generated.

The uses of identification of amines in daily life are;

- In agriculture, it serves as a good source of making herbicides as well as acting as emulsifiers.

- In medicine, it is used in the manufacture of some popular pain killers like demerol and morphine.

- In medicine, it is used in the manufacture of an anesthetic drug known as novocaine.

- In chemical industries, amines are used as lubricating oils and also as corrosion inhibitors in boilers.

Answer:

108 g

Explanation:

First we convert 6.00 moles of H₂ into moles of H₂O, using the stoichiometric coefficients of the reaction:

Then we can convert 6.00 moles of H₂O into grams, using the molar mass of water:

The answer is 108 grams of water.

Answer:

4NH3 + 3O2 → 2N2 + 6H2O

a. 2 moles

b. 4 moles

c. 3 moles

Explanation:

a. 4 mol NH3

4NH3 + 3O2 → 2N2 + 6H2O

  4      :     3    :    2     :    6

  4                                        (mol)

⇒[tex]n_{N2}[/tex] = 4 × 2 ÷ 4 = 2

b. 4 mol N2

c. 4.5 mol O2

4NH3 + 3O2 → 2N2 + 6H2O

4      :     3    :    2     :    6

           4.5                          (mol)

⇒[tex]n_{N2}[/tex] = 4.5 × 2 ÷ 3 = 3

[tex]\text{N}_{2}\text{O}_{3}=\text{nitrogen trioxide}\\\text{N}_{2}=\text{nitrogen}\\\text{O}_{2}=\text{oxygen}[/tex]

Answer: The element B will have ONE unpaired electron in the p orbital.

Explanation:

The electronic configuration of each given element is as follows.

Atomic number of calcium (Ca) is 20.

Ca: [tex]1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2}[/tex]

Atomic number of nitrogen (N) is 7.

N: [tex]1s^{2} 2s^{2} 2p^{3}[/tex]

Atomic number of boron (B) is 5.

B: [tex]1s^{2} 2s^{2} 2p^{1}[/tex]

Atomic number of argon (Ar) is 18.

Ar: [tex]1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6}[/tex]

Atomic number of bromine (Br) is 35.

Br: [tex][Ar] 4s^{2} 3d^{10} 4p^{5}[/tex]

Therefore, boron is the only element that have one unpaired electron in the p-orbital.

Thus, we can conclude that element B will have ONE unpaired electron in the p orbital.

Answer:  Molality = concentration = 0,33 mol/l

Explanation: Concentration c = n/V   and  n = m/M

M = 55,44 g/mol  and m = 1.9 g , V = 0.100 l

c = m/(MV)

Answer:

A. Mass

Explanation:

If it takes up space it has to have mass. Not everything that is matter has a color, certain length or width.

Answer:

Explanation:

Yes , it is a redox reaction .

3O₂ + 4Fe → 2Fe₂O₃.

Increase of oxidation number shows oxidation and decrease of oxidation number shows reduction .

Neutral atoms have zero oxidation number .

O₂ and Fe are neutral atoms so their oxidation number is zero.

In Fe₂O₃ , charge on each Fe atom is + 3 . So its oxidation number is + 3 . Similarly charge on each of O in this compound is - 2 . So its oxidation number is -2 .

Oxidation number of Fe increases from 0 to + 3 . Hence it is oxidized .

Oxidation number of O decreases from 0 to  -2  . Hence it is reduced  .

Fe is oxidized by O ( oxygen ) so oxygen is the oxidizing agent .

O is reduced  by Fe ( iron  ) so iron  is the reducing agent .

Answer: Element is P, phosphor

Explanation:  Phosphor has oxidation number -III and it has 15 protons. So it is possible to have 16 neutrons. Other elements having oxidation number -III are N and As which can not have an isotope with 16 neutrons.

Answer:

91.41 g of LiClO₃.

Explanation:

We'll begin by calculating the number of mole of O₂ that occupied 33.8 L. This can be obtained as follow:

22.4 L = 1 mole of O₂

Therefore,

33.8 L = 33.8 L × 1 mole / 22.4 L

33.8 L = 1.51 mole of O₂

Next, the balanced equation for the reaction.

2LiCl + 3O₂ —> 2LiClO₃

From the balanced equation above,

3 moles of O₂ reacted to produce 2 moles of LiClO₃.

Therefore, 1.51 mole of O₂ will react to produce = (1.51 × 2)/3 = 1.01 mole of LiClO₃.

Finally, we shall determine the mass of 1.01 mole of LiClO₃. This can be obtained as follow:

Mole of LiClO₃ = 1.01 mole

Molar mass of LiClO₃ = 7 + 35.5 + (3×16)

= 7 + 35.5 + 48

= 90.5 g/mol

Mass of LiClO₃ =?

Mass = mole × molar mass

Mass of LiClO₃ = 1.01 × 90.5

Mass of LiClO₃ = 91.41 g

Thus, 91.41 g of LiClO₃ were obtained from the reaction.

Answer:

it is already balanced

c5h12+8O2-->5CO2 +6h2o

Explanation:

Answer:

3.3 * 10^-12

Explanation:

The balanced equation of the reaction is;

2H+(aq) + Cu(s) ---------> H2(g) + Cu2+(aq)

Hence two electrons were transferred so n=2

E°cell = E°cathode - E°anode

E°cell = 0 V - 0.34 V

E°cell = - 0.34 V

Then;

E°cell = 0.0592/n log K

Substituting values;

- 0.34 = 0.0592/2 log K

- 0.34/0.0296 = log K

-11.486 = log K

K = Antilog (-11.486)

K = 3.3 * 10^-12

Answer:

14.4g

Explanation: