15 moles of NaOH are dissolved in 2. 0 L of solution. What is the molarity of the solution?

The volume of 7.5 M stock solution needed to produce 460 mL of a 0.6 M dilute solution is 36.8 mL.

To determine the volume of a 7.5 M stock solution needed to produce 460 mL of a 0.6 M dilute solution, we can use the dilution equation:

C1V1 = C2V2

Where C1 is the initial concentration (7.5 M), V1 is the volume of the stock solution needed, C2 is the final concentration (0.6 M), and V2 is the final volume (460 mL).

We want to solve for V1, so we can rearrange the equation:

V1 = (C2V2) / C1

Now, plug in the known values:

V1 = (0.6 M * 460 mL) / 7.5 M

V1 = 276 mL / 7.5

V1 ≈ 36.8 mL

So, approximately 36.8 mL of the 7.5 M stock solution is needed to produce 460 mL of a 0.6 M dilute solution.

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a) The hydrogen ion concentration of the solution is 7.3761 x 10^-5 moles per liter, b) the total surface area of the country is 6.9470 x 10^4 square miles.

(a) The hydrogen ion concentration of the solution is given as 0.000073761 moles per liter. To express this number in scientific notation, we need to move the decimal point to the right until there is only one non-zero digit to the left of the decimal point, and then multiply by 10 raised to the appropriate power.

In this case, we can move the decimal point five places to the right to obtain \(7.3761 \times 10^{-5\) moles per liter.

(b) The total surface area of the country is given as 69,470 square miles. To express this number in scientific notation, we need to convert it to a number between 1 and 10, multiplied by a power of 10.

We can do this by moving the decimal point to the left until there is only one non-zero digit to the left of the decimal point, and then multiplying by 10 raised to the appropriate power. In this case, we can move the decimal point four places to the left to obtain \(6.9470 \times 10^4\) square miles.

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Answer:

I THINK expose all six plants to green light

Explanation:

The reaction that is occurring in this question is:

H2 + I2 -> 2 HI

As we can see, the molar ratio for H2 and I2 is 1:1, one mole of H2 for each mole of I2, and the molar ratio between any reactant and HI is 1:2, which means the number of moles of product will be two times more than the number of moles of the reactants

We have:

3.0 moles of H2

2.0 moles of I2

1.0 moles of HI

According to the molar ratio, we should have the name number of moles of I2 and H2, which means that H2 is in excess and I2 is the limiting reactant with 2 moles

If we have 2 moles of I2, and the molar ratio of I2 and HI is 1:2, we should have 4 moles of HI being produced

Therefore the theoretical number of moles is 4.0 moles of HI from 2.0 moles of I2

Answer:

Explanation:

Here, we want to get the name for the given chemical formula

Looking at the formula, we look at the names of the element

The names are sodium and fluorine

We write the name of the metal element first, then the non-metal atom

Thus, we have it as

The net ionic equation for strontium hydroxide reacting with sulfuric acid is Sr(OH)₂ + H₂SO₄ → SrSO₄ + 2H₂O.

When a base reacts with an acid, a neutralization reaction takes place. In this reaction, strontium hydroxide (Sr(OH)₂) reacts with sulfuric acid (H₂SO₄) to create strontium sulfate (SrSO₄) and water (H₂O). The balanced molecular equation is given below:

Sr(OH)₂  + H₂ SO₄ → SrSO₄ + 2H₂O.

We can break down the reactants and products into ions and write the net ionic equation to show only the species involved in the reaction.

The complete ionic equation is:

Sr²⁺(aq) + 2OH⁻(aq) + 2H⁺(aq) + SO₄²⁻(aq) → SrSO₄(s) + 2H₂O(l)

By eliminating the spectator ions (ions that do not participate in the reaction), we get the net ionic equation:

Sr²⁺(aq) + 2OH⁻(aq) + 2H⁺(aq) + SO₄²⁻(aq) → SrSO₄(s) + 2H₂O(l)

Hence, the net ionic equation for strontium hydroxide reacting with sulfuric acid is Sr(OH)₂ + H₂SO₄ → SrSO₄ + 2H₂O.

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The daughter isotope which is present in the given radioactive decay is ²¹⁸₈₆R.

A daughter isotope is the isotope that remains following the radioactive decay of an original isotope.

In the question parent nuclei is given that ²²²₈₈R, and given chemical equation is:

²²²₈₈R → ²¹⁸₈₆R + ⁴₂He

In the above equation ²¹⁸₈₆R is the daughter nuclei and ⁴₂He is the emitted particle or rays.

Hence ²¹⁸₈₆R is the daughter isotope.

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Answer:

222 88Ra

Explanation:

Answer:

1.476 mol molecules

Explanation:

Answer:

Chemists make observations on the macroscopic a scale that lead to conclusions about microscopic features

Explanation:

Many important chemical observations are made on the macroscopic scale. This is because, many of the scientific equipments available are not presently able to provide direct evidence about microscopic processes. Evidences obtained from macroscopic observations could serve as important insights into the nature of certain microscopic processes.

This is evident in the study of the structure of the atom. Most of the evidences that led to the deduction of the atomic structure were obtained from macroscopic evidence but ultimately provided important information about the microscopic structure of the atom.

The statement that would best explain the observation of no liquid being evident in the sealed bottle is: b. Liquid and vapor are at equilibrium in the bottle.

When liquid and vapor are at equilibrium in a closed system, it means that the rate of condensation (liquid turning into vapor) is equal to the rate of vaporization (vapor turning into liquid). In this case, it appears that all the liquid has vaporized, and no liquid is evident. This suggests that the liquid and vapor have reached a state of equilibrium, where the amount of liquid remaining is negligible compared to the amount of vapor present.

