The heat of vaporization for water is 40. 7 kJ/mol. How much heat energy must 150. 0 g of water absorb to boil away completely? Use the formula mc001-1. Jpg. 339. 2 kJ 381. 6 kJ 610. 5 kJ 6,105 kJ.

The final pressure in the water bottle at 45 °C will be 1044 mm Hg, assuming the volume doesn't change using combined gas law.

Thus, the combined gas law can be used to estimate the final pressure  which is (P1 x V1) / T1 = (P2 x V2) / T2 where P1 is equal to 772 mm Hg, V1 is equal to 490.0 mL, and T1 is equal to 292.15 K. V2 is equal to V1 = 490.0 mL assuming the volume doesn't change, and the final temperature in Kelvin is equal to 318.15 K.

The equation of combined gas law can be rearranged to solve for P2 which is final pressure:

P2 = (P1 x V1 x T2) / (V2 x T1)

P2 = (772 mm Hg x 490.0 mL x 318.15 K) / (490.0 mL x 292.15 K)

P2 = 1044 mm Hg

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The first and last dissociation reactions of phosphoric acid are of no significance to pH buffering in cells because they occur at extreme pH values that are outside the physiological pH range.

Phosphoric acid (\(H_{3}PO_{4}\)) has three dissociation reactions:

1. \(H_{3}PO_{4}\) → \(H_{2}PO_{4}^{-}\) + H+
2. \(H_{2}PO_{4}^{-}\) → \(HPO_{4}^{-}\) + H+
3. \(HPO_{4}^{-}\)  → \(PO_{4}^{3-}\)  + H+

The first reaction occurs at a very low pH (pKa1 ≈ 2.15), and the last reaction occurs at a high pH (pKa3 ≈ 12.3). Cellular pH is maintained within a narrow physiological range (approximately 7.2-7.4). The second dissociation reaction (pKa2 ≈ 7.2) is the most relevant for cellular pH buffering since it occurs within this physiological pH range. Therefore, the first and last dissociation reactions of phosphoric acid are of no significance to pH buffering in cells as they do not contribute to maintaining the cellular pH within the required range.

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Rounding is a critical concept when dealing with medication. The rule of rounding is to round to the nearest whole number if the measurement is in milliliters per hour (ml/hr) or drops per minute (gtts/min). If the measurement is not in ml/hr or gtts/min, then round to the tenth place.

1. Order: Axid 0.3g p.o. ahs Supply: Axid 150mg capsules Give: The medication supply is given in milligrams, and the medication dosage is given in grams. Therefore, to convert 0.3g to mg, we need to multiply 0.3 by 1000 (since 1g = 1000mg).

Hence, 0.3g is equivalent to 300mg. Since the medication supply is given in 150mg capsules, we need to determine how many capsules are required to meet the dosage. 300mg/150mg per capsule = 2 capsules. Thus, the medication order is to give 2 capsules.

2. Order: Amoxcil 125mg p.o. q8h Supply: A bottle of Amoxcil powder says add 12ml to yield a solution of 50mg/ml Give: To calculate the milliliters of the Amoxcil solution to give, we need to know the volume of the medication needed to yield the required dosage.

Since the medication dosage is 125mg, we need to divide 125 by 50 to determine the milliliters of the solution required. 125mg/50mg per ml = 2.5ml. Hence, we need to give 2.5ml of the Amoxcil solution. The medication is in milliliters, so we need to round it to the nearest whole number. Therefore, the answer is 3 ml.

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The percent ionization of formic acid increases by 38.46% when potassium formate is added to the solution.

Here is the calculation of the percent ionization of an aqueous solution containing 0.13 M formic acid, HCOOH(aq), and 0.11 M potassium formate, HCOOK(aq). For formic acid, Ka = 1.8:

Initial concentrations:

HCOOH = 0.13 M

HCOOK = 0.11 M

Equilibrium concentrations:

HCOOH = 0.13 - x M

HCOOK = 0.11 + x M

H+ = x M

HCOO- = x M

Where x is the amount of formic acid that ionizes.

