what is the molality of a solution

Answer:

[tex]\rho _{oil}<\rho _{plastic}<\rho _{water}[/tex]

Explanation:

Hello,

In this case, since water and oil are immiscible due to the oil's nonpolarity and water's polarity, when mixed, the oil remains on the water since it is less dense than water. In such a way, for a plastic sunk in the oil and floating on the water (in middle of them) we can conclude that the plastic have a mid density, therefore, the required organization is:

[tex]\rho _{oil}<\rho _{plastic}<\rho _{water}[/tex]

Best regards.

Answer:

The 3rd one

Explanation:

Answer:

[tex]Ksp=1.2[/tex]

Explanation:

Hello,

In this case, as the saturated solution has 7.1 grams of sodium carbonate, the solubility product is computed by firstly computing the molar solubility by using its molar mass (106 g/mol):

[tex]Molar \ solubility=\frac{7.1gNa_2CO_3}{0.1L}*\frac{1molNa_2CO_3}{106gNa_2CO_3}=0.67M[/tex]

Next, as its dissociation reaction is:

[tex]Na_2CO_3(s)\rightleftharpoons 2Na^+(aq)+CO_3^{2-}(aq)[/tex]

The equilibrium expression is:

[tex]Ksp=[Na^+]^2[CO_3^{2-}][/tex]

And the concentrations are related with the molar solubility (2:1 mole ratio between ionic species):

[tex]Ksp=(2*0.67)^2*(0.67)\\\\Ksp=1.2[/tex]

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Answer:

Entropy change of ice changing to water at 0°C is equal to 57.1 J/K

Explanation:

When a substance undergoes a phase change, it occurs at constant temperature.

The entropy change Δs, is given by the formula below;

Δs = q/T

where q is the quantity of heat absorbed or evolved in Joules and T is temperature in Kelvin at which the phase change occur

From the given data, T = 0°C = 273.15 K, q = 15.6 KJ = 15600 J

Δs = 15600 J / 273.15 K

Δs = 57.111 J/K

Therefore, entropy change of ice changing to water at 0°C is equal to 57.1 J/K

The entropy change of ice changing to water will be "57.1 J/K".

The shift in what seems like a thermodynamic system's condition of confusion is caused by the transformation of heat as well as enthalpy towards activity. Entropy seems to be greater mostly in a network with a high quantity or measure of chaos.

According to the question,

Temperature, T = 0°C or,

                          = 273.15 K

Heat, q = 15.6 KJ or,

            = 15600 J

We know the formula,

Entropy change, Δs = [tex]\frac{q}{T}[/tex]

By substituting the values, we get

                                 = [tex]\frac{15600}{273.15}[/tex]

                                 = 57.11 J/K

Thus the above answer is correct.    

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Answer:

461.85 degrees Celsius

Density measures how tightly packed particles are.

If particles are tightly packed together, they will be more dense.

If they are loosely together, they will be less dense.

However, a common mistake is thinking that if something

is more dense it means that it's heavier.

However, that's not the case.

It has to do with how particles are packed in an object.

Answer:

[tex]V_2=1.17L[/tex]

Explanation:

Hello,

In this case, for dilution processes, we must remember that the amount of solute remains the same, therefore, we can write:

[tex]V_1C_1=V_2C_2[/tex]

Whereas V accounts for volume and C for concentration that in this case is %(m/v). In such a way, the final volume V2 turns out:

[tex]V_2=\frac{V_1C_1}{C_2}= \frac{0.551L*50.0\%}{23.5\%}\\ \\V_2=1.17L[/tex]

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Answer:

7.38 g/cm³ is the density of the metal

Explanation:

In a Face-centered cubic unit cell you have 4 atoms. Also, the edge length is √8×r (r is radius of the atom).

To solve this problem, we need first to calculate the volume of the unit cell and then, with molar mass calculate the mass of 4 atoms. As density is the ratio between mass and volume we can obtain this value.

