In the PhET simulation window, click the Macro menu in the top left corner of the screen. This view gives a view of the beaker at a macroscopic level (as your naked eye would see it). The Micro menu shows what happens to sugars and salts at the molecular level when they dissolve in water (note that you can use the arrows to switch to other type of solutes). Use both the Macro and Micro menus in the PhET simulation to help complete the following statements regarding solutions.

Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer.

Hints

Help

Reset

ions

solute

negatively

opposite

Partical Charges

positively

NaCl

sugar

CaCl2

1. Pure water contains only water molecules that interact strongly with each other due to their ___________, which are graphically depicted as δ+ and δ−.

2. Solutions are formed when a _______ like a salt or sugar becomes homogeneously distributed in a solvent like water, and this distribution can be viewed in the Micro view.

3. When salts dissolve, they separate into individua________that strongly interact with the water molecules.

4. Binary salts are made up of two elements at varying ratios, where one element is a charged cation, and the other is a_________ charged anion.

5. In the Micro view, each shake of the container releases 6 molecules of its respective solute, but 6 molecules of the salt________actually produce more ions in solution than 6 molecules of the salt__________.

6. Not all soluble molecules are salts, e.g., a covalent species like__________readily dissolves in water without forming ions.

7. The reason they dissolve is because their partially charged atoms are able to associate with the partially charged atoms of water molecules, and these attractive forces occur as long as they are between atoms with_________charges.

Answer:A and B

Explanation:a stats final stages of life depend on its mass and stars die when they run out of fuel

Answer:

[OH-] = 10^-4 M

Explanation:

pOH= 14- pH= 14-10=4

[OH-]= antilog (-4)= 10^-4M

Answer:

_5_ AsO2−(aq) + 3 Mn2+(aq) + _2_ H2O(l) → _5_ As(s) + _3_ MnO4−(aq) + _4_ H+(aq)

Explanation:

Step 1:

The unbalanced equation:

AsO2−(aq) + 3 Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)

Step 2:

Balancing the equation.

AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)

The above equation can be balanced as follow:

There are 3 atoms of Mn on the left side of the equation and 1 atom on the right side. It can be balance by putting 3 in front of MnO4− as shown below:

AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)

There are 12 atoms of O on the right side and a total of 3 atoms on the left side. It can be balance by putting 5 in front of AsO2− and 2 in front of H2O as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)

There are 4 atoms of H on the left side and 1 atom on the right side. It can be balance by putting 4 in front of H+ as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + 4H+(aq)

There are 5 atoms of As on the left side and 1 atom on the right side. It can be balance by putting 5 in front of As as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → 5As(s) + 3MnO4−(aq) + 4H+(aq)

Now the equation is balanced

The balanced redox reaction is

5AsO⁻₂(aq) + 3Mn²⁺(aq) + 2H₂O(l) → 5As(s) + 3MnO₄−(aq) + 4H+(aq)

The unbalanced redox reaction is :

AsO₂−(aq) + 3 Mn²⁺(aq) + H₂O(l) → As(s) + MnO₄−(aq) + H⁺(aq)

The above equation can be balanced  by ensuring the atom of the elements

on the left hand side is equal to those on the right hand side.

We have 3 atoms of Mn on the left side of the equation and 1 atom on the

right hand side.This will be balanced by putting 3 in front of MnO₄− as shown

below:

AsO₂−(aq) + 3 Mn²⁺(aq) + H₂O(l) → As(s) + MnO₄−(aq) + H⁺(aq)

We have 3 atoms of O on the left hand side and 12 atoms of O on the right

hand side. This is  balanced by putting 5 in front of AsO₂− and 2 in front of

H₂O as shown below:

5AsO₂−(aq) + 3 Mn²⁺(aq) + 2H₂O(l) → As(s) + MnO₄−(aq) + H⁺(aq)

We have 4 atoms of H on the left hand side and 1 atom of H on the right

hand side.We can balance by putting 4 in front of H⁺ as shown below:

5AsO₂−(aq) + 3 Mn²⁺(aq) + 2H₂O(l) → As(s) + MnO₄−(aq) + 4H⁺(aq)

We have 5 atoms of As on the left hand side and 1 atom of As on the right

hand side. We can balance by putting 5 in front of As as shown below:

5AsO₂−(aq) + 3 Mn²⁺(aq) + 2H₂O(l) → 5As(s) + MnO₄−(aq) + 4H⁺(aq)

The equation is therefore now balanced as the number of atoms of the

element on the left hand side are equal with those on the right hand side.