Ka = [H+][HCOO-] / [HCOOH]

1.8 = x^2 / 0.13 - x

x^2 + 2.34x - 2.34 = 0

(x + 4.68)(x - 0.5) = 0

x = 0.5

Therefore, the percent ionization of the formic acid solution is:

percent ionization = (x / [HCOOH]) * 100% = (0.5 / 0.13) * 100% = 38.46%

The presence of potassium formate in the solution increases the percent ionization of formic acid because potassium formate is a base. Bases react with acids to form water and a salt. In this case, potassium formate reacts with formic acid to form potassium hydrogen formate and water. This reaction removes some of the formic acid from the solution, which increases the concentration of the hydrogen ions and the percent ionization of the formic acid solution.

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The 85th percentile of blood pressure readings is approximately 137.64. a. To calculate the z-score for a blood pressure reading of 140, we can use the formula:

z = (x - μ) / σ

where x is the value (140 in this case), μ is the mean (121), and σ is the standard deviation (16).

Substituting the values into the formula:

z = (140 - 121) / 16

z ≈ 1.19 (rounded to two decimal places)

b. To find the proportion of Canadians with high blood pressure, we need to calculate the area under the normal distribution curve for values above 140. This can be done by finding the cumulative probability using the z-score.

Using a standard normal distribution table or a calculator, we can find that the cumulative probability corresponding to a z-score of 1.19 is approximately 0.881.

Therefore, the proportion of Canadians with high blood pressure is approximately 0.881 (rounded to four decimal places).

c. To find the proportion of Canadians with systolic blood pressure in the range from 100 to 140, we need to calculate the area under the normal distribution curve between these two values.

Using the z-scores corresponding to 100 and 140, we can find the cumulative probabilities for each value. The cumulative probability for a z-score of -1.25 (corresponding to 100) is approximately 0.105, and the cumulative probability for a z-score of 1.19 (corresponding to 140) is approximately 0.881 (as calculated in part b).

The proportion with systolic blood pressure between 100 and 140 is the difference between these two probabilities:

Proportion = 0.881 - 0.105 ≈ 0.776 (rounded to four decimal places)

d. The 85th percentile represents the value below which 85% of the blood pressure readings fall. To find the 85th percentile, we need to determine the z-score that corresponds to an area of 0.85 under the normal distribution curve.

Using a standard normal distribution table or a calculator, we can find that the z-score corresponding to an area of 0.85 is approximately 1.04.

To find the actual blood pressure reading at the 85th percentile, we can use the z-score formula:

x = μ + (z * σ)

Substituting the values:

x = 121 + (1.04 * 16)

x ≈ 137.64

Therefore, the 85th percentile of blood pressure readings is approximately 137.64.

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The difference in binding energy per nucleon between 23/11 Na and 23/12 Mg can be calculated by finding the total binding energy for each isobar and dividing it by the respective number of nucleons.

To calculate the difference in binding energy per nucleon between the isobars 23/11 Na and 23/12 Mg, we need to find the total binding energy for each isobar and then divide it by the respective number of nucleons.

The atomic mass of 23/11 Na is 23, which means it has 23 nucleons (protons and neutrons). The atomic number is 11, indicating it has 11 protons.

The atomic mass of 23/12 Mg is also 23, so it has 23 nucleons. However, the atomic number is 12, indicating it has 12 protons.

We can use the equation:

Binding Energy per Nucleon = (Total Binding Energy) / (Number of Nucleons)

To find the total binding energy, we can consult a table or use an approximate average value. Let's assume the average binding energy per nucleon for both elements is 8.5 MeV (million electron volts).

For 23/11 Na:

Binding Energy per Nucleon = (Total Binding Energy of Na) / (Number of Nucleons)

                         = (8.5 MeV) / (23 nucleons)

For 23/12 Mg:

Binding Energy per Nucleon = (Total Binding Energy of Mg) / (Number of Nucleons)

                         = (8.5 MeV) / (23 nucleons)

The difference in binding energy per nucleon can then be calculated by subtracting the value for Na from the value for Mg.

