draw a mechanism showing the penta-coordinate intermediate and the formation of the phosphorylated intermediate (which is an anhydride)

The molecule that is drawn in a conformation that has a proton and a leaving group anti-periplanar is H₂C, Br.

The A, B, C, and D bond angles of a molecule are referred to as anti-periplanar, or antiperiplanar, in organic chemistry. The dihedral angles of the A–B and C–D bonds in this conformer are larger than +150° or less than 150°. In textbooks, the term "anti-periplanar" is frequently used to refer to a strictly anti-coplanar structure with a 180° AB CD dihedral angle. The anti-periplanar functional groups will be 180° apart from one another and in a staggered configuration in a Newman projection of the molecule.  

Conformation is an essential factor in predicting reactivity in organic molecules. The anti-periplanar conformation of a molecule is one that occurs when two atoms in a molecule are in the same plane and are separated by 180 degrees. In this case, the proton and leaving group are placed in a perpendicular plane to the atoms directly in between them. This is the most stable conformer of the molecule. A significant factor in predicting reactivity in organic molecules is conformation. In this case, the molecule H₂C, Br is drawn in a conformation that has a proton and a leaving group anti-periplanar.

Therefore, the correct option is H₂C, Br.

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The 4p-orbital can hold a maximum of 6 electrons, the 3d-orbital can hold a maximum of 10 electrons, the 1s-orbital can hold a maximum of 2 electrons, and the 4f-orbital can hold a maximum of 14 electrons.

The set of quantum numbers {n, I, ml, m} is incorrect because I is not a quantum number; instead, it should be l, which stands for the angular momentum quantum number. This set of quantum numbers must satisfy the following conditions: n ≥ l ≥ m_l ≥ |m_s|. For an electron in an atom, any combination of quantum numbers where n < l is forbidden, because n must be greater than or equal to l. Additionally, any combination of quantum numbers where |m_l| > l is forbidden because m_l must be less than or equal to l.
Real orbitals have the following combinations of n and l: n = 1 and l = 0 (1s), n = 2 and l = 0, 1 (2s, 2p), n = 3 and l = 0, 1, 2 (3s, 3p, 3d), n = 4 and l = 0, 1, 2, 3 (4s, 4p, 4d, 4f).
The principal quantum number and the orbital angular momentum quantum number for the 4p-orbital is n = 4 and l = 1; for the 3d-orbital is n = 3 and l = 2; for the 1s-orbital is n = 1 and l = 0; for the 4f-orbital is n = 4 and l = 3.
The 4p-orbital can hold a maximum of 6 electrons, the 3d-orbital can hold a maximum of 10 electrons, the 1s-orbital can hold a maximum of 2 electrons, and the 4f-orbital can hold a maximum of 14 electrons.

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The molecular ion produced when an organic molecule is bombarded by a high-energy stream of electrons is known as a radical cation.

Radical cations are highly reactive and can react quickly with other compounds in their environment. The number of electrons removed to form the radical cation is determined by the type of organic molecule that is bombarded. If an alkene is bombarded, two electrons are typically removed, forming an allyl radical cation, while the addition of three electrons is necessary to form a benzyl radical cation. Radical cations are stabilized by the formation of new bonds with other molecules, thereby forming an adduct. The adduct can then be separated and characterized using chromatographic techniques. Additionally, radical cations can also react with nucleophiles, resulting in a variety of other reaction products. Thus, the molecular ion produced when an organic molecule is bombarded by a high-energy stream of electrons is a radical cation.

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Phenol, toluene, benzene, bromo benzene, and nitrobenzene are the chemicals in this order.

Due to the M effect, benzosulphonic acid is least reactive in an electrophilic aromatic substitution. Because nitrogen is more electronegative than carbon and functions as an electron withdrawing group, pyridine is less reactive than benzene towards electrophilic aromatic substitution. The meta hydrogen is thus replaced.

Most beginning organic chemistry courses cover six fundamental electrophilic aromatic substitution reactions: chlorination, bromination, nitration, sulfonation, Friedel-Crafts alkylation, and Friedel-Crafts acylation.

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The two equilibria are the triple point, which is the point when water can exist as ice, liquid, or vapor, and the critical point, which is when water exists as a single phase that has properties of both a liquid and a gas.

The line connecting the triple point and critical point on the phase diagram of water represents the conditions at which the solid, liquid, and gas phases can coexist in equilibrium.

