Calculate the mass percent of solute in each solution.
Calculate the mass percent of 3.87 g KCl
dissolved in 55.6 g H2O.

mass percent:
%

Calculate the mass percent of 16.5 g KNO3
dissolved in 848 g H2O.

mass percent:

The pH of the solution is 2.46. This indicates that the solution is acidic.

To calculate the pH of a solution of a weak acid such as HA, we need to use the equilibrium expression for the dissociation of the acid:

HA ⇌ H+ + A-

The acid dissociation constant, Ka, is given as[tex]4.2\times 10^{-6}[/tex], which is defined as:

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

At equilibrium, the concentration of HA that dissociates is much smaller than the initial concentration, so we can assume that the concentration of HA remaining is equal to its initial concentration, which is 0.22 M. Let x be the concentration of H+ ions that are formed when HA dissociates. Then the equilibrium concentration of A- ions is also x, since the acid is monoprotic.

Substituting these values into the expression for Ka, we get:

[tex]4.2\times 10^-6 = x^2 / (0.22 - x)[/tex]

Solving for x using the quadratic equation gives x = 3.451x10^-3 M.

The pH of the solution can be calculated using the formula:

pH = -log[H+]

Therefore, the pH of the solution is: pH = -log(3.451x10^-3) = 2.46

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Explanation:

To calculate the pH of a buffer with a known pKa for the conjugate acid, the Henderson-Hasselbach equation can be used to obtain the pH of the buffer:

[tex]pH = pK_a + log_{10}(\frac{[A^-]}{[HA]} )[/tex]

[A-] is the concentration of the conjugate base and [HA] is the concentration of the conjugate acid.

Yes, a chemical reaction will take place if a bar of silver is placed into a solution containing [tex]Cu^{2+}[/tex] ions.

The reaction that will occur is a type of single replacement reaction. The silver metal will react with the copper ions in the solution, causing the copper ions to be reduced and the silver metal to be oxidized. The overall balanced chemical equation for this reaction is:

Ag(s) + [tex]Cu^{2+}[/tex](aq) → Cu(s) + [tex]Ag^+[/tex](aq)

In this equation, Ag represents silver, Cu represents copper, and the (s) and (aq) notations indicate solid and aqueous states, respectively. The arrow indicates the direction of the reaction.

As a result of this reaction, a thin layer of copper metal will be deposited onto the surface of the silver bar, while silver ions will enter the solution. This reaction is often used in electroplating applications, where a layer of one metal is deposited onto the surface of another metal.

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A polar molecule is a type of chemical compound where there is an uneven distribution of electrons among the covalently bound atoms.

Thus, The term "polarity" refers to how unlike two molecules' electrical poles are from one another. If they are quite dissimilar, the species is considered to be a highly polar molecule.

Some chemical species, such chains of carbon molecules, are believed to be nonpolar molecules because they exchange electrons equally. A molecule is often classified as polar or nonpolar based on the sum of all of its bonds taken into account.

The atom with the higher electronegativity will often draw more electrons when it is linked to another atom. If the difference is small, a nonpolar bond can be used.

Thus, A polar molecule is a type of chemical compound where there is an uneven distribution of electrons among the covalently bound atoms.

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Calculating the temperature change of an item when a specific quantity of heat is added to or withdrawn from it may be done using the equation q = mcT. According to the equation, the amount of heat, q, is determined by the object's mass, m, its specific heat capacity, c, and the temperature change, T.

We may utilise the equation T = q/mc to find the solution for the temperature change, T. By dividing both sides of the equation by mc, this equation may be created from the original equation, q = mcT. mc/mc is equal to 1 on the right while q/mc is equal to T on the left. T thus equals q/mc.

To use this formula, we requireT thus equals q/mc. We must first establish the values of q, m, and c before we can use this equation. In joules, the quantity of heat, q, may be calculated. The object's mass, m, can be expressed in kilogrammes. A table of specific heat capacities contains the object's c specific heat capacity. Once these numbers have been established, we may enter them into the equation and find the value of T.

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A pressure of 12.5 atm is required to pop a kernel of popcorn. It's worth noting that this is a very high pressure, and it's the build-up of this pressure that causes the kernel to "pop".