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12,005 barium atoms would have to be laid side by side to span a distance of 2.60 mm

Barium can be defined as a chemical element with the symbol Ba and atomic number 56.

To find the number of barium atoms that would have to be laid side by side to span a distance of 2.60 mm.

we need to divide the distance by the radius of one barium atom:

2.60 mm / 217 pm = 12,005

Therefore, 12,005 barium atoms would have to be laid side by side to span a distance of 2.60 mm.

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NI3 has the highest molar mass = 394.707 grams

Molecular mass = Mass of the atom x the number of atoms

Atomic mass of H = 1.00 amu

O = 15.99 amu

P = 30.97 amu

= 3(Mass of H) + 1(Mass of P) + 4(Mass of O)

= 3(1)+ 1(30.97) + 4(15.99)

= 3 + 30.97 + 63.96

= 97.93 gram

= 1 (mass of N) + 3(mass of I)

= 1(14.0067)+3(126.9)

= 394.7067 grams

= 1(atomic mass of Fe) + 3(atomic mass of Cl)

= 1(55.845)+3(35.45)

=  55.845 + 106.35

= 162.195 grams

KCl = 1(mass of K) + 1(mass of Cl)

= 1(39.09) + 35.45

= 74.54 gram

Out of these NI3 has the highest molar mass = 394.707 grams

Hence NI3 is a correct answer.

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Answer:

The balanced chemical equation for the reaction you provided is: CaC2 + 2H2O → Ca(OH)2 + C2H2.

According to this equation, for every mole of CaC2 that reacts with 2 moles of H2O, one mole of C2H2 is formed. So if 1.25 moles of C2H2 are formed, it means that 1.25 moles of CaC2 reacted with 2.5 moles of H2O.

Therefore, 2.5 moles of water are consumed when 1.25 moles of C2H2 are formed.

Steps:

Here are the steps I took to determine how many moles of water are consumed when 1.25 moles of C2H2 are formed:

1. First, I looked at the balanced chemical equation you provided: CaC2 + 2H2O → Ca(OH)2 + C2H2.

2. From this equation, I could see that for every mole of CaC2 that reacts with 2 moles of H2O, one mole of C2H2 is formed. This is known as the stoichiometry of the reaction.

3. Since you mentioned that 1.25 moles of C2H2 are formed, I used the stoichiometry of the reaction to determine that 1.25 moles of CaC2 must have reacted with 2.5 moles of H2O (since for every mole of CaC2 that reacts, 2 moles of H2O are consumed).

4. Therefore, I concluded that 2.5 moles of water are consumed when 1.25 moles of C2H2 are formed.

I hope this explanation helps! Let me know if you have any further questions.

The reactivity of a substance with oxygen is a chemical, not a physical property. The reason it is called a chemical property is that it relies on its electron configuration to determine how it behaves around other substances.

A chemical property is a property of any material that becomes apparent during or after a chemical reaction; that is, any quality that can only be determined by changing the chemical properties of a substance.

Oxygen is a very reactive element, is highly paramagnetic and readily combines with other elements. One of the most important chemical properties of oxygen is that it promotes combustion. Even at room temperature, oxygen binds to elements and forms e.g. rust.

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Aluminum + hydrochloride acid yield aluminum chloride + hydrogen gas therefore the balanced equation is 2Al+6HCl→2AlCl₃+3H₂.

This is referred to as the process in which one or more substances which are called the reactants, are converted to one or more different substances known as the products and it involves the exchange of electrons between elements.

In this scenario, when Aluminum and hydrochloride acid reacts they form aluminum chloride and hydrogen gas which is the product and the equation can be seen above.

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By using ideal gas law , water at 23°C and 745 mm Hg of total pressure, the volume it would have taken up would have been **0.87 L**.

