PLEASE HELPPPPPPPPPPPPPPPP

Using Hess's law we found:

1) By adding reaction 10.2 with the reverse of reaction 10.1 we get reaction 10.3:

KOH(aq) + HCl(aq)  → H₂O(l) + KCl(aq)   ΔH  (10.3)

2) The ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy (ΔH).    

The reactions of dissolution (10.1) and neutralization (10.2) are:

KOH(s) → KOH(aq)   ΔHsoln    (10.1)

KOH(s) + HCl(aq) → H₂O(l) + KCl(aq)     ΔHneut     (10.2)

1) According to Hess's law, the total change in enthalpy of a reaction resulting from differents changes in various reactions can be calculated as the sum of all the enthalpies of all those reactions.      

Hence, to get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq)    (10.3)

We need to add reaction 10.2 to the reverse of reaction 10.1

KOH(s) + HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq) + KOH(s)

Canceling the KOH(s) from both sides, we get reaction 10.3:

KOH(aq) + HCl(aq)  → H₂O(l) + KCl(aq)    (10.3)

2) The change in enthalpy for reaction 10.3 can be calculated as the sum of the enthalpies ΔHsoln and ΔHneut:

[tex] \Delta H = \Delta H_{soln} + \Delta H_{neut} [/tex]

The enthalpy of reaction 10.1 (ΔHsoln) changed its sign when we reversed reaction 10.1, so:

[tex] \Delta H = \Delta H_{neut} - \Delta H_{soln} [/tex]

Therefore, the ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy ΔH.

Learn more here:

I hope it helps you!

Answer:

go get the stuff.

Explanation:

AnswerExplanation:I finsished d

Answer:

17.55L

Explanation:

0.48mol : 11.7

0.72mol :   x

0.48x = 8.424

x = 17.55

Answer:

D

Explanation:

D Is The Answer Fella, My Head Hurt Really Bad

Since nonmetals gain electrons, the correct question to ask about group 16 elements is; "Will group 16 elements gain electrons to bond with group 2 in an XY format?"

Group 16 elements are divalent and they form divalent negative ions. The periodic table is arranged in groups and periods. The elements in the same group have the same number of valence electrons. All elements in group 2 have six valence electrons.

If a wants to form a compound with the non metals of group 16, the correct question to ask is;"Will group 16 elements gain electrons to bond with group 2 in an XY format?"

Learn more: https://brainly.com/question/14281129

Answer:

we heat up because the component with lower boiling evaporates first,

leaving the other behind

Answer: 1.324L

Explanation: use Boyles law, sorry so late!

The reduction half equation of the galvanic cell is is;  3Cu2+(aq)  + 3e ------> 3Cu(s).

A galvanic cell is a cell that produces chemical energy by a spontaneous chemical reaction. The equation of the reaction is; 2Cr(s) + 3Cu2+(aq) → 2Cr3+ (aq) + 3Cu(s). We can see that chromium was oxidized and copper was reduced.

Since reduction occurs at the cathode, it follows that the reduction half equation is;  3Cu2+(aq)  + 3e ------> 3Cu(s).

Learn more about galvanic cell; https://brainly.com/question/25606438

Iodine (I) is oxidized, and bromine (Br) is reduced. The correct option is the last option - lodine (I) is oxidized, and bromine (Br) is reduced.

To determine which elements are oxidized and reduced,

First, we will define the terms Oxidation and Reduction

Oxidation is simply defined as the loss of electrons. It can also be defined as increase in oxidation number.

Reduction is defined as the gain of electrons. It can also be defined as decrease in oxidation number.

The given chemical equation is

Br₂(l) + 2NaI(aq) → I₂(s) + 2NaBr(aq)

Oxidation number of Bromine decreased from 0 to -1.

Therefore, Bromine is reduced.

Oxidation number of Iodine increased from -1 to 0.

Therefore, Iodine is oxidized.

Oxidation number of sodium did not change.

Therefore, Sodium is neither oxidized nor reduced.

Hence, Iodine (I) is oxidized, and bromine (Br) is reduced. The correct option is the last option - lodine (I) is oxidized, and bromine (Br) is reduced.

Learn more here: https://brainly.com/question/12913997

The temperature of an object will increase if the kinetic energy of the particle increases.

Kinetic energy is an energy that is said to be in motion. According to the kinetic molecular theory of ideal gas, the particles of the gas are usually moving in constant random motion and they exert no force on each other.