To calculate the pressure required to pop a kernel of popcorn, we need to use the ideal gas law, which relates the pressure (P), volume (V), temperature (T), and number of moles of gas (n) in a system:

PV = nRT

Where R is the universal gas constant. We can assume that the temperature inside the popcorn kernel just before it pops is the same as the temperature in the microwave (100°C or 373 K). We also know that the only gas inside the kernel is water vapor, which has a molecular weight of 18 g/mol.

To calculate the number of moles of water vapor that is produced during the popping of the kernel, we can use the mass of water in the popcorn:

n = m/M

Where m is the mass of water (16.7 g) and M is the molar mass of water (18 g/mol). This gives us:

n = 16.7 g / 18 g/mol = 0.9286 mol

Now we can rearrange the ideal gas law to solve for pressure:

P = nRT / V

Substituting the known values, we get:

P = (0.9286 mol)(0.0821 Latm/molK)(373 K) / 3.1 L

P = 12.5 atm

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The new volume of the gas is 56.6 liters when the temperature is raised to 325 K and the pressure is lowered to 1.2 atm.

PV = nRT

Where R is the ideal gas constant. Since the number of moles is constant in this problem, we can simplify the ideal gas law to:

P1V1/T1 = P2V2/T2

Where the subscripts 1 and 2 refer to the initial and final states of the gas, respectively.

We can now plug in the given values for the initial state of the gas:

P1 = 2.3 atm

V1 = 25 L

T1 = 299 K

And the given values for the final state of the gas:

P2 = 1.2 atm

T2 = 325 K

We can then solve for V2:

P1V1/T1 = P2V2/T2

(2.3 atm)(25 L)/(299 K) = (1.2 atm)V2/(325 K)

V2 = (2.3 atm)(25 L)(325 K)/(1.2 atm)(299 K)

V2 = 56.6 L (rounded to three significant figures)

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The relationship  between the density (d) of a gas and the rate at which the gas diffuses is r = k/√d.

option B.

Graham's law of diffusion, also known as Graham's law of effusion, states that the rate of diffusion or effusion of a gas is inversely proportional to the square root of its gas density.

In other words, lighter gases diffuse or effuse faster than heavier gases at the same temperature and pressure.

Mathematically, this law is written as;

r ∝ 1/√d

r = k/√d

where;

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Answer:700 ml

Explanation:

How much water would I need to add to 500 mL of a 2.4 M KCl solution to make a 1.0 M solution? Need to add 1200 – 500 = 700 mL.

From the image that has been shown, the element is most likely aluminium.

Because the third electron is being taken from a different energy level than the first and second electrons, the quantum leap in ionization energy for aluminum occurs after IE3. In particular, the 2p subshell loses the third electron whereas the 3s subshell loses the first two electrons.

It requires less energy to remove the third electron than it took to remove the second electron because the 2p subshell is at a lower energy level than the 3s subshell.

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The correct answer is (a) The half-life of your candy is 4 days.

It appears that the disappearance of your candy follows first-order kinetics. This is because the rate at which you are eating the candy is proportional to the amount you have left, which is a characteristic of first-order reactions.

We can calculate the half-life of the candy as follows: t1/2 = (ln2)/k

where t1/2 is the half-life, k is the rate constant, and ln is the natural logarithm.

Assuming that the candy is consumed completely after 14 days, we know that the remaining amount after 8 days is 1/4 of the initial amount.  The fraction of candy remaining after 6 days (i.e., the half-life) is:

1/2 = (1/4)e^(-k*8)

Solving for k, we get:

k = ln(2)/8 = 0.0866 day^-1

Substituting this value for k in the half-life equation, we get:

t1/2 = (ln2)/0.0866 day^-1 ≈ 8.0 days

hence, the correct option is  (a) The half-life of your candy is 4 days.

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The volume of 0.360 M KOH that is needed to neutralize 10.4 mL of 0.185 M H₂SO₄ is 10.68 ml.

Neutralization is a chemical reaction in which acid and base react to form salt and water. Hydrogen (H⁺) ions and hydroxide (OH⁻ ions) react with each other to form water.

The strong acid and strong base neutralization have a pH value of 7.

The beaker gets warm which indicates that the reaction between acid and base is an exothermic reaction releasing heat energy into the surroundings.