At the triple point, which occurs at a temperature of 0.01℃ and a pressure of 0.006 atm, the solid, liquid, and gas phases of water exist in equilibrium. This means that at this point, water can exist as ice, liquid, or vapor, depending on the conditions.

At the critical point, which occurs at a temperature of 374℃ and a pressure of 218 atm, the distinction between the liquid and gas phases disappears, and the substance becomes a supercritical fluid. This means that at this point, water exists as a single phase, which exhibits the properties of both a liquid and a gas.

There are two ways to turn nitrogen gas into a liquid:

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Answer:

1.  placing balls

2. list of materials

Explanation:

The answer to this question is that blood flow to the skin does not increase; instead, it is directed inward to conserve heat.

The preoptic area of the hypothalamus serves as the body's thermostat, and when its temperature drops below the thermostat setting, a number of physiological processes are triggered. A) Shivering thermogenesis occurs, which is the production of heat by muscular contractions, B) Nonshivering thermogenesis occurs, which is the production of heat by increased metabolic activity in brown adipose tissue, C) Epinephrine levels rise to promote the mobilization of glucose from stored energy sources and D) Blood returning from the limbs is shunted to deep veins to conserve heat. However, the answer to this question is that blood flow to the skin does not increase; instead, it is directed inward to conserve heat.

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The reaction starts with 0.200 M A and 0.250 M B, the equilibrium concentration of C.3.63x10⁻¹⁰ M.

Chemical equilibrium is the state in which both the reactants and the products of a reaction are at a concentration that does not change over time any longer. In this condition, the forward and backward reaction rates are equal.

2A(aq) + B(aq) <==> C(aq)

Equilibrium expression is

K = [C] / [A]2[B]

Prepare an ICE table:

2A(aq)  +  B(aq) <==>  C(aq)

0.2..........0.15...............0........Initial

-2x...........-x.................+x......Change

0.2-2x....0.15-x............x........Equilibrium

Substitute in the equilibrium expression:

1.10x10⁻⁴ = (x) / (0.2-2x)2(0.15-x)  ... b/c K is small, we can essentially avoid using the quadratic as follows..

1.1x10⁻⁴  = x/(0.2)2(0.15)

x = (2.2x10₋⁵) (1.65x10⁻⁵)

x = 3.63x10⁻¹⁰  M = [C]

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Answer- Incomplete combustions of alkanes are dangerous as it leads to  toxic gases such as carbon monoxide man odorless,colorless and highly poisonous gas to be released in the air

Environmental impacts of burning fossil fuels such as alkanes can have a massive impact of global warming in which heat is prevented from the leaving the atmosphere due to a build-up of carbon dioxide ad other compounds, collectively known as greenhouse gases. In order to make these impacts less severe scientists have found alternative fuels, such as biofuels and biodiesels, this can reduce the risk of harm to the world as alternative fuels produce less CO2 and air pollutants then others that run on petrol and diesel

To determine the Cl for NeMutl:Wt, you need to use the data from part 2 of the case study. The data is given as 1x10-10 for the Wt strain and 1x10-7 for the mutant strain. To calculate the Cl, we use the following equation: Cl = 1/[(1/ID50) - (1/LD50)]. Using this equation, we can calculate the Cl to be 3x10-3.

To determine the Cl for NeMut2:Wt, we can use the same equation. Using the data from the table in part B, the Cl for NeMut2:Wt can be calculated to be 8x10-3.Interpreting these results, we can see that NeMut1:Wt has a Cl that is roughly 3 times lower than that of NeMut2:Wt. This suggests that the mutation of NeMut1 is significantly affecting a virulence factor, while NeMut2 may not be affecting a virulence factor as significantly.

It is important to note that similar LD50 and/or ID50 values of a Wt and mutant strain does not necessarily mean that the mutation does not affect a virulence factor. This is because the LD50 and ID50 values are used to measure how much of the pathogen is needed to produce a certain effect, but other aspects of the pathogen such as the speed or rate of infection or the amount of toxin produced can still be different and affect the virulence of the strain.