Volume = a³

a = √8×r

(r = 198x10⁻¹²m)

a = 5.6x10⁻¹⁰ m

Volume = 1.756x10⁻²⁸ m³

1m = 100cm → 1m³ = (100cm)³:

1.756x10⁻²⁸ m³× ((100cm)³ / 1m³) =

There are 4 atoms of gold:

4 atoms × (1mol / 6.022x10²³ atoms) = 6.64x10⁻²⁴ moles of gold

As 1 mole weighs 195.08g:

6.64x10⁻²⁴ moles of gold × (195.08g / mol) =

1.296x10⁻²¹g / 1.756x10⁻²² cm³ =

The density of the metal is 7.40 g/cm³

In cubic crystal system, face-centered cubic FFC is the name given to sort of atom arrangement observed in which structure is made up of atoms organized in a cube with a portion of an atom in each corner and six extra atoms in the center of each cube face.

It is expressed by using the formula:

[tex]\mathbf{\rho = \dfrac{Z \times M}{N_A\times a^}}[/tex]

where;

For face-centered cubic FFC;

The edge length  [tex]\mathbf{a =2 \sqrt{2}\times r }[/tex]

[tex]\mathbf{a =2 \sqrt{2}\times 198 \ pm }[/tex]

[tex]\mathbf{a =560.0285 \ pm }[/tex]

a = 5.60 × 10⁻⁸ cm

Replacing it into the previous equation, we have:

[tex]\mathbf{\rho = \dfrac{4 \times 195.8}{6.022 \times 10^{23} \times( 5.60 \times 10^{-8} )^3}}[/tex]

[tex]\mathbf{\rho = 7.40\ g/cm^3 }[/tex]

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Answer:

Atomic no = 12 = Mg

Explanation:

It is given that,

The atomic number of two elements that are represented by letter Q and R are 9 and 12.

We need to write the electronic configuration of R. Atomic number shows the number of protons in atom.

For R, atomic number = 12

Its electronic configuration is : 2,8,2

It has two valance electrons in its outermost shell. The element is Magnesium (Mg).

Answer:

-138.9 kJ/mol

Explanation:

Step 1: Convert 235.8°C to the Kelvin scale

We will use the following expression.

K = °C + 273.15 = 235.8°C + 273.15 = 509.0 K

Step 2: Calculate the standard enthalpy of reaction (ΔH°)

We will use the following expression.

ΔG° = ΔH° - T.ΔS°

ΔH° = ΔG° / T.ΔS°

ΔH° = (-936.92kJ/mol) / 509.0K × 0.51379 kJ/mol.K

ΔH° = -3.583 kJ (for 1 mole of balanced reaction)

Step 3: Convert -9.9°C to the Kelvin scale

K = °C + 273.15 = -9.9°C + 273.15 = 263.3 K

Step 4: Calculate ΔG° at 263.3 K

ΔG° = ΔH° - T.ΔS°

ΔG° = -3.583 kJ/mol - 263.3 K × 0.51379 kJ/mol.K

ΔG° = -138.9 kJ/mol

Answer:

The best reducing agent is Zn(s)

Explanation:

A reducing agent must to be able to reduce another compound, by oxidizing itself. Consequently, the oxidation potential must be high. The oxidation potential of a compound is the reduction potential of the same compound  with the opposite charge. Given the reduction potentials, the best reducing agent will be the compound with the most negative reduction potential. Among the following reduction potentials:

Cd₂⁺(aq)|Cd(s) ⇒ -0.40 V

Zn²⁺(aq)|Zn(s) ⇒ -0.76 V

Ni²⁺(aq)|Ni(s) ⇒‑0.25 V

Cu²⁺(aq)|Cu(s) ⇒ +0.34 V

Ag⁺(aq)|Ag(s) ⇒ +0.80 V

The most negative is Zn²⁺(aq)|Zn(s) ⇒ -0.76 V

From this, the most reducing agent is Zn. Zn(s) is oxidized to Zn²⁺ ions with the highest oxidation potential (0.76 V).