In the limit of low pressures and high temperatures, the ideal gas law describes a relationship between a gas's pressure P, volume V, and temperature T such that the molecules of the gas move practically independently of one another. It can be derived from the kinetic theory of gases and is predicated on the following premises: (1) the gas is made up of numerous molecules that move randomly and in accordance with Newton's laws of motion; (2) the volume of the molecules is negligibly small in comparison to the volume occupied by the gas; and (3) no forces act on the molecules other than elastic collisions that last for a negligibly short period of time.

The ideal gas law can be used to resolve this issue. PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is temperature in Kelvin, is the formula for the ideal gas law.

P = 760 mmHg and T = 273 K are the STP (Standard Temperature and Pressure) conditions.

It is possible to compute the amount of dry gas at STP as follows:

V1 = nRT/P

where V1 is the dry gas volume at STP.

n/V = P/RT

The moles in a gas per unit volume are denoted by the ratio n/V.

In the case of dry gas, n/V is equal to (760 mmHg)/(62.36 L.mmHg-1.K⁻¹x 273 K) = 0.0282 mol/L.

The formula is (21 mmHg)/(62.36 L.mmHg-1.K⁻¹ x 296 K) = 0.00089 mol/L for water vapour.

The following formula can be used to determine the total number of moles per unit volume of gas at 745 mmHg and 296 K:

n/V = P/RT

n/V for total gas is equal to (0.0257 mol/L)/745 mmHg/(62.36 L.mmHg-1.K⁻¹ x 296 K).

The following formula can be used to get the number of moles per unit volume of dry gas at 745 mmHg and 296 K:

n/V for dry gas at 745 mmHg and 296 K equals (Ptotal - Pvapour)/PSTP for dry gas at STP.

where P vapour is the water vapour pressure at 23 degrees Celsius, which is 21 mmHg.

(0.0282 mol/L) x (745 - 21)/760 equals n/V for dry gas at 745 mmHg and 296 K.

= 0.0265 mol/L

The following formula can be used to get the volume of dry gas at 745 mmHg and 296 K:

V2 = nRT/P

where V2 is the dry gas volume at 745 mmHg and 296 °C.

V2 is equal to 0.0265 mol/L times 62.36 L.mmHg-1.K-1 x 296 K and 745 mmHg.

= **0.87 L**

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.

The pH after the following volumes of titrant have been added 0 ml 20 mL 59. 1 mL 60. 0 mL 71. 4 mL 73. 4 mL are 11.89, 11.89, 8.45, 8.45, 7.98, 8.95 respectively.

The reaction between NH3 and HCl can be represented by the following equation: NH3 + HCl → NH4+ + Cl-

To calculate the pH after different volumes of titrant have been added, we need to determine the amount of titrant that has reacted with the analyte and the resulting concentration of the products.

A. 0 mL of titrant (initial state)

At the start, there is no titrant added to the analyte, so the concentration of NH3 is 0.050 M. NH3 is a weak base, so we can use the Kb expression to calculate the concentration of OH-:

\(Kb = [NH4+][OH-] / [NH3]\)

\(1.8 * 10^{-5} = x^2 / (0.050 - x)\)

initial concentration of NH3 is much greater than the initial concentration of HCl, we can assume that the concentration of NH3 does not change significantly during the titration.

\(Kb = x^2 / 0.050\\x = \sqrt{Kb * 0.050} = 1.3 * 10^{-3} M\)

The concentration of OH- is equal to \(1.3 * 10^{-3} M\), so we can calculate the pH:

\(pH = 14 - pOH = 14 - (-log[OH-]) = 11.89\)

Therefore, the pH at the start of the titration is 11.89.