Given,

Concentration of KOH = 0.36M

Concentration of H₂SO₄ = 0.185M

Volume of H₂SO₄ = 10.4 ml

In the reaction, one mole of H₂SO₄ needs 2 moles of KOH

2 × Concentration of H₂SO₄ × Volume of H₂SO₄ = Concentration of KOH × Volume of KOH

2 × 0.185 × 10.4 = 0.360 × Volume of KOH

Volume of KOH = 10.68 ml

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The side containing HCl will be the anode.

The [H+] at the HCl electrode will decrease until all the HCO3- ions on the other side of the cell are converted to H2CO3.

In a galvanic cell, the anode is the electrode where oxidation occurs, while the cathode is the electrode where reduction occurs.

While the reduction of H+ ions to make hydrogen gas takes place at the cathode, the process at the anode involves the oxidation of hydrogen to create H+ ions. Since HCl is a source of H+ ions, the side that contains HCl will serve as the anode. As a result, claim three is accurate.

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The concept molarity is used here to determine the pH after adding 12.6 mL of the base. The term molarity is an important method which is used to calculate the concentration of a solution. Here the pH is 0.501.

The term molarity is defined as the number of moles of the solute dissolved per litre of the solution. It is also called the molar concentration. It is represented as 'M' and its unit is mol / L.

Molarity is given as:

M = Number of moles / Volume of solution in liters

'n' of HCN = 36.9 × 1 L / 1000 mL × 0.423 = 0.0156 mol

'n' of Ba(OH)₂ = 12.6 × 1L / 1000 mL × 0.382 = 0.0048 mol

Excess H⁺ = 0.0108

Total volume = 36.9 + 12.6 = 49.5 mL = 0.0495 L

Concentration of H⁺ = 0.0156 / 0.0495 = 0.315

So pH is:

pH = - log[H⁺]

pH = - log[ 0.315] = 0.501

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To draw a Bohr Rutherford diagram for carbon-12, we would have the symbol C in the center, with 2 electrons in the first energy level and 4 electrons in the second energy level, arranged in pairs. Each energy level would be represented by a circle around the nucleus, with the appropriate number of electrons placed in the circles.

A Bohr Rutherford diagram is a visual representation of the electron arrangement in an atom. Carbon-12 has 6 protons and 6 neutrons, which means it has 6 electrons as well since it is a neutral atom. In a Bohr Rutherford diagram, the nucleus is represented by the symbol for the element and the protons and neutrons are shown as small circles inside the symbol. The electrons are represented as dots or circles around the symbol, in the order of increasing energy levels.

In the case of carbon-12, the first energy level can hold a maximum of 2 electrons, and the second energy level can hold a maximum of 8 electrons. Therefore, the first energy level will have 2 electrons, and the remaining 4 electrons will go in the second energy level.

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To determine which sample has the greatest mass, we need to use the molar mass of each substance to calculate the mass of each sample. The molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol).

The molar mass of camphor (C10H16O) is approximately 152.23 g/mol. Therefore, 0.37 mol of camphor has a mass of:

mass = moles x molar mass = 0.37 mol x 152.23 g/mol = 56.34 g

The molar mass of ammonia (NH3) is approximately 17.03 g/mol. Therefore, 4.2 mol of ammonia has a mass of:

mass = moles x molar mass = 4.2 mol x 17.03 g/mol = 71.43 g

The molar mass of krypton (Kr) is approximately 83.80 g/mol. Therefore, 9.3 mol of krypton has a mass of:

mass = moles x molar mass = 9.3 mol x 83.80 g/mol = 779.34 g

The molar mass of iodine vapor (I2) is approximately 253.81 g/mol. Therefore, 4.0 mol of iodine vapor has a mass of:

mass = moles x molar mass = 4.0 mol x 253.81 g/mol = 1015.24 g

The molar mass of formaldehyde (CH2O) is approximately 30.03 g/mol. Therefore, 1.6 mol of formaldehyde has a mass of:

mass = moles x molar mass = 1.6 mol x 30.03 g/mol = 48.05 g

Therefore, the sample with the greatest mass is 4.0 mol of iodine vapor, which has a mass of 1015.24 g.

Answer: 4.0 mol of iodine vapor, 12

Explanation: took the test

Exploring a Natural Wonder: The Crystal Clear Waters of Lake Serenity

Introduction:

Nestled in the picturesque landscape of our region lies a hidden gem, Lake Serenity. This calm body of water is not only a source of breathtaking beauty, but also a vital natural resource that deserves our attention. In this article, we'll delve into the reasons why people should be captivated by Lake Serenity, explore the factors contributing to its presence in our area, and shed light on the changing dynamics affecting its offerings.