Also, the temperature varies directly proportional to the average kinetic energy of the gas particles. As a result, when the kinetic energy of the particles increases, the temperature will also increase.

Learn more about the kinetic theory of gas here:

https://brainly.com/question/9949658?referrer=searchResults

Molar mass of Argon, Ar = 39.95 g Mass of 1 atom, m = 39.95 / (6.023 * 10^23) = 6.64 * 10^(-23) g =6.64 * 10^(-26) Kg Now, Kinetic Energy, KE = (1/2) * m * v^2 where velocity,v = 650 m/s Mass of 1 atom, m = 6.64 *...

- BRAINLIEST answerer

Answer:

1.40×10-20

Explanation:

Oh calculate the mass of 1 atom of argon the molar mass of argon 39.95g that is 6.0²×10²³atoms maybe...?

From the dilution formula, we have that at constant value of the solute, the volume of the solution is inversely proportional to the concentration.

The correct responses are;

Reasons:

The given parameters are;

The concentration of the KCl solution = 4% m/v

Taking the solution as solution of KCl

The volume of the solution in container 1 = Two times less than the volume of KCl solution.

[tex]V_{container \, 1} = \displaystyle \mathbf{ \frac{1}{2} \cdot V_{4\% \, solution}}[/tex]

The volume of the solution in container 2 = Two times larger compared to the volume of KCl solution.

[tex]V_{container \, 2} = \mathbf{\displaystyle 2 \times V_{4\% \, solution}}[/tex]

The volume of the solution in container 3 = Two times larger than the container two solution volume.

[tex]V_{container \, 3} = \displaystyle \mathbf{ 2 \times V_{container \, 2}}[/tex]

Therefore;

[tex]V_{container \, 3} = \displaystyle 2 \times 2 \times V_{4\% \, solution} = \mathbf{4 \times V_{4\% \, solution }}[/tex]

Part A Required:

a. To select the container that represent the dilution of the 4% solution to 2%

Solution:

The dilution formula is; C₁·V₁ = C₂·V₂

Therefore;

[tex]\displaystyle V_1 = \mathbf{\frac{C_1 \cdot V_1}{C_2}}[/tex]

C₁ =4%, C₂ = 2%, we get;

[tex]\displaystyle V_1 = \frac{4 \cdot V_1}{2} = 2 \cdot V_1[/tex]

The volume of the container that represents a 2% dilution is container 2

which is filled with a solution of a volume two times larger compared to the

KCl solution.

Part B:

Required:

The container diluted to a 1% m/v KCl solution.

Solution;

Using the dilution formula, we have;

C₁ = 4%, C₂ = 1%

Therefore;

[tex]\displaystyle V_1 = \frac{C_1 \cdot V_1}{C_2}[/tex]

[tex]\displaystyle V_1 = \frac{4 \cdot V_1}{1} = \mathbf{4 \cdot V_1}[/tex]

The volume of the solution is four times the volume of the 4% KCl solution, which is equivalent to the volume in container 3.

Possible parts of the question are;

Select the container that represents the dilution of the 4% (m/v) KCl solution to obtain the solutions that follows;

Part A: a 2% (m/v) KCl solution

Part B: a solution that is a 1% (m/v) KCl solution

Please see attached drawings

Learn more here:

https://brainly.com/question/11493179

Answer:

-85 °C

Explanation:

O and S are in the same group( Group 16). Since S is below O it's atomic mass is higher than O. So molar mass of H2S is higher than H2O. The strength of Vanderwaal Interactions ( London dispersion forces) increases when the molar mass increases. However, only H2O can form H bonds with each other. This is because electronegativity of O is higher than S and therefore H in H2O has a higher partial positive charge than H of H2S.

H bond dominate among these 2 types of forces so the strength of attractions between molecules is higher in H2O than H2S. Therefore more energy should be supplied for H2O to break inter

molecular forces and convert from solid to liquid state than H2S. So mpt of H2O must be higher than that of H2S.

This is the answer i hope it helps

A sample of gas has Pi = 0.768 ATM, Vi = 10.5 L, and Ti = 300 K. What is the final pressure if VF = 7.85 L and T and f = 250 K?

Answer:

Subduction Process where the oceanic plate subducted under the continental plates because it denser than the Continental plate.

Answer: International Union of Pure and Applied Chemistry

The International Union of Pure and Applied Chemistry (IUPAC), established in 1919, is the international body that represents chemistry and related sciences and technologies.