B. 20 mL of titrant

After adding 20 mL of 0.025 M HCl, the volume of the solution is 50 mL (30 mL NH3 + 20 mL HCl). The moles of HCl added is:

moles of HCl = volume x concentration = 0.020 L x 0.025 mol/L = 5 x 10^-4 mol

Since the reaction is a 1:1 reaction, the moles of NH3 remaining is equal to the moles of HCl added.

concentration of NH3 = moles of NH3 / volume of NH3 = (0.050 mol/L x 0.030 L - 5 x 10^-4 mol) / 0.030 L = 0.048 mol/L

Since the concentration of NH3 has decreased, we need to recalculate the concentration of OH- using the new concentration of NH3:

\(Kb = [NH4+][OH-] / [NH3]\\1.8 * 10^{-5} = x^2 / (0.048 - x)\)

Solving for x, we get:

\(x = 1.3 * 10^{-3} M\)

The concentration of OH- is still \(x = 1.3 * 10^{-3} M\), so we can calculate the pH:

pH = 14 - pOH = 14 - (-log[OH-]) = 11.89

Therefore, the pH after adding 20 mL of titrant is still 11.89.

Similarly for  C. 59.1 mL of titrant

The pH after adding 59.1 mL of titrant is 8.45.

D. 60 mL of titrant

The pH after adding 60 mL of titrant is 8.45.

E. 71.4 mL of titrant

The pH after adding 71.4 mL of titrant is 7.98.

F. 73.4 mL of titrant

The pH after adding 73.4 mL of titrant is 8.95.

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Answer:

A. the electric motor

Explanation:

A cell is a biological molecule which is the basic and functional unit of life. Cells undergo series of processes to function appropriately. ATP is an acronym for adenosine triphosphate, which is the source of energy for various cell processes.

In the given mechanical system, the electric motor provides the energy required energy to drive the system. Therefore, the electric motor has the same major function of providing energy for the system as the ATP in a cell.

Answer:

evaporation takes place

The amount of a sample of 100 grams of a radioactive isotope that would remain after 732 days would be 14.0625 grams.

Given, the Half-life of the radioactive isotope = 273 days.Time elapsed = 732 days.Initial quantity or sample = 100 grams. Let's determine how many half-lives have passed since 732 days: Number of half-lives = (time elapsed) / (half-life)= 732 / 273 ≈ 2.683

Half-life #1: After the first half-life of 273 days, the sample will be halved. Therefore, after 273 days, the quantity remaining will be 1/2 * 100g = 50g

Half-life #2: After the second half-life of 273 days, the sample will be halved again. Therefore, after 546 days, the quantity remaining will be 1/2 * 50g = 25gHalf-life #3: After the third half-life of 273 days, the sample will be halved again.

Therefore, after 819 days, the quantity remaining will be 1/2 * 25g = 12.5gHowever, the time elapsed from 819 days to 732 days is 87 days. This time interval is less than the half-life. As a result, it is critical to calculate the amount that would be left over after 732 days using a different method. Let us consider the remaining amount from 819 days (12.5g) as the new initial quantity for the remaining 87 days. The half-life of the radioactive isotope is 273 days.

Therefore, the rate of decay for each day will be: Rate of decay per day = (1/2)^(1/273)≈ 0.002540401Therefore, the amount of the sample remaining after 87 days (or 0.3195 half-lives) can be calculated using the following formula: Q = Q0(0.5)^(t/h)where Q0 is the original quantity, Q is the remaining quantity after time t, and h is the half-life of the isotope. Q = 12.5g × (0.5)^(0.3195)Q ≈ 6.5625g

Therefore, the total amount of the sample remaining after 732 days can be found by adding up the amounts of the sample remaining from each half-life: Total remaining = 50g + 25g + 6.5625gTotal remaining ≈ 81.5625 the amount of a sample of 100 grams of a radioactive isotope that would remain after 732 days would be 14.0625 grams.

After 732 days, the sample would have decayed by three half-lives (819 days) and an additional 87 days. As a result, 81.5625g of the sample will remain after 732 days. Therefore, 100g - 81.5625g = 18.4375g of the sample would have decayed in 732 days.

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The aromatic region in the spectrum of biphenyl is different from the starting material. The starting material, which is likely a substituted benzene, would show a single peak in the aromatic region. Biphenyl, on the other hand, would show two peaks in the aromatic region.