Answer:

(R)-but-3-en-2-ylbenzene

Explanation:

In this reaction, we have a very strong base (sodium ethoxide). This base, will remove a hydrogen producing a double bond. We know that the reaction occurs through an E2 mechanism, therefore, the hydrogen that is removed must have an angle of 180º with respect to the leaving group (the "OH"). This is known as the anti-periplanar configuration.

The hydrogen that has this configuration is the one that placed with the dashed bond (red hydrogen). In such a way, that the base will remove this hydrogen, the "OH" will leave the molecule and a double bond will be formed between the methyl and the carbon that was previously attached to the "OH", producing the molecule (R) -but-3- en-2-ylbenzene.

See figure 1

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Answer:

Bakelite is a polymer made up of the monomers phenol and formaldehyde. This phenol-formaldehyde resin is a thermosetting polymer.

Answer: The monomers of bakelite are formaldehyde and phenol

Explanation:

Answer:

Lone pairs cause more repulsion than bond pairs

Explanation:

A lone pair takes up more space around the central atom than bond pairs of electrons. This is because, a lone pair is attracted to only one nucleus while bond pairs are attracted to two nuclei.

Hence the repulsion between lone pairs is far greater than the repulsion between bond pairs or repulsion between a lone pair and a bond pair. The presence of a lone pair therefore distorts a molecule away from the ideal shape predicted on the basis of the valence shell electron pair repulsion theory.

Lone pairs are found to decrease the observed bond angles in a molecule.

Answer:

-195.8º < -191.5º < 100º

Explanation:

Water, or H20, starts boiling at 100ºC.

Nitrogen, or N2, starts boiling at -195.8ºC.

Carbon monoxide, or C0, starts boiling at -191.5ºC.

When we place these in order from decreasing boiling point:

-195.8º goes first, then -191.5º, and 100º goes last.

Answer:

therefore, N2, CO, H20

Decreasing boiling point

Explanation:

the bond existing in H2O is hydrogen bond

bond existing in N2 is covalent bond, force existing is dipole-dipole-interaction

bond existing in CO is covalent bond , force existing between is induced -dipole- induced dipole-interaction

hydrogen bond is the strongest , followed by dipole-dipole-interaction and induced -dipole- induced dipole-interaction

the stronger the bond , the higher the boiling point

therefore, N2, CO, H20

-------------------------------------->

Decreasing boiling point

Answer:

  30. 5 planes are shown

  31. 1 plane

  32. CEF

  33. on line AB

  34. E or F

   35. ABCD or BCEF or CDEF or ACEF

Explanation:

30. Each of the surfaces of the rectangular pyramid is a plane. There are 5 planes.

__

31. 3 points define one plane only.

__

32. The only points shown on the same line segment are points E, F, and C.

__

33. If G is to be collinear with A and B, it must lie on line AB.

__

34. The only points shown that are not on plane N are points E and F. Either of those will do.

__

35. There are three planes that have 4 points shown on them. The four points that are on the same plane are any of ...

Plane ACEF is not shown on the diagram, but we know that those 4 points are also coplanar. (Any point not on line CE, together with the three points on that line, will define a plane with 4 coplanar points.)

Answer:

Below

Explanation:

Let n be the quantity of matter in the Calcium Bromide

● n = m/ M

M is the atomic weight and m is the mass

M of CaBr2 is the sum of the atomic wieght of its components (2 Bromes atoms and 1 calcium atom)

M = 40.1 + 2×79.9

● 0.422 = m/ (40.1+2×79.9)

●0.422 = m/ 199.9

● m = 0.422 × 199.9

● m = 84.35 g wich is 88.4 g approximatively

88.4 g approximatively is  the mass in grams of 0.442 moles of calcium bromide, CaBr2 ,therefore option (d) is correct.

Mass is the amount of matter that a body possesses. Mass is usually measured in grams (g) or kilograms (kg) .

To calculate mass in grams of 0.442 moles of calcium bromide, CaBr2,

Let n be the quantity of matter in the Calcium Bromide

M is the atomic weight and m is the mass

M of CaBr2 is the sum of the atomic weight of its components

Mass of  Ca = 40.1 , Mass of Br = 79.9

M = 40.1 + 2×79.9

  0.422 = m/ (40.1+2×79.9)

  0.422 = m/ 199.9

  m = 0.422 × 199.9

  m = 84.35 g which is 88.4 g approximatively .