Why should you care?

Lake Serenity is an irresistible attraction for nature lovers and wanderers alike. Its crystal clear waters provide a haven for a diverse array of flora and fauna, making it an ecosystem of immense ecological importance. From the beautiful sight of water lilies gently swaying in the breeze to the playful dances of fish beneath the glittering surface, Lake Serenity offers a refuge from the chaos of modern life. Connecting to this source not only provides a sense of calm, but also allows us to connect with nature on a deeper level, promoting personal well-being and enhancing a sense of environmental stewardship.

Local connection:

The presence of Lake Serenity in our area can be attributed to a combination of geological and hydrological factors. Thousands of years ago, glacial movements shaped the landscape and created a basin that eventually filled with water to create this beautiful lake. Located in an area rich in geological diversity, the lake owes its remarkable clarity to the surrounding rock formations, which act as natural filters, purifying the water and maintaining its pristine quality.

Change in dynamics:

While the beauty of Lake Serenity remains unblemished, it is not immune to the changing dynamics affecting our environment. Climate change and human activity pose challenges to the delicate balance of this natural resource. Rising temperatures and altered precipitation patterns have the potential to disrupt the lake's ecosystem and affect the growth and survival of its aquatic inhabitants. In addition, human interference, pollution and uncontrolled recreational activities can introduce harmful substances into the water, threatening the purity and long-term viability of this precious resource.

Protective and sustainable practices:

We recognize the value of Lake Serenity and it is critical for us to be actively involved in conservation efforts. By adopting sustainable practices, we can ensure the longevity of this natural resource for future generations to cherish. Responsible waste management, controlled development and the promotion of ecological recreational activities are essential steps to protect the integrity of the lake. A collaborative effort involving local communities, environmental organizations and government bodies can play a key role in implementing measures to protect and preserve Lake Serenity.

Conclusion:

Lake Serenity is a testimony to the amazing wonders that nature bestows upon us. Its crystal clear waters and vibrant ecosystem are a reminder of the importance of protecting our natural resources. By appreciating and respecting this unique resource, we not only benefit personally, but also contribute to the protection of our environment. Let's accept the responsibility to protect Lake Serenity and ensure that its pristine beauty remains a source of inspiration for generations to come.

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The biological process in an organism that produces methane is called methanogenesis.

Methanogenesis,  can as well be regarded as the biomethanation, which can be consiered as a form of anaerobic respirationwhereby there is a utilization of the carbon as the terminal electron acceptor, resulting in the production of methane.

It shoud be noted that Methanogenesis do follows the process of anaerobic respiration here there is a generation of methane as the final product of metabolism in theprocess the organic matter such as glucose is oxidized to CO2, however in hydrogenotrophic methanogenesis, H2 is oxidized to H+.

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complete qustion;

What is biological process in an organism that produces methane​

Answer:

Pb3O4

Explanation:

Element lead oxygen

Perc. 90.65%. (100-90.65)=9.35%

mol ratio. 90.65/207. 9.35/16

mol. 0.437923 0.584375

sim. rat.

0.437923/0.439723 0.584375/0.437923

ratio. 1 1.3

multiply the ratios by 3 (1×3). (1.3×3)

3 : 3.9

therefore the formula is Pb3O4.

I multiplied through by 3 so I can get a whole value/ a value easy to estimate.

I hope this helps.

The estimated standard reduction potential for the [tex][Au(SCN)_4]^-[/tex] to Au half-reaction is 0.746 V.

The standard reduction potential is described as the tendency for a chemical species to be reduced, and is measured in volts at standard conditions.

The half-reaction for the reduction of  to Au is given as :

[tex][Au(SCN)_4]^-[/tex] + 3e– → Au

K = 1/[[tex][Au(SCN)_4]^-[/tex]]

K  = 1/2.86×1045

We apply the Nernst equation to find  the standard reduction potential for the half-reaction:

[tex][Au(SCN)_4]^-[/tex]

E° = E° for (cell) - (0.0592 V/n) log(K)

E° for (cell) = E°(reduction) - E°(oxidation)

E° for (cell) = 1.52 V - 0 V = 1.52 V

Therefore, E° = 1.52 V - (0.0592 V/3) log(2.86×1045)

E° = 1.52 V - 0.774 V

E° = 0.746 V

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#complete question:

If the formation constant of [tex][Au(SCN)_4]^-[/tex] is 2.86×10^45 and the reduction potential of Au3+ to Au is 1.52 V under standard conditions, calculate an estimate of the standard reduction potential for the half-reaction below:

[tex][Au(SCN)_4]^-[/tex] (aq) + e– → [tex][Au(SCN)_3]^-[/tex](aq)

In the formula for the equilibrium concentration of H2O, solve for y and add the answer. The difference between the equilibrium concentration of H2O and the concentration of H2O produced in the first stage is then the contribution to [H2O] from the second ionisation step.