The difference in the aromatic region can be attributed to the presence of two aromatic rings in biphenyl. Each ring has its own set of hydrogen atoms, which results in two separate peaks. The peak corresponding to the hydrogens on the ortho and para positions (H-2, H-3, H-5, H-6) will appear at a higher field (lower ppm) due to deshielding from the adjacent ring. The peak corresponding to the hydrogens on the meta positions (H-1, H-4) will appear at a lower field (higher ppm) due to shielding from the adjacent ring.

As for how many hydrogen atoms should be integrated for in the spectrum of biphenyl, there should be 10 hydrogen atoms integrated for, 4 on one ring and 6 on the other.

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Answer:

A. Coal

Explanation:

\( Volume_{rock} \) = water displaced by rock

Water displaced by rock = volume of water after rock is dropped into the cylinder - volume of water before the rock was dropped into the water

Water displaced by the rock = 180 ml - 150 ml = 30 ml

\( Volume_{rock} = 30 ml \)

Density of rock:

40 grams => 30 ml

x grams => 1 ml

Cross multiply

1*40 = 30*x

40 = 30x

40/30 = 30x/30

1.3 = x

Density of rock = 1.3 g per 1 ml

Recall: 1 ml = 1 cm³

Therefore,

Density of the rock = 1.3 g/cm³

1.3 g/cm³ falls within the range of 1.1 - 1.4 g/cm³

Therefore, the rock is identified as Coal.

The Alka-Seltzer tablet contains 59.5% sodium hydrogen carbonate (NaHCO3) by weight.

The interaction between sodium bicarbonate and citric acid, which occurs when a 0.123-g Alka-Seltzer tablet is dissolved in water at 27 degrees Celsius, will result in the production of 0.0248-g of carbon dioxide under your particular conditions of pressure and temperature. When students first dissolve the pill in pure water, there is an excess of sodium bicarbonate, making the acid (H+) the limiting reactant. There is a maximum amount of NaHCO3 that can react. Carbon dioxide gas is created when an acid and alka - seltzer combine. It fizzed and bubbled when you added the tablet to the vinegar or water. It will take less time to complete and will fizz more vigorously if the reaction is progressing more quickly.

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Answer: no

Explanation:

It aint got the dingdong and besides whos the mommyy??

the sun produces high frequency waves like gamma rays.

Answer:

400 joules sisisiss

Explanation:

Answer:

rabbit is one of the sweet animal

in the world

Answer:

Given above are noble gases and other noble gases would be xenon, radon, oganesson, krypton.

hope it helps!

Answer: A

Explanation: just did it

Actually the main difference:

Homogeneous mixture is a mixture consisting of constituent substances that are mixed evenly.  As a result, each part of the mixture has the same properties

 In contrast to heterogeneous mixtures where the constituent substances are not mixed evenly.  Thus, there are parts of the mixture that have different properties.

Differences between Homogeneous and Heterogeneous Mixtures in detail

1. Differences Between Homogeneous and Heterogeneous Mixtures Based on Their Definitions

Homogeneous is a mixture that is uniform in all its parts and forms a single phase.  An example of a homogeneous mixture is air.  In addition, homogeneous mixtures can also be commonly referred to as solutions.

Meanwhile, heterogeneous mixture is a mixture that is not similar or not uniform, and the formation of two or more phases, as well as the existence of a clear boundary between the phases.  Examples of heterogeneous mixtures include oil and water, a mixture of lime and sand, then a mixture of iron powder and carbon, and many others.

2. Differences Between Homogeneous and Heterogeneous Mixtures Based on Their Characteristics

The characteristics of this homogeneous mixture are in the form of constituent particles that cannot be distinguished from one another, have almost the same color, and have a similar taste.  Not only that, substances that have been mixed have the same ratio, have the same concentration level, and are in the form of solids, gases, and liquids.

And this mixture cannot be separated, if you use a mechanical method, but you can separate it when you use a more difficult method.  The example of such separation is similar to that of distillation.  That is one of the differences between homogeneous and heterogeneous mixtures based on their characteristics.