Read more on https://brainly.com/question/25544428

Answer:

They are probably archeabacteria because the colony lives in an extreme environment.

Explanation:

Archeabacteria are singled cell microorganisms that live in extreme environment. They are found in hot springs, salt lakes, oceans, soils and marshlands. They posses different shapes like rods, spheres, spiral and plates. Thermophiles, halophiles, and methanogens are the three types of archeabacteria.

Eubacteria are microorganisms that are found in most of the earth's habitats like soil, water, etc. They are have different shapes like cocci, bacilli, filaments, vibro,etc. They do not live in extreme environment unlike the archeabacteria. This is the major difference between the archeabacteria and eubacteria.

Both archeabacteria and eubacteria

are prokaryotes. Archeabacteria can both be autotrophic or heterotrophic and can live in places without oxygen. Some eubacteria are autotrophs and some are heterotrophs.

Answer:

They are probably archaebacteria because the colony lives in an extreme environment.

Explanation:

Answer: The monomer of nylon-6 is caprolactam.

Explanation:

From the question, we are asked to draw the cyclic monomer from the structure of Nylon-6.

Two repeating monomer units can be seen in the polymeric structure given in the problem.

From this we can see and conclude that The monomer of nylon-6 is caprolactam.

attached is a diagrammatic representation of the Structure (Nylon-6 monomer).

cheers i hope this helps!!!!

Final answer:

Nylon-6 is synthesized from the cyclic monomer epsilon-caprolactam through a process called ring-opening polymerization, unlike other nylons that are produced by condensation polymerization. Epsilon-caprolactam features a 6-carbon ring with an amide linkage, making it unique.

Explanation:

Nylon-6 is a unique type of polymer that is derived from a cyclic monomer known as epsilon-caprolactam. Unlike the Nylon 6,6 polymer, which is made from condensation polymerization of hexanedioic acid and 1,6-diaminohexane, Nylon-6 utilizes a process called ring-opening polymerization. This process involves breaking open the ring structure of the epsilon-caprolactam to form linear chains, which then link together head-to-tail to form the polymer. This is a key difference as it does not release a small molecule (like water or hydrogen chloride) during the polymerization process, making it a growth polymerization.

Epsilon-caprolactam, the monomer for Nylon-6, is a 6-carbon cyclic amide. To deduce and draw its cyclic monomer, one should recognize that the repeating unit in Nylon-6 consists of 6 carbons in a chain with an amide linkage connecting the ends. Thus, depicting epsilon-caprolactam involves drawing a ring structure containing 5 methylene (-CH2-) groups and one carbonyl group (=O) adjacent to an NH group, forming a lactam.

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

Final answer:

The question involves ranking solutions of weak acids and bases by their pH. After assessing the ionization and hydrolysis of the compounds, the order from lowest to highest pH is 0.1 M NH4Br, 0.1 M NaCN, 0.1 M KF, and 0.1 M KBr.

Explanation:

To rank the solutions in order of increasing pH, we must consider the ionization of each compound in water and the corresponding Ka or Kb values. A lower Ka or a higher Kb value usually results in a higher pH.

0.1 M KF: KF will form from the dissolution of a weak acid (HF) and a strong base. The resulting F- ions will hydrolyze to form OH- ions, creating a basic solution with a pH higher than 7 but not as high as that of a strong base.

Therefore, the order from lowest to highest pH should be b, a, c, d.

Answer:

13.279

Explanation:

24g/56.1056g/mol= 0.42 (mol KOH)

0.42 mol KOH/ 2.25 L= 0.19 M KOH

-log(0.19)= pOH= 0.72

pH=14-pOH

pH=14-0.72=13.279

sorry it's not the best explanation but since you're on a time crunch i hope this helps

A chemist prepared a solution of KOH by completely dissolving 24.0 grams of solid KOH in 2.25 liters of water at room temperature. The pH of the solution that the chemist prepared, to the nearest thousandth is 13.279.

pH is a numerical indicator of how acidic or basic aqueous or some other liquid solutions are. The phrase, which is frequently used during chemistry, biology, or agronomy, converts hydrogen ion concentrations.

