If the solubility of o2 at 0.370 atm and 25 °c is 15.3 g/100 g h2o, what is the solubility of o2 at a pressure of 2.40 atm and 25 °c?

Answer:  the actual answer is

step 1: A tree absorbs the carbon from the atmosphere into its leaves for photosynthesis

step 2: A caterpillar gets the carbon by eating the tree's leaves

step 3: A bird gets the carbon by eating the caterpillar

step 4: The bird flies into a building and dies instantly. It falls to the ground

step 5: The bird decomposes and the carbon is added to the atmosphere

Explanation:

The temperature of the gas will be 324K or [tex]\rm \bold{51^\cdot C}[/tex] when volume reaches 211 mL.

Charle's law stated that the volume occupied by fixed amount of gas is directly proportional to the temperature,while  pressure is constant.

 [tex]\rm \bold{\frac{V_1}{T_1} =\frac{V_2}{T_2}}[/tex]

Where,

[tex]\rm \bold V_1[/tex] is initial volume = 178mL

[tex]\rm \bold {V_2}[/tex] is final volume = 211mL

[tex]\rm \bold{ T_1}[/tex] is initial temperature = [tex]\rm \bold{0.00^\cdot C}[/tex] = 273K

[tex]\rm \bold {T_2}[/tex] is final temperature = ?

Put the values,

[tex]\rm \bold{\frac{178}{273} =\frac{211}{T_2}}\\\\\rm \bold{T_2=324 K}[/tex]

Hence, we can conclude that the temperature of the gas will be 324K or [tex]\rm \bold{51^\cdot C}[/tex] when volume reaches 211 mL.

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In the given reaction, label the species as acid, base, conjugate acid, and conjugate base. (a) Acid: NH4+, Base: H2O, Conjugate Acid: H3O+, Conjugate Base: NH3. (b) Acid: CN-, Base: H2O, Conjugate Acid: HCN, Conjugate Base: OH-

(a) In the reaction: NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq), we have NH4+ as the acid and H2O as the base. After the reaction, NH3 becomes the conjugate base and H3O+ becomes the conjugate acid.

(b) In the reaction: CN-(aq) + H2O(l) ⇌ HCN(aq) + OH-(aq), we have CN- as the acid and H2O as the base. After the reaction, HCN becomes the conjugate acid and OH- becomes the conjugate base.

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It will require  for 20 % of U-238 atoms in a sample of U-238 to decay.

Further Explanation:

Radioactive decay involves stabilization of unstable atomic nucleus and is accompanied by the release of energy. This emission of energy can be in form of different particles like alpha, beta and gamma particles.

Half-life is time period in which half of the radioactive species is consumed. It is denoted by .

The expression for half-life is given as follows:

[tex]\lambda = \dfrac{{0.693}}{{{t_{{\text{1/2}}}}}}[/tex]                    …… (1)

Where,

[tex]{t_{{\text{1/2}}}}[/tex] is half-life period

[tex]\lambda[/tex] is the decay constant.

The half-life period for decay of U-238 is [tex]4.5 \times {10^9}{\text{ yrs}}[/tex].

Substitute [tex]4.5 \times {10^9}{\text{ yrs}}[/tex] for [tex]{t_{{\text{1/2}}}}[/tex]  in equation (1).

[tex]\begin{aligned}\lambda&= \dfrac{{0.693}}{{4.5 \times {{10}^9}{\text{ yrs}}}} \\&= 1.54 \times {10^{ - 10}}{\text{ yr}}{{\text{s}}^{ - 1}} \\\end{galigned}[/tex]  

Since it is radioactive decay, it is first-order reaction. Therefore the expression for rate of decay of U-238is given as follows:

[tex]\lambda = \dfrac{{2.303}}{t}\log \left( {\dfrac{a}{{a - x}}} \right)[/tex]

                                                …… (2)

Where,

[tex]\lambda[/tex] is the decay or rate constant.

t is the time taken for decay process.

a is the initial amount of sample.

x is the amount of sample that has been decayed.

Rearrange equation (2) to calculate t.

[tex]t = \dfrac{{2.303}}{\lambda }\log \left( {\dfrac{a}{{a - x}}} \right)[/tex]                                                                                          …… (3)

Consider 100 g to be initial amount of U-238. Since 20 % of it is decayed in radioactive process, 20 g of U-238 is decayed and therefore 80 g of the sample is left behind.

Substitute 100 g for a, 80 g for (a–x) and [tex]1.54 \times {10^{ - 10}}{\text{ yr}}{{\text{s}}^{ - 1}}[/tex] for [tex]\lambda[/tex] in equation (3).