3. Difference between Homogeneous and Heterogeneous Mixtures Based on Their Properties

The nature of this homogeneous mixture has the property that if every part of a homogeneous mixture is often the same, both in terms of color, taste, and comparison.  An example of the nature of the homogeneous mixture is a spoonful of sugar dissolved in water.

While the nature of a heterogeneous mixture is a mixture of two or more substances, which have properties, the constituent substances are not the same or the alias is not uniform.  So that the two mixed substances can still be distinguished by the particles.

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Water's ability to act as a good solvent is best explained by its polar nature. Water molecules are composed of two hydrogen atoms covalently bonded to an oxygen atom.

The oxygen atom has a higher electronegativity than the hydrogen atoms which gives the water molecule a slightly negative charge on the oxygen side and a slightly positive charge on the hydrogen side.

This polarity allows water molecules to interact with other polar molecules, forming hydrogen bonds and allowing them to dissolve a variety of substances. The hydrogen bonds form between the oxygen of one molecule and the hydrogen of another, allowing water molecules to surround and interact with the molecules of the substance being dissolved.

This polarity also allows water molecules to move freely making them highly mobile, allowing them to form a homogeneous solution with the dissolved substances. This ability of water to dissolve a variety of substances is what makes it a good solvent.

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Answer:

Independent variable

Explanation:

Answer:

Explanation:

Sodium Chloride = Na⁺ + Cl⁻ = NaCl

Lithium bromide = Li⁺ + Br⁻ = LiBr

Magnesium fluoride = Mg ⁺²  +  F⁻ = Mg₂F.

Potassium Oxide = K⁺ + O⁻² = K₂O

Calcium sulphide  = Ca⁺² + S⁻² = CaS

Aluminum Iodide = Al⁺³ + I⁻ = Al I₃

Barium bromide = Ba⁺² + Br⁻ = BaBr₂

Aluminum sulphide = Al⁺³ + S⁻ = Al₂S₃

Calcium Phosphide = Ca⁺² + P⁻³ = Ca₃P₂

Lithium Nitride = Li⁺ + N⁻³ = Li₃N

Magnesium Oxide = Mg⁺² + O⁻² = MgO

Aluminum Fluoride = Al⁺³ + F⁻ = AlF₃

Lithium Oxide = Li⁺ + O⁻² = Li₂O

Beryllium iodide = Be⁺² + I⁻ = BeI₂

Answer: I'm in no way like an expert but I've heard water goes from an elevated area to a lower area. So it should be arrows that start at a more elevated to less elevated. Like white to brown, and orange to yellow, and stuff like that. I hope that helped? Anyway, have a lovely day!

Explanation:

The required pH of the solution is 0.5655.

A chemist dissolves 171 mg of pure nitric acid in enough water to make up 100 mL of solution. We are to calculate the pH of the solution. We know that nitric acid is a strong acid, so we can assume that it dissociates completely in water.The chemical equation for the dissociation of nitric acid in water is given by:HNO3(aq) + H2O(l) → H3O+(aq) + NO3-(aq)We know that the concentration of nitric acid is given by:Concentration = mass / molar massVolume = 100 mL = 0.100 LTherefore, the concentration of the nitric acid solution is given by:Concentration = 171 mg / 63.01 g mol-1 × 0.100 L= 0.2712 MThe concentration of H3O+ is equal to the concentration of the nitric acid. Thus, [H3O+] = 0.2712 MThe pH of the solution can be calculated using the formula:pH = -log[H3O+]pH = -log[0.2712]pH = 0.5655 (rounded to four significant figures)Therefore, the pH of the solution is 0.5655.

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A brownfield is a property, the expansion, redevelopment, or reuse of which may be complicated by the presence or potential presence of a hazardous substance, pollutant, or contaminant.

A “brownfield” generally refers to a parcel of land that was previously used for industrial purposes and which is contaminated by low concentrations of hazardous chemicals.

A brownfield development requires more work and investment upfront: existing structures may have to be demolished, materials must be removed, and developers may have to engage in extensive environmental cleanup to remove pollutants.

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