The hydrogen ion concentration in pure water, which has a pH of 7, is 107 gram-equivalents per liter, making it neutral.

24g/56.1056g/mol= 0.42 (mol KOH)

0.42 mol KOH/ 2.25 L= 0.19 M KOH

-log(0.19)= pOH= 0.72

pH=14-pOH

pH=14-0.72

   =13.279

To know more about pH, here:

https://brainly.com/question/27945512

#SPJ2

Answer:

the answer 37

Explanation:

Answer:

False

Explanation:

Metals are made up of atoms bonded to each other by metallic bonds. Such bonds have a sea of delocalized electrons, which are free to flow. Electricity requires an unobstructed flow of electrons in order to be able to be conducted. Because all metals provide this inherently, then all metals can conduct electricity, making the statement false.

Hope this helps!

In a neutral atom, there are equal numbers of electrons and protons, which gives the atom its neutral charge. Neutrons do not factor into the atom's charge, and allotropes are different forms of an element, not part of its atomic structure.

In a neutral atom, there are equal numbers of electrons and protons. The number of protons in an atom's nucleus determines the atomic number, and the atom's charge is neutral because the positive charge of the protons is balanced by the negative charge of the electrons. Neutrons, on the other hand, do not carry an electronic charge and are not necessarily equal in number to the protons or electrons. Concerning allotropes, they are different forms of the same element and it's not directly related to this question about atomic structure.

https://brainly.com/question/7328982

#SPJ6

In neutral atoms, there are equal numbers of electrons and protons. Each element, when neutral, has a characteristic number of electrons that matches its atomic number (the number of protons). Neutrons are electrically neutral and their number does not affect the atom's charge.

In all neutral atoms, there are equal numbers of electrons and protons. This is because the number of protons, which carry a positive charge, and the number of electrons, which carry a negative charge, balance each other in a neutral atom, resulting in an overall charge of zero.

Each element, when electrically neutral, has a characteristic number of electrons equal to its atomic number, which is the number of protons it contains. For example, in a carbon-12 atom (which is neutral), there are six protons and six electrons, yielding a net charge of zero.

It's important to note that while neutrons also exist within an atom, they are electrically neutral. Therefore, the number of neutrons does not affect the atom's charge and doesn't necessarily match the number of protons or electrons.

https://brainly.com/question/3288162

#SPJ6

Answer:

The answer is 4.8659

Explanation:

Given:

[C₆H₅COOH]=0.11M

[C₆H₅COO]=2*0.2=0.4M

Ka=6.3x10⁻⁵

First, calculate the pKa:

[tex]pKa=-logKa=-log(6.3x10^{-5} )=4.2007[/tex]

The pH is:

[tex]pH=pKa+log\frac{C6H5COO]}{[C6H5COOH]} =4.2007+log\frac{0.4}{0.11} =4.7614[/tex]

Like the volume is 5L, the volume of C₆H₅COO is x, then, the volume of C₆H₅COOH is 5-x

[tex]4.7614=4.2007+log\frac{0.4x}{0.11*(5-x)}[/tex]

[tex]0.5607=log\frac{0.4x}{0.11*(5-x)}[/tex]

Solving for x:

x=2.49L=2490mL of C₆H₅COO

2510mL of C₆H₅COOH

The milimoles of C₆H₅COOH and C₆H₅COO is:

nC₆H₅COOH=(0.11*2510)-50=226.1mmol

nC₆H₅COO=(0.4*2490)+50=1046mmol

The pH is:

[tex]pH=4.2007+log\frac{1046}{226.1} =4.8659[/tex]

Answer:

4.86

Explanation:

The pH of a solution is a measure of the molar concentration of hydrogen ions in the solution and as such is a measure of the acidity or basicity of the solution.

Please kindly check attachment for the step by step solution of the given problem.

Answer:

0.2 mol/kJ

Explanation:

Methane is stored at a temperature of 111.4K, [tex]T_c[/tex]. The heat source to vapourization of methane is ambient air which is at 300 K. [tex]T_H[/tex]

Estimate the vaporization rate at the efficiency of heat engine 60% of its carnot value.