[tex]\begin{aligned}t &= \dfrac{{2.303}}{{1.54 \times {{10}^{ - 10}}{\text{ yr}}{{\text{s}}^{ - 1}}}}\log \left( {\dfrac{{100{\text{ g}}}}{{80{\text{ g}}}}} \right)\\&= 1.449 \times {10^9}{\text{ yrs}}\\\end{aligned}[/tex]  

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Radioactivity

Keywords: half-life, t, a, x, a – x, 1.449*10^9 yrs, U-238, decay constant, radioactivity, half-life period.

When the concentration of b is doubled in a chemical reaction that is zero order in a, one-half order in b, and second order in c, the reaction rate will increase by a factor of √2 (approximately 1.414), with other reactant concentrations held constant.

The question you're asking relates to the rate of a chemical reaction and how it changes with varying concentrations of reactants. Specifically, there is a reaction where the rate is zero order in a, one-half order in b, and second order in c. According to the given reaction orders, the rate expression would be:

Rate = k [a]0[b]1/2[c]2

Since the reaction is zero order in a, changing the concentration of a does not affect the rate. However, since it is one-half order in b, if the concentration of b is doubled, the rate will increase by a factor of the square root of 2. This is because:

New Rate = k [a]0(2[b])1/2[c]2 = k [a]0[b]1/2 × √2 [c]2
= Rate × √2

Therefore, when the concentration of b is doubled, and a and c remain constant, the reaction rate will increase by a factor of √2 (approximately 1.414).

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Without specific enthalpy change (ΔH) values for the reactions in question, we cannot accurately calculate the heat transfer per mole of ethanol produced. However, if these values were provided, the general formula qp = ΔH(T2-T1)/T1*T2 could be used to determine this, where qp represents heat transfer at constant pressure, T1 and T2 are the initial and final temperatures respectively.

This question pertains to the field of thermochemistry and the role of heat transfer in chemical reactions. The process in question involves the transformation of ethylene gas and steam into ethanol, or C2H5OH. We would need the specific heat, or enthalpy change (ΔH), values for these reactions to calculate the heat transfer per mole of ethanol produced. Since these values haven't been provided, we can't provide a definite answer.

Generally, in such questions, when ΔH values and temperatures are given, one formula that can be applied is q_p = ΔH(T2-T1)/T1*T2 where T1 is initial temperature, T2 is final temperature and qp represents heat transfer at constant pressure.

This formula can be used to estimate the heat transfer associated with the process per mole of ethanol produced, once the ΔH value is known. It's also important to remember that if the reaction is exothermic (releases heat), the ΔH value would be negative, and if it's endothermic (absorbs heat), the ΔH value would be positive.

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Answer:

You need to know the speed, direction (displacement) and time to describe the velocity of an object.

Empirical formula is the simplest whole number ratio of elements in a compound. Molecular formula is n times the empirical formula, where n is integers 1,2,...  Therefore, the molecular formula of the compound C[tex]_{10}[/tex]H[tex]_{10}[/tex]O[tex]_5[/tex].

Empirical formula gives the proportion of the element that is present in a compound. It does not give information about the actual number or the arrangements of the atoms.

n= molecular formula÷ empirical formula

Molar masses of  C[tex]_2[/tex]H[tex]_2[/tex]O

C = 12.0107

H = 2.016

O = 15.999

Molar masses of  C[tex]_2[/tex]H[tex]_2[/tex]O =2(12.0107)+2(2.016)+15.999

                                         =24.0214+4.032+15.999

                                         = 44.0524

252.22/44.0524 = 5

C[tex]_2[/tex]H[tex]_2[/tex]O ×5=C[tex]_{10}[/tex]H[tex]_{10}[/tex]O[tex]_5[/tex]

The molecular formula is C[tex]_{10}[/tex]H[tex]_{10}[/tex]O[tex]_5[/tex]

Therefore, the molecular formula of the compound C[tex]_{10}[/tex]H[tex]_{10}[/tex]O[tex]_5[/tex].

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For this we have to consider the reactive series of metals in which metals are arranged according to their reactivity order.

Oxidation of a metal can be defined as ability to lose electrons to become positively charged metal and a metal will lose its electrons by reacting with a metal that is having low chemical reactivity as a higher reactive metal can lose its electrons to a low reactive metal. So here a metal should be one that is more reactive than tin so that it can lose its electrons to tin to become positively charged and less reactive than iron so that it will not give its electrons to iron to become positively charged. If we look into reactive series of metal than we will find that nickel is the example of a metal that is more reactive than tin and less reactive than iron.

Nickel and cadmium is the metal that can be oxidized with a sn2+ solution, but not with a fe2+ solution.

An ionic chemical reaction occurs on the surface of metal in the presence of oxygen.

Metals that can be oxidized in solution must have a weaker oxidizing agent or be less reactive than tin.

With solution, the same metals cannot be oxidized.