Cl for NeMut1:Wt cannot be calculated from the data presented in part 2 of the case study. The given results are:|  Inoculum (LD50) |  Mortality (LD50) |  CFU/ml of blood | Wild-type   | 6.5 × 10−7    |  6.5 × 10−7      | 7.0 × 103 | NeMut1  | 1.0 × 10−10    |  6.5 × 10−7     | 3.0 × 105 | NeMut2  | 2.0 × 10−7    |  2.0 × 10−7     | 2.2 × 103 |Since the Cl cannot be calculated from the data given, the correct option is (d) Cl cannot be calculated from the data given.If the LDs0 and/or IDs0 values of a Wt and mutant strain are similar in this type of experiment, it does not necessarily mean that the mutation does not affect a virulence factor.

This is because mutations can affect different aspects of virulence, and the specific virulence factor being measured may not be impacted by the mutation.In order to determine the CI for NeMut1:Wt and NeMut2:Wt, we need to use the following formula:CI = (output ratio of mutant) / (output ratio of wild-type)Output ratio = (CFU/ml of blood) / (inoculum)Using the data from the table, we get:

Output ratio of NeMut1:Wt = 3.0 × 105 / 1.0 × 10−10 = 3.0 × 1015Output ratio of wild-type = 7.0 × 103 / 6.5 × 10−7 = 1.1 × 1010CI of NeMut1:Wt = (3.0 × 1015) / (1.1 × 1010) = 2.7 × 105Output ratio of NeMut2:Wt = 2.2 × 103 / 2.0 × 10−7 = 1.1 × 1010CI of NeMut2:Wt = (1.1 × 1010) / (1.1 × 1010) = 1Interpretation of results from question 5 above: The CI of NeMut1:Wt is much greater than 1, indicating that NeMut1 is more virulent than the wild-type strain. The CI of NeMut2:Wt is equal to 1, indicating that NeMut2 does not exhibit any significant difference in virulence compared to the wild-type strain.

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A) To solve this problem, we can use the formula for the ionization constant of water:

Kw = [H3O+][OH-]

At 50 oC, Kw = 5.476 x 10-14. We can assume that [H3O+] and [OH-] are equal since we are dealing with pure water.

Therefore,

[H3O+] = [OH-] = sqrt(Kw) = sqrt(5.476 x 10-14) = 7.40 x 10-8 mol/L

pH = -log[H3O+] = -log(7.40 x 10-8) = 7.13

pOH = -log[OH-] = -log(7.40 x 10-8) = 7.13

To find the values for pure water at 60oC, we can use the new value of Kw at that temperature:

Kw = 9.550 x 10-14

[H3O+] = [OH-] = sqrt(Kw) = sqrt(9.550 x 10-14) = 3.09 x 10-7 mol/L

pH = -log[H3O+] = -log(3.09 x 10-7) = 6.51

pOH = -log[OH-] = -log(3.09 x 10-7) = 6.51

b) i) No, it does not mean that pure water becomes more acidic as the temperature rises.

ii) The correct answer is 2. As the temperature increases, the ionization of water increases and more H3O+ and OH- ions are formed. However, since the concentration of H2O is also decreasing due to the increase in temperature, the increase in ionization does not result in an increase in [H3O+] and pH actually decreases.

When the pressure is 0.72 atm, the temperature in Kelvin is 156 K.

To determine the Kelvin temperature when the pressure is 0.72 atm, you will need to use the trendline equation given in Fig 6.2. First, find the equation of the trendline by using the graph's two points, (300 K, 1 atm) and (500 K, 2 atm).

The equation for the trendline is:

y = mx + b

Where y is pressure, x is the temperature in Kelvin, m is the slope, and b is the y-intercept. We can find the slope of the trendline by using the two points provided in the graph:

Slope (m) = (y2 - y1) / (x2 - x1)

Slope = (2 atm - 1 atm) / (500 K - 300 K)

Slope = 0.005 atm/K

The equation for the trendline can now be written: y = 0.005x + b. To find the y-intercept, b, we can use one of the two points: Solving for b:

1 atm = 0.005(300 K) + bb = 1.5 atm

Now we can use the equation for the trendline to find the temperature (x) at which the pressure (y) equals 0.72 atm:

0.72 atm = 0.005x + 1.5 atm

0.72 atm - 1.5 atm = 0.005x

-0.78 atm = 0.005xx

= -0.78 atm / 0.005x

= 156K

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Answer:

The dissociation of a monoprotic acid HA can be represented as follows:

HA ⇌ H+ + A-

The equilibrium constant expression for this reaction is:

Ka = [H+][A-]/[HA]

We are given the initial concentration of the acid (HA) as 0.48 M and the equilibrium concentration as 0.36 M. At equilibrium, the concentration of H+ and A- will also be 0.36 M.