[tex]pH = -\log[H^{+}] \\\\\implies \log[H^{+}] = -pH\\\\\implies [H^{+}] = 10^{-pH}\\\\\implies [H^{+}] = 10^{-4.83} = 0.000015[/tex]

[MnCl6] 3- is high spin and has five unpaired electrons while [Mn(CN)6] 3- has only two unpaired electrons.

A complex may be low spin or high spin depending on the kind of ligand attached to the central metal atom/ion. If the ligand is a weak field ligand, the complex may be high spin (maximum number of unpaired electrons). If the complex is low spin, there are few unpaired electrons (minimum number of unpaired electrons). In that case, the ligand is a strong field ligand.

In the octahedral geometry,  [MnCl6] 3- is high spin and has five unpaired electrons since the chloride ion is a weak field ligand. On the other hand  [Mn(CN)6] 3- has only two unpaired electrons because the cyanide ion is a strong field ligand.

Learn more: https://brainly.com/question/6111443

The freezing point of a solution prepared by dissolving biphenyl (C12H10) into naphthalene is  67.99oC.

Freezing point can be obtained from;

ΔT = K m i

ΔT = freezing point depression

m = molality of the solution = 1.506 m

i = Van't Hoff factor = 1 for molecular substances

K = Freezing constant of naphthalene = 8.15 oC/m

ΔT = 8.15 oC/m × 1.506 m × 1 = 12.27oC

Freezing point of pure naphthalene = 80.26 °C

ΔT =  Freezing point of pure naphthalene - Freezing point of solution

Freezing point of solution =  Freezing point of pure naphthalene - ΔT

Freezing point of solution =  80.26 °C - 12.27oC = 67.99oC

Learn more:

When the pressure of the equilibrium mixture of SO2, O2, and SO3 is doubled at constant temperature, the Kp is also doubled.

The equation of the reaction is given by; 2SO2 + O2 → 2SO3. The Kp of the reaction is obtained from;

Kp = pSO3^2/pSO2^2 . pO2

Since the Kp depends on the individual partial pressures of each of SO2, O2, and SO3 at equilibrium, if the pressure of the equilibrium mixture is doubled, the Kp is also doubled.

Learn more: https://brainly.com/question/11324711

Vectors may be plasmids. An example of one of several useful features of commercially prepared cloning vectors is MULTIPLE CLONING SITES.

The pUC18 vector is a widely used standardized cloning vector for replication in Escherichia coli.

A multiple cloning site can be defined as a short DNA fragment observed in genetically engineered plasmids.

These DNA fragments (multiple cloning sites) contain twenty (20) or more sites where restriction enzymes can cut in order to generate recombinant DNA molecules.

Learn more in:

https://brainly.com/question/14020637

Answer:

44

Explanation:

Answer:

qrtyuioplkjhgfdssssssazxcvbn

Answer:

Answer D. Picture II shows a chemical change, because the same substance changes form

Explanation:

This is the temperature that water molecules slow down enough to stick to each other and form a solid crystal

Answer:

D Im 90% Sure

Explanation:

If its not right i owe you one I did this one before

Answer:

Explanation:

It''s chemical formula is Na3PO4

The time taken for 0.240 mole sample of radiuim to decay to 1.88×10¯³ mole is 79.8 days

We'll begin by calculating the number of half-lives that has elapsed.

Original amount (N₀) = 0.240 mole

Amount remaining (N) = 1.88×10¯³ mole

N = 1/2ⁿ × N₀

1.88×10¯³ = 1/2ⁿ × 0.240

Cross multiply

1.88×10¯³ × 2ⁿ = 0.240

Divide both side by 1.88×10¯³

2ⁿ = 0.240 / 1.88×10¯³

2ⁿ = 128

2ⁿ = 2⁷

Thus, 7 half-lives has elapsed

Number of half-lives (n) = 7

Half-life (t½) = 11.4 days

t = n × t½

t = 7 × 11.4

Therefore, the time taken for 0.240 mole sample of radiuim to decay to 1.88×10¯³ mole is 79.8 days

Learn more: https://brainly.com/question/13266270

Answer:

[tex]\blue{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}[/tex]

[tex]\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}[/tex]

Answer:

Explanation:

Hydrodistillation is a traditional method for the extraction of bioactive compounds from plants. In this method, plant materials are packed in a still compartment then water is added in sufficient amount and brought to a boil. ... The vapor mixture of water and oil is condensed by indirect cooling with water.