Calculate the vaporization rate from the given data by relation shown below:

[tex]Vaporization rate = \frac{Q_c}{[\frac{\delta H_n^{lv}}{W}]} ..........(1)[/tex]

here,

[tex]Q_c[/tex] is the heat at temperature [tex]T_c, \delta H_n^{lv}[/tex] is the phase transition enthalpy of methane and W is the work

Calculate [tex]Q_c[/tex] from the equation shown below:

[tex]Q_c = Q_n (1 - \eta_{HE}) .............(2)[/tex]

where Q_n is the  heat at temperature of [tex]T_n[/tex] and [tex]\eta_{HE}[/tex] is the efficiency of heat engine

calculate [tex]Q_H[/tex] from the relation shown below:

[tex]Q_H = \frac{W}{\eta_{HE}} ..........(3)[/tex]

calculate the heat engine efficiency from the given carnot engine efficiency as shown below

[tex]\eta_{HE} = 0.6 \times \eta_{carnot} ............(4)[/tex]

here, [tex]\eta_{carnot}[/tex] is the carnot engine efficiency

[tex]\eta{carnot} = 1 - \frac{T_c}{T_H}[/tex]

substituting the values of temperature, we have

[tex]= 1 - \frac{111.4K}{300K}\\= 0.629\\[/tex]

substitute values of [tex]\eta_{carnot}[/tex] in equation 4, we get

[tex]\eta_{HE} =0.6 \times \eta_{carnot}\\ = 0.6 \times 0.629\\ = 0.3774\\\\[/tex]

check the attached file for additional solution

Answer:

49.544 g.

Explanation:

The balanced equation of reaction is given below;

Hg(ClO3)2 (aq) + Na2S (aq) --------> 2 NaClO3 (aq) + HgS (s).

So, the parameters given in the question are; Mass of Hg(ClO3) = 118.08 and the mass of Na2S = 16.642 g.

Therefore, the first thing we are going to be looking at is the reactant which is the limiting reagent from the number of moles

(1). For Hg(ClO3), the molar mass = 367.5 g/mol. Therefore, the number of moles, n = mass/ molar mass.

Number of moles = 118.08/ 367.5.

Number of moles = 0.321 moles.

(2). For Na2S, the molar mass = 78.05 g /mol.

The number of moles = 16.643 / 78.05.

The number of moles= 0.213.

Therefore, the limiting reagent = Na2S.

This means that the excess reagent is Hg(ClO3).

==> In the balanced equation of reaction above, the solid precipitate = HgS.

Hence, the mass of HgS formed = 0.213 × 232.6 g/mol. = 49.544 g.

This problem involves calculating the mass of a precipitate from a doubly replacement reaction. Sodium sulfide is the limiting reactant and by using stoichiometry, the moles of sodium sulfide will amount to the same moles of HgS, the precipitate. Multiplying the moles of HgS by its molar mass gives the mass of the precipitate.

This question requires a two-step process that starts with writing balanced chemical equations for the reaction and then using stoichiometry to calculate the mass of the precipitate. The balanced reaction is 2Na2S + Hg2(ClO3)2 -> 2NaClO3 + Hg2S2. Using stoichiometry and details given in the question, you can calculate the molar mass of the substances and find the limiting reactant.

Mercury(II) chlorate (Hg2(ClO3)2) has a molar mass of about 511.63 g/mol. Sodium sulfide (Na2S) has a molar mass of about 78.04 g/mol. Using stoichiometry, it's clear that sodium sulfide is the limiting reactant.

Using the moles of sodium sulfide calculated, you find that the reaction will produce the same number of moles of HgS as a precipitate. By multiplying this number by the molar mass of HgS (232.66 g/mol), you obtain the mass of the precipitate that will be formed. Deduce the exact values yourself for a better understanding of this chemical reaction involving stoichiometry and solubility product.

https://brainly.com/question/30218216

#SPJ3

Answer:

Entropy change is zero

Explanation:

∆S=S2-S1=0

We're S is entropy

Answer:

c. 3.00 M HCl

Explanation:

From dilution formula

C1V1 = C2V2

C1=?, V1= 10.0ml, C= 1.5, V2= 20.0ml

Substitute and Simplify

C1×10= 1.5×20

C1= 3.00M

Answer: The atomic number and mass number of the new element formed is 89 and 228 respectively.