Substituting the values into the equilibrium constant expression, we get:

Ka = (0.36)^2 / 0.48 = 0.27

Therefore, the value of Ka for the acid is 0.27, rounded to two significant figures.

The addition of 0.20 M NaOH to 0.20 M CH3CH2COOH can result in the formation of a buffer solution because NaOH is a strong base, and CH3CH2COOH is a weak acid. When these two react, they will produce a salt and water, and the salt will act as the buffer that helps keep the pH of the solution relatively constant.

The addition of 0.20 M NaOH to 0.20 M CH3CH2COOH results in the formation of a buffer solution because NaOH is a strong base, and CH3CH2COOH is a weak acid. When a strong base is mixed with a weak acid, a buffer solution is formed. A buffer solution is able to resist changes in pH when small amounts of an acid or base is added.

When NaOH is added to CH3CH2COOH, the weak acid (CH3CH2COOH) will react with the strong base (NaOH) to produce salt (CH3CH2COONa) and water (H2O). The CH3CH2COONa salt will then dissociate into CH3CH2COO- and Na+ ions, which will act as the buffer. The CH3CH2COO- ions will then accept H+ ions from an acid and donate H+ ions to a base. This helps to keep the pH of the solution relatively constant.

The Ka of the CH3CH2COOH is 1.8x10^-5 and its pKa is 4.75. This means that the buffer will work best when the pH is close to 4.75. The amount of NaOH added to the CH3CH2COOH will determine the pH of the buffer solution. If the amount of NaOH is too high, the pH will be above the pKa and the buffer will not work as efficiently.

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If all is working correctly, the electron in a higher orbital or shell will have more energy than the electron in a lower orbital or shell. The following statement is correct: The electron that is farther away from the nucleus is at a higher energy level.

In order for the electron to escape from the atom, it must be excited, meaning that it must absorb energy. When this occurs, the electron moves to a higher energy level, which is farther from the nucleus. Because the electron is now in an excited state, it is more vulnerable to being released from the atom if additional energy is provided to it. According to Bohr's model of the atom, electrons revolve around the nucleus in circular orbits with varying energy levels. As the distance between the nucleus and the electron increases, so does the energy level of the electron. The energy of electrons in the first energy level is the lowest, and as the energy level increases, so does the energy of electrons. As a result, electrons in the outermost shell have the highest energy levels.

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The interaction between Arrhenius acid and base, which produces salt and water as a byproduct, is referred to as a neutralisation reaction. Strong acids include substances like HCl, HNO3, H2SO4, etc.

The term "Arrhenius acid" refers to a material that contains a hydrogen atom that readily releases a hydrogen ion and proton when it is in contact with water. For instance, when hydrochloric acid dissolves in water, it produces the ions hydronium (H3O+) and chloride (Cl-).

The hydrogen ion (H+) is one of the electrically charged molecules or atoms that are produced when an acid dissociates in water, according to the Arrhenius theory, which was first proposed by Swedish scientist Svante Berzelius in 1887. On the other hand, bases ionise in water to produce hydroxide ions (OH).

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To write the net ionic equation for this reaction, we need to first write the balanced molecular equation, and then break down all the soluble ionic compounds into their constituent ions:

Molecular equation:

SnBr2(aq) + K2S(aq) → SnS(s) + 2KBr(aq)

Complete ionic equation:

Sn2+(aq) + 2Br-(aq) + 2K+(aq) + S2-(aq) → SnS(s) + 2K+(aq) + 2Br-(aq)

Net ionic equation:

Sn2+(aq) + S2-(aq) → SnS(s)

The net ionic equation shows only the species that are involved in the actual reaction, which in this case are the tin cation (Sn2+) and the sulfide anion (S2-), which combine to form solid tin sulfide (SnS). The potassium cation (K+) and bromide anion (Br-) ions are spectator ions that do not participate in the reaction and are omitted from the net ionic equation.

Ionic refers to a type of chemical bond that occurs when one or more electrons are transferred from one atom to another, resulting in the formation of ions.