When might the second step's contribution be disregarded? When the second step's contribution is significantly less than the first step's contribution, the second step's participation might be overlooked.

When the contribution of the second step to the contribution of the first step is much less than 1, this happens. Calculating the proportion of the second step's contribution to the first step's contribution can be used to calculate this.

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Answer: 0.0920 M/s

Explanation: For every 1 mol of Cl2 used up, 2 mol of NOCl is produced. Thus, the rate of formation of NOCl is double the rate of Cl2 loss, which is 2*(4.60*10^-2) = 9.20*10^-2 M/s, or 0.0920 M/s.

Answer:

V=mass/Density

Explanation:

Since density is measured as grams/mL or grams/cm^3 and we are interested in converting mass to volume, we need to rewrite the equation.
Therefore, in order to convert the gram to mL, you just need to multiply the mass of the substance in grams with its density to get the volume in liters.

Answer:

Volume equals mass divided by density.

and

Mass equals density times volume.

Hope this helps :)

Pls brainliest...

The correct answer is option C) 468.

We can use the following formula to determine the effective (rms) velocity of H2S at 26.0°C:

v(rms) = sqrt((3 * R * T) / (M))

Where:

- v(rms) is the root mean square velocity

- R is the gas constant (8.314 J/(mol·K))

- T is the temperature in Kelvin (26.0 °C = 26.0 + 273.15 = 299.15 K)

- M is the molar mass of H2S (which can be calculated by summing the atomic masses of hydrogen (H) and sulfur (S))

The molar mass of hydrogen is approximately 1.008 g/mol and sulfur has an atomic mass of approximately 32.06 g/mol.

Molar mass of H2S = (2 * molar mass of H) + molar mass of S

               = (2 x 1.008 g/mol) + 32.06 g/mol

               = 34.076 g/mol

Now we plug the values ​​into the formula and solve for v(rms):

v(rms) = sqrt((3 * 8.314 J/(mol K) * 299.15 K) / (34.076 g/mol))

Calculating this expression gives us:

v(rms) ≈ 467.78 m/s

Rounding this value to the nearest whole number, the root mean square velocity of H2S at 26.0°C is approximately 468 m/s.

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The correct answer is: The absorption of 1.20 × 10² kJ of heat. The given enthalpy change for this reaction is AHixn = +824.2 kJ. This value represents the energy change that occurs during the reaction when 2 moles of [tex]Fe_2O_3[/tex] react to form 4 moles of Fe and 3 moles of [tex]O_2[/tex].

The given reaction shows the production of oxygen gas from iron oxide:

[tex]2 Fe_2O_3 4 Fe + 3O_2[/tex]

We are given that the formation of 14.0 g of [tex]O_2[/tex] occurs, and we need to determine the amount of heat absorbed or released. We can first calculate the number of moles of [tex]O_2[/tex] formed:

molar mass of [tex]O_2[/tex] = 32 g/mol

moles of [tex]O_2[/tex] = 14.0 g / 32 g/mol

                   = 0.4375 mol

From the balanced equation, we can see that for every 3 moles of [tex]O_2[/tex]formed, 824.2 kJ of heat is absorbed.

Therefore, the amount of heat absorbed for the formation of 0.4375 mol of O2 can be calculated using a proportion:

(824.2 kJ / 3 mol) x 0.4375 mol = 120.1 kJ

Thus, the formation of 14.0 g of [tex]O_2[/tex] results in the absorption of 120.1 kJ of heat, which is closest to the answer option "the absorption of 1.20 × 10² kJ of heat."