Thus, the correct option is Nickel and cadmium.

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To prepare a 300.0 mL of a 1.50 M KBr solution, you need 53.55 grams of KBr, calculated by multiplying the required moles (0.450 moles) by the molar mass of KBr (119.00 g/mol).

To find the mass of KBr needed to make a 300.0 mL (0.300 L) of a 1.50 M solution, we apply the formula:

Molarity (M) = moles of solute / liters of solution

First, calculate the moles of KBr required:

Next, convert moles to grams using the molar mass of KBr (119.00 g/mol):

Therefore, you would need 53.55 grams of KBr to make a 300.0 mL of a 1.50 M KBr solution.

The answer is C falling water hope this helps

Electricity at a hydroelectric dam is produced by falling water, which turns turbines to generate electricity. The ultimate source of this energy is solar energy that drives the hydrological cycle.

The energy source that produces electricity at a hydroelectric dam is C. falling water. This process involves the conversion of potential energy stored in the water behind the dam into kinetic energy as the water flows down. The kinetic energy turns the blades of turbines, which then drive a generator to produce electricity. The ultimate source of the electrical energy produced by a hydroelectric plant is solar energy, which drives the hydrological cycle that lifts water to high elevations through evaporation and precipitation.

Answer:

Electrons will be gained

Answer: C) develop a way to adjust its density so that it can be above and below 1.0 gram per cubic centimeter.

Explanation: Submarine is a kind of special ship that can float above and below water. We know that an object can only float over water when its density is less than the density of water and it would sink if its density is greater than the density of water.

A submarine has tanks in which water is filled if we want to sink it and the water from the tanks is taken out with the help of water pumps to rise the submarine. These water tanks in the submarine helps to adjust its density.

The over all density is increased by filling the water tanks to sink the submarine and the density is decreased by removing the water from the tanks to rise the submarine.

So, option third or C is correct.

Explanation:

It is known that density is the amount of mass divided by volume.

Mathematically,     Density = [tex]\frac{mass}{volume}[/tex]

It is given that mass is 13.0 kg or 13000 g (as 1 kg = 1000 g). And, volume is 1650 [tex]cm^{3}[/tex].

Therefore, calculate the density as follows.

                    Density = [tex]\frac{mass}{volume}[/tex]

                                 = [tex]\frac{13000 g}{1650 cm^{3}}[/tex]    

                                 = [tex]7.87 g/cm^{3}[/tex]

Thus, we can conclude that density of the given iron plate is [tex]7.87 g/cm^{3}[/tex].

Final answer:

n-Heptane has a higher viscosity than n-hexane due to its longer carbon chain and greater molecular weight, which leads to stronger dispersion forces and hence more resistance to flow.

Explanation:

The viscosity of a liquid is a measure of its resistance to deformation or flow. In comparing the viscosities of n-hexane (CH₃(CH₂)₄CH₃) and n-heptane (CH₃(CH₂)₅CH₃), we must consider their molecular weights and the strength of their intermolecular forces. Because n-heptane has a longer carbon chain and greater molecular weight than n-hexane, it also has stronger dispersion forces. These dispersion forces increase with the molecular surface area, so the larger n-heptane molecules experience more significant intermolecular attractions. As a result, n-heptane has a higher viscosity compared to n-hexane.

Answer: The element needs to react with other element by gaining an electron to complete its valence electron.

Explanation: The element which exhibit 7 valence electrons are halogens. They readily react with other element, for example: Hydrogen and Sodium, in order to gain an electron to complete their outermost shell.

Answer: The molality of solution is 1.94 m.

Explanation:

We are given:

Mass of sulfuric acid = 17.75 grams

Volume of solution = 100 mL

Density of solution = 1.1094 g/mL

To calculate the mass of solution, we use the equation:

[tex]Density=\frac{Mass}{Volume}[/tex]

Putting values in above equation, we get:

[tex]1.1094g/mL=\frac{\text{Mass of solution}}{100mL}[/tex]

[tex]\text{Mass of solution}=110.94g[/tex]

Mass of solvent = Mass of solution - Mass of solute

Mass of solvent = 110.94 - 17.75 = 93.19 g

To calculate the molality of solution, we use the equation:

[tex]Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}[/tex]

Where,

[tex]m_{solute}[/tex] = Given mass of solute [tex](H_2SO_4)[/tex] = 17.75 g

[tex]M_{solute}[/tex] = Molar mass of solute [tex](H_2SO_4)[/tex] = 98 g/mol

[tex]W_{solvent}[/tex] = Mass of solvent = 93.19 g

Putting values in above equation, we get:

[tex]\text{Molality of }H_2SO_4=\frac{17.75\times 1000}{98\times 93.19}[/tex]

[tex]\text{Molality of }H_2SO_4=1.94m[/tex]

Hence, the molality of solution is 1.94 m.