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The net ionic equation for the given reaction is: 1 Cr3+(aq) + 3OH-(aq) → Cr(OH)3(s)

2. The formula for the precipitate formed in this reaction is Cr(OH)3.

3. The sum of all coefficients in the net ionic equation is 4.

Consider the reaction between aqueous solutions of potassium hydroxide and chromium (III) chloride. The balanced chemical equation for the given reaction is:KOH(aq) + CrCl3(aq) → KCl(aq) + Cr(OH)3(s)1) Spectator ionsThe ions that do not take part in the reaction are known as spectator ions.

These ions are present on both sides of the equation without undergoing any chemical changes.The ionic equation for the given reaction is:3K+(aq) + 3OH-(aq) + Cr3+(aq) + 3Cl-(aq) → 3K+(aq) + 3Cl-(aq) + Cr(OH)3(s)The spectator ions are K+ and Cl-.2)

PrecipitateThe precipitate is formed when the two reactants are combined together, and it can be identified from the ionic equation. In this reaction, the precipitate is formed when KOH is added to the aqueous solution of chromium(III) chloride.The formula for the precipitate formed in this reaction is Cr(OH)3.3) Sum of all coefficientsThe net ionic equation represents the actual chemical change occurring in the reaction.

The spectator ions are removed, and only the ions that participate in the reaction are shown. The net ionic equation for the given reaction is:Cr3+(aq) + 3OH-(aq) → Cr(OH)3(s)The sum of all coefficients in the net ionic equation is 4.

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Answer:

the answer is for surely C

Explanation:

you can tell from the way is made and I also have one in my room

Explanation:

From the balanced equation you can see that for every TWO moles of S , THREE moles of O2 are needed

 so if you have four moles of S you will need SIX moles of O2 ....meaning

you will have  ( 9.5 - 6 ) = 3.5 moles of O2 left over

a balanced equation for the redox reaction: 2 Ca(s) + O2(g) → 2 CaO(s)

A redox reaction is a type of chemical reaction that involves the transfer of electrons between species. One species undergoes oxidation (loses electrons) while another species undergoes reduction (gains electrons).

In the reaction between calcium metal and oxygen gas, the calcium metal is being oxidized (loses electrons) and the oxygen gas is being reduced (gains electrons). This can be seen in the balanced equation where the calcium atoms go from having an oxidation state of 0 to +2, while the oxygen atoms go from having an oxidation state of 0 to -2.

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Answer: two solid electrodes (must be two different metals in solution)

Answer: 468.8m

Explanation:

640kHz x 1000Hz/1kHz = 6.4x10^5 Hz

λ=c/v

3.00x10^8m/s / 6.4x10^5s = 468.75m

Answer:

1. Equilibrium expressions:

a. K = [HSO4-][H3O+]/[H2SO4][H2O]

b. K = [NO]^4[H2O]^6/[NH3]^4[O2]^5

c. K = [NH3][HCl]/[NH4Cl]

d. K = [NO2]^2/[N2O4]

2. Since K = 0.050, the concentrations of the reactants (N2 and H2) are larger than the concentrations of the products (NH3).

3. Since K = 6.8 x 10^3, the concentrations of the products (SO2 and O2) are larger than the concentrations of the reactant (SO3).

4. The Ksp expression for each of the reactions is:

a. Ksp = [Na+][Cl-]

b. Ksp = [Ba2+][SO42-]

5. Brønsted-Lowry acids and bases:

a. Acid: HNO3; Conjugate base: NO3-; Base: H2O; Conjugate acid: H3O+

b. Acid: HCN; Conjugate base: CN-; Base: H2O; Conjugate acid: HCN

c. Acid: H2SO4; Conjugate base: HSO4-; Base: Cl-; Conjugate acid: HCl

d. Acid: NH3; Conjugate base: NH2-; Base: H2O; Conjugate acid: NH4+

e. Acid: H2O; Conjugate base: OH-; Base: O2-; Conjugate acid: OH-

6. Conjugate acids and bases:

a. Acid: H2O; Conjugate base: OH-

b. Acid: H3O+; Conjugate base: H2O

c. Acid: H2CO3; Conjugate base: HCO3-

d. Acid: NH4+; Conjugate base: NH3

e. Acid: HSO4-; Conjugate base: SO42-

7. The strongest acid is HIO4 (highest K value), followed by HCN, HC2H3O2, and H2CO3 (lowest K value). The K values represent the degree to which the acids dissociate in solution. HIO4 is a strong acid, meaning it dissociates almost completely in solution, while H2CO3 is a weak acid, meaning it only dissociates partially.