Thus ,88.4 g approximatively is  the mass in grams of 0.442 moles of calcium bromide, CaBr2 , hence option (d) is correct .

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Answer:

The balanced chemical reaction is given as:

[tex]Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)[/tex]

Explanation:

When aqueous cesium sulfate and aqueous barium perchlorate are mixed together it gives white precipitate barium sulfate and aqueous solution od cesium perchlorate.

The balanced chemical reaction is given as:

[tex]Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)[/tex]

According to reaction, 1 mole of cesium sulfate reacts with 1 mole of barium perchlorate to give 1 mole of a white precipitate of barium sulfate and 2 moles of cesium perchlorate.

Answer:

D. Reduction takes place at the cathode, which is negatively charged.

Explanation:

In an electrolytic cell there are two electrodes; the cathode and the anode. The anode is the positive electrode while the cathode is the negative electrode. Oxidation occurs at the anode while reduction occurs at the cathode.

At the anode, species give up electrons and become positively charged ions while at the cathode species accept electrons and become reduced.

Answer:

The amount of heat to absorb is 6,261 J

Explanation:

Calorimetry is in charge of measuring the amount of heat generated or lost in certain physical or chemical processes.

The total energy required is the sum of the energy to heat the ice from -20 ° C to ice of 0 ° C, melting the ice of 0 ° C in 0 ° C water and finally heating the water to 60 ° C.

So:

Being the sensible heat of a body the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous), the expression is used:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial).

In this case, m= 10 g, specific heat of the ice= 2.1 [tex]\frac{J}{g*C}[/tex] and ΔT=0 C - (-20 C)= 20 C

Replacing: Q= 10 g*2.1 [tex]\frac{J}{g*C}[/tex] *20 C and solving: Q=420 J

The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:

Q= m* ΔHfusion

In this case, being 1 mol of water= 18 grams: Q= 10 g*[tex]6.0 \frac{kJ}{mol} *\frac{1 mol of water}{18 g}[/tex]= 3.333 kJ= 3,333 J (being kJ=1,000 J)

In this case the expression used in the first step is used, but being: m= 10 g, specific heat of the water= 4.18 [tex]\frac{J}{g*C}[/tex] and ΔT=60 C - (0 C)= 60 C

Replacing: Q= 10 g*4.18 [tex]\frac{J}{g*C}[/tex] *60 C and solving: Q=2,508 J

Finally, Qtotal= 420 J + 3,333 J + 2,508 J

Qtotal= 6,261 J

The amount of heat to absorb is 6,261 J

The amount of heat to absorb is 6,261 J.

Heat required to raise the temperature of ice from -20 °C to 0 °C.

The formula for specific heat is used to calculate the amount of heat

Q = c * m * ΔT

Where,

Q =heat exchanged by a body,

m= mass of the body

c= specific heat

ΔT= change in temperature

Given:

m= 10 g,

specific heat of the ice= 2.1

ΔT=0 C - (-20 C)= 20 C

On substituting the values:

Q= 10 g*2.1  *20 C

Q=420 J

Heat required to convert 0 °C ice to 0 °C water.

The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:

Q= m* ΔHfusion

Heat required to raise the temperature of water from 0 °C to 60 °C

m= 10 g,

Specific heat of the water= 4.18  

ΔT=60 C - (0 C)= 60 C

On substituting:

Q= 10 g*4.18  *60 C

Q=2,508 J

Thus, Qtotal= 420 J + 3,333 J + 2,508 J

Qtotal= 6,261 J

The amount of heat to absorb is 6,261 J

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Answer:

The answer is o2

Explanation:

I took the test

Answer: C. the same amount of energy as

Explanation:

A reversible reaction is a chemical reaction where the reactants form products that, in turn, react together to give the reactants back.

Reversible reactions will reach an equilibrium point where the concentrations of the reactants and products will no longer change.