Explanation:

Beta decay is defined as the process in which beta particle is emitted. In this process, a neutron gets converted to a proton and an electron.  The released beta particle is also known as electron.

In this process, the atomic number of the daughter nuclei gets increased by a factor of 1 but the mass number remains the same.

[tex]_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta[/tex]

The chemical equation for the beta decay of Ra-228 follows:

[tex]_{88}^{228}\tyextrm{Ra}\rightarrow _{89}^{228}\textrm{Ac}+_{-1}^0\beta[/tex]

Hence, the atomic number and mass number of the new element formed is 89 and 228 respectively.

Answer:

The weight of aspirin lost is  [tex]W =0.03267(V_A +V_B) N[/tex]

The percent yield is lowered by [tex]A = \frac{Amount \ of \ aspirin \ obtained - Lost Amount }{Expected\ Amount\ of \ Aspirin } *\frac{100}{1}[/tex]  

Explanation:

From the question we are told that

       Since the solubility of aspirin in water is 1 g per 300 mL it implies that after crystallization the solution would contain 1 g for every 300mL of water  at 25°

     Let assume that the volume of the solution is [tex]V_A[/tex]

          The the aspirin lost after filtration would be

                       [tex]= \frac{1}{300} * V_A[/tex]

                       [tex]=\frac{V_A}{300}g[/tex]

Let assume that you used water of volume [tex]V_B[/tex] to wash the crystallized aspirin then the lost during washing would be

                   [tex]= \frac{1}{300} * V_B[/tex]

                   [tex]= \frac{V_B}{300}[/tex]

So the total loss is

                 [tex]= \frac{V_A}{300} + \frac{V_B}{300}[/tex]

               [tex]\frac{V_A +V_B}{300}g[/tex]

So the weight of aspirin lost denoted by W is

                [tex]W = \frac{V_A +V_B}{300} *9.8[/tex]

                     [tex]W =0.03267(V_A +V_B) N[/tex]

Let denote How much was your percent yield lowered by this loss by A

    So

             [tex]A = \frac{Amount \ of \ aspirin \ obtained - Lost Amount }{Expected\ Amount\ of \ Aspirin } *\frac{100}{1}[/tex]  

Answer: a) [tex]Cl_2[/tex] is the limiting reagent

b)  0.27 g of [tex]AlCl_3[/tex] will be produced.

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} Al=\frac{0.80g}{27g/mol}=0.030moles[/tex]

[tex]\text{Moles of} Cl_2=\frac{0.23g}{71g/mol}=0.003moles[/tex]

[tex]2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)[/tex]

According to stoichiometry :

3 moles of [tex]Cl_2[/tex] require = 2 moles of [tex]Al[/tex]

Thus 0.003 moles of [tex]Cl_2[/tex] will require=[tex]\frac{2}{3}\times 0.003=0.002moles[/tex]  of [tex]Al[/tex]

Thus [tex]Cl_2[/tex] is the limiting reagent as it limits the formation of product and [tex]Al[/tex] is the excess reagent.

b) As 3 moles of [tex]Cl_2[/tex] give = 2 moles of [tex]AlCl_3[/tex]

Thus 0.003 moles of [tex]Cl_2[/tex] give =[tex]\frac{2}{3}\times 0.003=0.002moles[/tex]  of [tex]AlCl_3[/tex]

Mass of [tex]AlCl_3=moles\times {\text {Molar mass}}=0.002moles\times 133g/mol=0.27g[/tex]

Thus 0.27 g of [tex]AlCl_3[/tex] will be produced.

Answer:

P = 2.25 atm

Explanation:

Given data:

Mass of dry ice = 10 g

Volume of container = 2.5 L

Temperature = 25°C (25+273=298 K)

Pressure inside container = ?

Solution:

First of all we will calculate the number of moles of dry ice.

Number of moles = mass / molar mass

Number of moles = 10 g/ 44 g/mol

Number of moles = 0.23 mol

Now we will determine the pressure.