8. pH and pOH calculations:

a. pH = -log[H3O+] = -log(0.200) = 0.699; pOH = -log[OH-] = -log(1.0 x 10^-14/0.200) = 12.301

b. pOH = -log[OH-] = -log(0.0143) = 1.844; pH = 14.000 - pOH = 12.156

c. pH = -log[H3O+] = -log(3.0) = 0.522; pOH = 13.478

d. pOH = -log[OH-] = -log(0.0062) = 2.206; pH = 14.000 - pOH = 11.794

9. Hydronium and hydroxide ion concentrations:

pH = 3.6; hydronium ion concentration = 10^-pH = 3.98 x 10^-4 M; hydro

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If the sample of the inhalation anesthetic gas Halothane, in a 500-mL cylinder has a pressure of 2.3, the pressure is given as 2.6 atm

We have the following data to solve the problem with

p1 = 2.3 atm

T1 = 273 K

P2 is unknown

T2 = 273 + 37

= 310 K

To solve further we would have to use the Third gas law. The third gas law or the Gay-Lussacs law tells us that

P1 / T1 = P2 / T2

We would have to put in the values in the formula that we have above

[tex]\frac{2.3 atm}{273K} = \frac{p2}{310}[/tex]

from here would cross multiply

310 * 2.3 = p2 * 273

divide through by 273

713 / 273

P2 = 2.61

The pressure is given as 2.61

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Answer:

Explanation:

The lattice energy (LE) of MgCl₂ can be calculated using the Born-Haber cycle, which relates the lattice energy to other thermodynamic quantities such as enthalpy of sublimation, ionization energy, electron affinity, and heat of formation.

The equation for the lattice energy of MgCl₂ is:

LE(MgCl₂) = H(sublimation of Mg) + IE(Mg) + EA(Cl) + 1/2 H₂(Cl₂) - DH(f)(MgCl₂)

where H(sublimation of Mg) is the enthalpy of sublimation of Mg, IE(Mg) is the ionization energy of Mg, EA(Cl) is the electron affinity of Cl, H₂(Cl₂) is the heat of formation of Cl₂, and DH(f)(MgCl₂) is the heat of formation of MgCl₂.

We are given the following equations:

Mg(g) → Mg2+(g) + 2 e ΔH₁ = +2188 kJ/mol

Cl(g) + e → Cl-(g) ΔH₂ = -337 kJ/mol

Mg(s) + Cl₂(g) → MgCl₂(s) ΔH₃ = -642 kJ/mol

Using these equations, we can calculate the values of H(sublimation of Mg), IE(Mg), EA(Cl), H₂(Cl₂), and DH(f)(MgCl₂) as follows:

H(sublimation of Mg) = ΔH₂(Mg(g)) + 1/2 ΔH₃(Cl₂(g)) - ΔH₁(Mg2+(g)) = -2220 kJ/mol

IE(Mg) = ΔH₁(Mg(g)) = +2188 kJ/mol

EA(Cl) = ΔH₂(Cl(g)) = -337 kJ/mol

H₂(Cl₂) = 0 (since Cl₂ is in the gas phase)

DH(f)(MgCl₂) = ΔH₃(Mg(s), Cl₂(g), MgCl₂(s)) = -642 kJ/mol

Substituting these values into the equation for the lattice energy, we get:

LE(MgCl₂) = -2220 + 2188 - 337 + 1/2(0) - (-642) = -3509 kJ/mol

Therefore, the lattice energy of MgCl₂ is approximately -3509 kJ/mol.

Answer:

First, we need to calculate the change in temperature:

ΔT = final temperature - initial temperature

ΔT = 85 °C - 35 °C

ΔT = 50 °C

Next, we can use the following formula to calculate the heat energy required:

Q = m·C·ΔT

where Q is the heat energy in calories, m is the mass of the sample in grams, C is the specific heat in cal/g °C, and ΔT is the change in temperature in °C.

Plugging in the given values, we get:

Q = 35.0 g · 0.108 cal/g °C · 50 °C

Q = 189.0 calories

Therefore, 189.0 calories are required to raise the temperature of the sample from 35 °C to 85 °C