[tex]A+B\rightleftharpoons C+D[/tex]

Thus if forward reaction is exothermic i.e. the heat is released , the backward reaction will be endothermic i.e. the heat is absorbed and in same amount.

The amount of energy released will be equal and opposite in sign to the energy absorbed in that reaction.

Answer:

C.) the same amount of energy as

Explanation:

I got it correct on founders edtell

Answer:

[tex][H^+]=0.000123M[/tex]

[tex]pH=3.91[/tex]

Explanation:

Hello,

In this case, dissociation reaction for acetic acid is:

[tex]CH_3COOH\rightleftharpoons CH_3COO^-+H^+[/tex]

For which the equilibrium expression is:

[tex]Ka=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}[/tex]

Which in terms of the reaction extent [tex]x[/tex] could be written as:

[tex]1.74x10^{-5}=\frac{x*x}{[CH_3COOH]_0-x}=\frac{x*x}{0.0010M-x}[/tex]

Thus, solving by using a solver or quadratic equation we obtain:

[tex]x_1=0.000123M\\\\x_2=-0.000141M[/tex]

And clearly the result is 0.000123M, which also equals the concentration of hydronium ion in solution:

[tex][H^+]=0.000123M[/tex]

Now, the pH is computed as follows:

[tex]pH=-log([H^+])=-log(0.000123)\\\\pH=3.91[/tex]

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Answer:

d. the conjugate base.

Explanation:

The general reaction of a weak acid, HA, with a strong base YOH, is:

HA + YOH → A⁻ + H₂O + Y⁻

Where A⁻ is the conjugate base of the weak acid and Y⁻ usually is a strong electrolyte.

That means after he complete reaction you don't have weak acid nor strong base, just conjugate base that will be in equilibrium with water, thus (Strong electrolyte doesn't change pH:

A⁻ + H₂O ⇄ HA + OH⁻

As the equilibrium is producing OH⁻, the pH of the solution is being affected for the conjugate base

Right option:

Answer:

a

Explanation:

answer is a on edg

The given question is incomplete.

The complete question is:

When methane is burned with oxygen, the products are carbon dioxide and water. If you produce 9 grams of water and 11 grams of carbon dioxide from 16 grams of oxygen, how many grams of methane were needed for the reaction?

Answer: 4 grams of methane were needed for the reaction

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

{tex]CH_4+2O_2\rightarrow CO_2+H_2O[/tex]

Given:  mass of oxygen = 16 g

Mass of carbon dioxide = 11 g

Mass of water = 9 g

Mass of products = Mass of carbon dioxide + mass of water = 11 g  +9 g = 20 g

Mass or reactant = mass of methane + mass of oxygen = mass of methane + 16 g

As mass of reactants = mass of products

mass of methane + 16 g= 20 g

mass of methane  = 4 g

Thus 4 grams of methane were needed for the reaction

Answer:

0.44

Explanation:

We know that the pH of any acid solution is given by the negative logarithm of its hydrogen ion concentration. Hence, if I can obtain the hydrogen ion concentration of any acid, I can obtain its pH.

For the acid, HNO3, [H^+] = [NO3^-]= 0.36 M

pH= -log [H^+]

pH= - log[0.36]

pH= 0.44

Answer:

a. Cyclohexanone

Explanation:

The principle of IR technique is based on the vibration of the bonds by using the energy that is in this region of the electromagnetic spectrum. For each bond, there is a specific energy that generates a specific vibration. In this case, you want to study the vibration that is given in the carbonyl group C=O. Which is located around 1700 cm-1.

Now, we must remember that the lower the wavenumber we will have less energy. So, what we should look for in these molecules, is a carbonyl group in which less energy is needed to vibrate since we look for the molecule with a smaller wavenumber.

If we look at the structure of all the molecules we will find that in the last three we have heteroatoms (atoms different to carbon I hydrogen) on the right side of the carbonyl group. These atoms allow the production of resonance structures which makes the molecule more stable. If the molecule is more stable we will need more energy to make it vibrate and therefore greater wavenumbers.

The molecule that fulfills this condition is the cyclohexanone.

See figure 1

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