PV = nRT

P = nRT/V

P = 0.23 mol × 0.0821 atm.L/mol.K  × 298 K / 2.5 L

P = 5.63 atm  / 2.5

P = 2.25 atm

Answer:

The equilibrium constant for the reversible reaction = 0.0164

Explanation:

At equilibrium the rate of forward reaction is equal to the rate of backwards reaction.

The reaction is given as

A ⇌ B

Rate of forward reaction is first order in [A] and the rate of backward reaction is also first order in [B]

The rate of forward reaction = |r₁| = k₁ [A]

The rate of backward reaction = |r₂| = k₂ [B]

(Taking only the magnitudes)

where k₁ and k₂ are the forward and backward rate constants respectively.

k₁ = 0.010 s⁻¹

k₂ = 0.0610 s⁻¹

|r₁| = 0.010 [A]

|r₂| = 0.016 [B]

At equilibrium, the rate of forward and backward reactions are equal

|r₁| = |r₂|

k₁ [A] = k₂ [B] (eqn 1)

Note that equilibrium constant, K, is given as

K = [B]/[A]

So, from eqn 1

k₁ [A] = k₂ [B]

[B]/[A] = (k₁/k₂) = (0.01/0.0610) = 0.0163934426 = 0.0164

K = [B]/[A] = (k₁/k₂) = 0.0164

Hope this Helps!!!

Final answer:

To find the empirical formula of the gaseous hydrocarbon, assume 100 grams of the compound. Since it is 11% hydrogen by mass, it contains 11 grams of hydrogen. The remaining mass is carbon. Divide the moles of each element by the smallest number of moles to get the empirical formula CH₂. The molecular formula cannot be determined without the molar mass.

Explanation:

Empirical Formula Calculation

To find the empirical formula, assume 100 grams of the gaseous hydrocarbon. Since it is 11% hydrogen by mass, this means it contains 11 grams of hydrogen. The remaining mass, 89 grams, must be carbon. The molar mass of hydrogen is 1 g/mol, and the molar mass of carbon is 12 g/mol. Divide the moles of each element by the smallest number of moles to get the empirical formula. In this case, 11 g H / 1 g/mol = 11 mol H and 89 g C / 12 g/mol = 7.42 mol C. Dividing both by the smallest number of moles, we get the empirical formula CH₂.

To determine the molecular formula, you'll need the molar mass of the compound. However, this information is not provided in the question. Without the molar mass, it is not possible to determine the molecular formula.

The intrinsic chemical properties of a reactant can affect the rate of reaction, as seen in the example of food spoiling more quickly on a kitchen counter than in a refrigerator due to temperature differences.

The option that exemplifies how the intrinsic chemical properties of a reactant affect the rate of reaction is Food spoils more quickly on a kitchen counter than in a refrigerator. This is because the temperature difference between the kitchen counter and the refrigerator affects the rate of chemical reactions. Chemical reactions typically occur faster at higher temperatures, so food spoils more quickly when exposed to the higher temperature of the kitchen counter compared to the lower temperature inside the refrigerator.

https://brainly.com/question/8592296

#SPJ12

Answer:

The expected freezing point of a 1.75 m solution of ethylene glycol is -3.26°C.

Explanation:

[tex]\Delta T_f=T-T_f[/tex]

[tex]\Delta T_f=K_f\times m[/tex]

[tex]m=\frac{\text{mass of solute}}{\text{Molar mass of solute}\times {Mass of solvent in kg}}[/tex]

where,

[tex]T[/tex] = Freezing point of solvent

[tex]T_f[/tex] = Freezing point of solution

[tex]\Delta T_f[/tex] =depression in freezing point

[tex]K_f[/tex] = freezing point constant

m = molality

we have :

Mass of ethylene glycol = 59.0 g

Molar mass of ethylene glycol = 62.1 g/mol

Mass of solvent i.e. water = 543 g = 0.543 kg ( 1 g = 0.001 kg)

[tex]K_f[/tex] =1.86°C/m ,

[tex]m =\frac{59.0 mol}{62.1 g/mol\times 0.543 kg}=1.75 m[/tex]

[tex]\Delta T_f=1.86^oC/m \times 1.75m[/tex]

[tex]\Delta T_f=3.26^oC[/tex]

Freezing point of pure water = T =  0°C

Freezing point of solution = [tex]T_f[/tex]

[tex]\Delta T_f=T-T_f[/tex]

[tex]T_f=T-\Delta T_f=0^oC-3.26^oC=-3.26^oC[/tex]

The expected freezing point of a 1.75 m solution of ethylene glycol is -3.26°C.