To calculate the molality of the sulfuric acid solution, we convert the volume to mass, find the number of moles of sulfuric acid, and divide by the mass of the solvent in kilograms.

To calculate the molality of a solution, we need to find the number of moles of solute and the mass of the solvent.

In this case, we are given the mass of sulfuric acid (17.75 g) and the volume of the solution (100.0 ml). First, we need to convert the volume to mass by multiplying it by the density of the solution (1.1094 g/ml). 100.0 ml x 1.1094 g/ml = 110.94 g.

To find the number of moles of sulfuric acid, we divide the mass by the molar mass of sulfuric acid (98.09 g/mol). 17.75 g / 98.09 g/mol ≈ 0.1808 mol.

Finally, we can calculate the molality by dividing the moles of solute by the mass of the solvent in kilograms. 0.1808 mol / 0.11094 kg = 1.63 mol/kg.

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Both the ball-and-stick and space-filling models are 3D molecular representations, with the former displaying the bonds and the latter showing the relative atom sizes and general shape.

A ball-and-stick model of a molecule provides much the same information as a space-filling model. Both are used to represent the three-dimensional structure of a molecule in different ways. The ball-and-stick model shows atoms as spheres and bonds as sticks, highlighting the bond angles and lengths between atoms.

In contrast, the space-filling model represents atoms by spheres that are sized proportionally to the atoms' radii, depicting the actual volume occupied by the atoms without showing the bonds explicitly.

While the ball-and-stick model may exaggerate the space between the atoms, it is beneficial for understanding the geometry of the chemical bonds. The space-filling model, however, allows for visualization of the molecule's general shape and the relative sizes of the atoms. Both models are simplifications and do not fully capture subtle variations in bond lengths and angles.

The number of moles of 1-propanol in 25.00 mL is 0.334 (to three significant digits). The molarity of the diluted 1-propanol solution in a 100.0 mL flask is 3.34 M (to three significant digits).

To calculate the number of moles of 1-propanol in 25.00 mL, we use the density of 1-propanol (0.803 g/mL) to find the mass:

Mass = Volume × Density = 25.00 mL  0.803 g/mL = 20.075 g

Next, we convert this mass to moles using the molar mass of 1-propanol (60.09 g/mol):

Moles of 1-propanol = Mass / Molar Mass = 20.075 g / 60.09 g/mol

This calculation yields 0.334 moles of 1-propanol (to three significant digits).

For the molarity of the diluted solution, we take into account that the total volume is now 100.0 mL. Since molarity is moles per liter (mol/L), we must first convert the volume from mL to L:

Volume in L = 100.0 mL (1 L / 1000 mL) = 0.100 L

Molarity = Moles of 1-propanol / Volume in L = 0.334 mol / 0.100 L

The molarity of the solution is 3.34 M (to three significant digits).

Answer:

The mass can not be neither created nor destroyed but transformed by chemical reactions or physical transformations in an isolated recipient.

Explanation:

Hello,

In case, no matter the carried out experiment, the law of conservation of mass always leads the same: the mass can not be neither created nor destroyed but transformed by chemical reactions or physical transformations in an isolated recipient.

In such a way, we must consider that any system closed to every form of transport of matter, will show a no change in its mass as time goes by, since the system's mass cannot change neither by additions nor withdrawals. Therefore, the quantity of mass is conserved over time.

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Answer is: the final pressure of the gas is closest to 3,17 atm.

p₁ = 3,2 atm.
T₁ = 310 K.
V₁ = 6 L.
p₂ = ?
T₂ = 410K.
V₂ = 8,0 L.
Use combined gas law - the volume of amount of gas is proportional to the ratio of its Kelvin temperature and its pressure. 
p₁V₁/T₁ = p₂V₂/T₂.
3,2 atm · 6,0 L ÷ 310 K  = p₂ · 8,0 L ÷ 410 K.
0,0619 = 0,0195p₂.
p₂ = 3,17 atm.

Set the crucible's lid slightly off-center to allow air to enter while keeping the magnesium oxide from escaping.

A crucible is a cup-shaped piece of laboratory equipment used to keep chemical compounds contained while they are heated to extremely high temperatures.

Crucibles come in a variety of sizes and are usually packaged with a crucible cover (or lid).

Keep the lid on the crucible while cooling to prevent moisture from the atmosphere from interacting with the anhydrous salt, especially if the lab is humid. As a result, the mass of water will be too low.

The most important apparatus because it will be used to obtain the final precipitate, which will tell us how much salt is in the solution.

The lid is used to cover the crucible so that the heated precipitates do not oxidize when they come into contact with air.

Thus, it is important that the lid should be kept correctly.

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oxides and carbonates, grad point..