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a. The balanced equation is:

As₂S₃ + 8HNO₃ + 8NO + 24H+ + 12H₂O → 2H₃AsO₄ + 16NO + 3S₈.

b. 289,455 coulombs of electricity is required for the reduction of 1.5 mol to NO.

a. As₂S₃ + 8HNO₃ → 2H₃AsO₄ + 8NO + 3S₈

Balance the sulfur atoms by adding 3S8 on the right-hand side.

As₂S₃ + 8HNO₃ → 2H₃AsO₄ + 8NO + 3S₈

Next, balance the nitrogen atoms by adding 8NO on the left-hand side.

As₂S₃ + 8HNO₃ + 8NO → 2H₃AsO₄ + 16NO + 3S₈

Finally, balance the hydrogen and oxygen atoms by adding 24H⁺ and 12H₂O on the left-hand side.

As₂S₃ + 8HNO₃ + 8NO + 24H+ + 12H₂O → 2H₃AsO₄ + 16NO + 3S₈

The balanced equation is:

As₂S₃ + 8HNO₃ + 8NO + 24H+ + 12H₂O → 2H₃AsO₄ + 16NO + 3S₈

b) From the balanced equation, it can be seen that the reduction of 1 mol of NO requires the transfer of 2 electrons.

2NO + 2e- → N₂O

Thus, the reduction of 1.5 mol of NO requires the transfer of:

1.5 mol NO × 2 mol e-/mol NO = 3 mol e-

One mole of electrons represents a charge of 96,485 coulombs (1 Faraday). Therefore, the number of coulombs required for the reduction of 1.5 mol of NO is:

3 mol e- × 96,485 C/mol e- = 289,455

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The polar bond is stronger than a nonpolar bond as there is an extra force of attraction present in the polar bonds like positive and negative charges.

Polarity is a concept that describes the difference in the electron density of atoms in a molecule.

The polarity of an atom increase with the increase in electron density difference.

When the electrons are not distributed equally during the bond formation such a type of bond is called a polar bond.

When the electrons are equally distributed during the bond formation such a type of bond is called a nonpolar bond.

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Answer:

90%

Explanation:

If you messed up 100 burgers, you only make 90.

90/100=.90

If the temperature is increased to 30.0 °C, an additional 39.9 g of solute can be dissolved in the remaining water, for a total of 62.5 g solute in 100 g of water at 30.0 °C.

To solve this problem, we can use the fact that solubility generally increases with temperature. We can first find the amount of solute that has already been dissolved in the 51.0 g of water at 20.0 °C, and then calculate how much more solute can be dissolved in the additional amount of water at 30.0 °C.

From the table, we see that the solubility of the solute at 20.0 °C is 44.3 g/100 g H₂O. This means that 51.0 g of water can dissolve:

(44.3 g solute / 100 g water) × (51.0 g water) = 22.6 g solute

at 20.0 °C.

Next, we can use the solubility at 30.0 °C to find out how much more solute can be dissolved in the remaining water at this temperature. From the table, we see that the solubility of the solute at 30.0 °C is 81.4 g/100 g H₂O.

This means that 100 g of water can dissolve 81.4 g of solute at 30.0 °C. To find out how much solute can be dissolved in the remaining water at 30.0 °C, we can set up the following proportion:

(81.4 g solute / 100 g water) = (x g solute / (100 - 51) g water)

where x is the amount of solute that can be dissolved in the remaining water at 30.0 °C.

Solving for x, we get:

x = (81.4 g solute / 100 g water) × (49.0 g water) = 39.9 g solute

Therefore, if the temperature is increased to 30.0 °C, an additional 39.9 g of solute can be dissolved in the remaining water, for a total of:

22.6 g + 39.9 g = 62.5 g solute in 100 g of water at 30.0 °C.

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The experiments showed that the cubic cuprites for Cu[tex]_2[/tex]O and the face-centered cubic (fcc) crystal structure of copper (Cu) metal.

X-ray Diffraction (XRD), Energy Disper-sive X-ray Fluorescence (EDXRF), Scanning Electron Microscope (SEM), and Transmission Electron Microscope (TEM) were used to characterise the Cu/Cu[tex]_2[/tex]O nanoparticles. The experiment showed that the cubic cuprites for Cu2O and the face-centered cubic (fcc) crystal structure of copper (Cu) metal. Hydrazine was used as the reducing agent and copper sulphate pentahydrate as the precursor in a modified chemical reduction process to create copper/copper oxide (Cu/Cu2O) nanoparticles in an aqueous media.

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