Answer: starting on the left:

1. 0.1M HCI

2. 0.001M HCI

3. 0.00001M HCI

4. distilled water

5. 0.00001 NaOH

6. 0.001M NaOH

7. 0.1M NaOH

Explanation:

The solutions arranged from most acidic to most basic would be: 0.1 M HCl, 0.001 M HCl, 0.00001 M HCl, Distilled Water, 0.00001 M NaOH, 0.001 M NaOH, and finally 0.1 M NaOH.

The pH of a solution depends on the concentration of hydronium ions (H+) and hydroxide ions (OH-). In this case, a lower concentration means less acidity or basicity. The solutions arranged in order from most acidic to most basic would be:

https://brainly.com/question/37216311

#SPJ2

Answer:

The solution is basic

Explanation:

Low pH means that a solution is acidi while low pOH means that a solution is basic.

Answer:

carbon dioxide

Explanation:

Answer:

85.0 kJ

Explanation:

Calculate the amount of heat needed to melt 160. g of solid octane (C8H18 ) and bring it to a temperature of 99.2 degrees c. Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol.

To melt 160 g of octane and bring it's temperature to 99.2°C

(from literature)

Heat of fusion of Octane = 20.740 kJ/mol

Melting point of octane = -57°C

Boiling point of Octane = 125.6 °C

Molar mass of octane = 114.23 g/mol

Heat capacity of octane = 255.68 J/K.mol

So, it is evident that Octane is still a liquid at 99.2°C.

So, the required heat is the heat required to melt octane and raise the temperature of Octane liquid to 99.2°C

First, we convert the mass of octane given to number of moles as the heat parameters provided by literature are given in molar units.

Number of moles = (mass)/(Molar mass)

Number of moles of octane = (160/114.23) = 1.401 moles

Heat required to melt the octane = nL = (1.401×20.740) = 29.05674 kJ

Heat required to raise the temperature of already melted octane from its melting temperature of -57°C to 99.2°C

= nCΔT

n = 1.401 moles

C = 255.68 J/K.mol

ΔT = (99.2 - (-57)) = 156.2°C (same as a temperature difference of 156.2 K)

Heat required to raise the temperature of already melted octane from -57°C to 99.2°C

= (1.401×255.68×156.2)

= 55,952.04 J = 55.952 kJ

Total heat required to melt the 160 g of Octane and raise its temperature to 99.2°C

= 29.05674 + 55.952 = 85.01 kJ = 85.0 kJ

Hope this Helps!!!

The heat needed to melt the solid octane and bring it to the desired temperature involves a two-step process - melting and then heating. You use the latent heat of fusion and specific heat capacity of octane to calculate the total heat required.

The amount of heat needed to melt the solid octane and bring it to the desired temperature involves a two-step process – melting, and heating. The process should be looked at in separate steps, similar to how the enthalpy of combustion is used to calculate the heat produced in the combustion of 1.00 L of isooctane in the reference provided. Note that different substances have different heat capacities and heat of fusion, so the associated values (likely given in the problem or textbook) for octane specifically would need to be used to get the final answer.

First, you would need to know the latent heat of fusion, which is the amount of energy required to change one kilogram of the substance from solid to liquid without changing its temperature. Then, you would multiply the mass of the octane by the latent heat of fusion to find out how much energy is required for it to go from solid to liquid.

Once the octane is in liquid form, we then need to heat it to the desired temperature. To do this, you use the specific heat capacity formula – the amount of energy required to raise one kilogram of the substance by one degree Celsius. Multiply the mass of the material by the specific heat capacity of the octane and the change in temperature (final temperature – initial temperature).

The added heat required for both of these processes would give you your final answer.

https://brainly.com/question/30320641

#SPJ11

Answer:

pH = 11.7

Explanation:

pOH= -log [OH]=-log[0.05]

=2.3

pOH+pH= 14

pH= 14-2.3= 11.7

Answer:

22.7

Explanation:

First, find the energy released by the mass of the sample. The heat of combustion is the heat per mole of the fuel:

ΔHC=qrxnn

We can rearrange the equation to solve for qrxn, remembering to convert the mass of sample into moles:

qrxn=ΔHrxn×n=−3374 kJ/mol×(0.500 g×1 mol213.125 g)=−7.916 kJ=−7916 J

The heat released by the reaction must be equal to the sum of the heat absorbed by the water and the calorimeter itself:

qrxn=−(qwater+qbomb)

The heat absorbed by the water can be calculated using the specific heat of water:

qwater=mcΔT

The heat absorbed by the calorimeter can be calculated from the heat capacity of the calorimeter:

qbomb=CΔT

Combine both equations into the first equation and substitute the known values, with ΔT=Tfinal−20.0∘C:

−7916 J=−[(4.184 Jg ∘C)(600. g)(Tfinal–20.0∘C)+(420. J∘C)(Tfinal–20.0∘C)]

Distribute the terms of each multiplication and simplify:

−7916 J=−[(2510.4 J∘C×Tfinal)–(2510.4 J∘C×20.0∘C)+(420. J∘C×Tfinal)–(420. J∘C×20.0∘C)]=−[(2510.4 J∘C×Tfinal)–50208 J+(420. J∘C×Tfinal)–8400 J]

Add the like terms and simplify:

−7916 J=−2930.4 J∘C×Tfinal+58608 J

Finally, solve for Tfinal:

−66524 J=−2930.4 J∘C×Tfinal

Tfinal=22.701∘C

The answer should have three significant figures, so round to 22.7∘C.

          A 0.500 g sample of C7H5N2O6 is burned in a calorimeter containing 600g of water. From the given information, the final temperature of the reaction is 22.71° C

From the given parameters:

From the listed parameters, the first step is to determine the number of moles of the sample by using the relation:

[tex]\mathbf{number \ of \ moles = \dfrac{mass}{molar \ mass}}[/tex]

[tex]\mathbf{number \ of \ moles = \dfrac{0.500 \ g}{213 \ g/mol}}[/tex]

number of moles of C7H5N2O6 = 0.00235 moles

However, the heat of combustion is the amount or quantity of heat energy released when one mole of a sample is burned.

Mathematically;

[tex]\mathbf{\Delta H_c =\dfrac{q_{rxn} }{ n}}[/tex]

where;

[tex]\mathbf{ q_{rxn}}[/tex] = quantity of heat released

n = number of moles

Making  [tex]\mathbf{ q_{rxn}}[/tex] the subject of the formula:

[tex]\mathbf{ q_{rxn}= \Delta H_{rxn} \times n }[/tex]

[tex]\mathbf{ q_{rxn}=-3374 \ kJ/moles \times 0.00235 \ moles }[/tex]

[tex]\mathbf{ q_{rxn}=-7.9289 \ kJ }[/tex]

[tex]\mathbf{ q_{rxn}=-7928.9 \ J }[/tex]

The heat released [tex]\mathbf{ q_{rxn}}[/tex] = heat absorbed by bomb calorimeter + heat absorbed by water

∴

[tex]\mathbf{ q_{rxn}= -(m\times c_{bc} \Delta T + c_{water} \times \Delta T)}[/tex]

[tex]\mathbf{-7928.9 \ J = -( 420 J/^0C \times \Delta T + 600 \ g \times 4.18 \ J/g ^0 C\times \Delta T)}[/tex]

[tex]\mathbf{7928.9 = (2928 \Delta T )}[/tex]

[tex]\mathbf{ \Delta T = \dfrac{7928.9}{2928} }[/tex]

[tex]\mathbf{ \Delta T =2 .71 ^0 \ C}[/tex]

Since ΔT = [tex]\mathbf{T_2 - T_1}[/tex]

2.71° C = T₂ - 20° C

T₂ = 20° C + 2.71° C

T₂ = 22.71° C

Therefore, we can conclude that the final temperature of the reaction is:

22.71° C

Learn more about calorimeter here:

https://brainly.com/question/10987564?referrer=searchResults