The vapor state is favored when equilibrium is established because the pressure exerted by the vapor phase reaches a point where it equals the vapor pressure of the liquid at that temperature. At this equilibrium point, no further net condensation or vaporization occurs, resulting in the absence of visible liquid in the sealed bottle.


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Conversion of Ethyne into Ethanol:

The ethanal can be prepared by the ethyne by treating the ethyne with mercuric sulphate in presence of acid. First the mercury attacks on the reactant ethyne and forms a cyclic structure. Then water attacks on this cyclic structure forming a mercury cation. Then by the removal of hydride alcohol forms.

Hydration of Ethyne:

Alkynes readily combine with water in the presence of acid (usually sulfuric acid) and mercury(II) salts (usually the sulfate is used) to form carbonyl compounds, in a process known as Kucherov's reaction. In the case of acetylene (ethyne) the product is acetaldehyde (ethanal), while other alkynes form ketones.

The most likely cause for the increase in pressure observed in the container as the reaction reaches equilibrium An increase in the temperature of the container is the most likely cause for the increase in pressure observed in the container as the reaction reaches equilibrium.

This is due to the fact that when a reaction reaches equilibrium, the rate of the forward and reverse reactions are equal. As the temperature of the container increases, the rate of the forward reaction increases, and the rate of the reverse reaction decrease.

This leads to a shift in the equilibrium towards the products, which in turn leads to an increase in the pressure of the gases in the container due to the increase in the number of gas molecules in the container.

Additionally, as the temperature increases, the molecules of the gas will move faster and collide more often, which also increases the pressure. To summarize, an increase in the temperature of the container is the most likely cause for the increase in pressure observed in the container as the reaction reaches equilibrium.

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The substance in the image above would be classified as a mixture of elements (option E).

A compound is a substance formed by chemical bonding of two or more elements in definite proportions by weight.

On the other hand, a mixture is made when two or more substances are combined, but they are not combined chemically.

According to this question, an image is shown with two different substances or elements as distinguished by coloration (white and purple). These elements are combined but not chemically bonded, hence, is a mixture.

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They are called alkaline earth metals because they form alkaline hydroxides, when reacted with water

The alkaline earth metals are located in Group 2 of the periodic table, which is immediately below the alkali metals in Group 1.

They share some similar properties with the alkali metals, such as low density and softness, and they also have some properties that are more similar to the earth metals, such as higher melting and boiling points, and a lower reactivity.

alkaline earth metals form compounds with oxygen that are basic (alkaline) in nature.

They also have a relatively low ionization energy, which means they lose electrons readily, and thus are able to form cations.

Alkali metals are a group of elements in the periodic table that have a single valence electron and are highly reactive.

Earth metals, on the other hand, are elements that are found in the earth and are relatively unreactive.

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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 gas having higher pressure at room temperature, is P = 6.11 atm.

What is the ideal gas law?

The general gas equation, often known as the ideal gas law, is the equation of state for a hypothetical ideal gas. It has a number of limitations, but it provides a decent approximation of the behavior of numerous gases under various circumstances.

According to the ideal gas equation,

PV =nRT

For O₂,

Number of moles of oxygen gas = 5.00 g /32 g/mol = 0.156 moles

P =?

V = 5 L

n =  0.156 moles

T = 25 + 273 = 298 K

R = 0.082 atmLK-1mol-1

P = nRT/V

P =  0.156 moles × 0.082 atmLK-1mol-1 × 298 K/5 L

P = 0.76 atm

For He,

Number of moles of He = 5/4 g/mol = 1.25 moles

P =?

V = 5 L

n = 1.25 moles

T = 25 + 273 = 298 K

R = 0.082 atmLK-1mol-1

P = nRT/V

P =1.25 moles × 0.082 atmLK-1mol-1 × 298 K/5 L

P = 6.11 atm

Hence the gas having higher pressure at room temperature is He.
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the molarity of the H₂SO₄ solution if it takes 40.0 ml of H₂SO₄ to neutralize 0.364 g of Na₂C0₃ is 0.0858

Calculation of the molarity of  H₂SO₄:

As per equation: H₂SO₄  + Na₂C0₃  ------> Na₂C0₄  +  H₂O  + CO₂

   1 mole of  H₂SO₄ reacts with  1 mole of Na₂C0₃

   1000   ml  1M   H₂SO₄        =    106 g. of Na₂C0₃  

     40  ml  1M  H₂SO₄       =    364 g. of Na₂C0₃

     1000   ml  1M   H₂SO₄   =     9.1 g.of Na₂C0₃

    __________________________________________

  1000 x 0.364

  ----------------------    =  9.1 g. of Na₂C0₃

        40

   Now  106 g. of Na₂C0₃   =    1 mole of  H₂SO₄

  Hence , 9.1 g. of Na₂C0₃  =0.0858 moles of H₂SO₄

  --------------------------------------------------------------------

          9.1 x1

        -------------                = 0.0858 moles of H₂SO₄

            106

0.0858 moles of H₂SO₄

The amount of moles of a solute in a liter of solution is known as its molarity. A capital M is used to represent molarity.

A solution in chemistry is a combination of two or more compounds in which neither ingredient undergoes a chemical transformation. For instance, salt water is a solution that contains both salt and water (the solvent) (the solute). The concentration of a material in a solution is its amount of dissolution. In other words, it refers to the number of ingredients that have been added to your drink. In molarity, concentration is often expressed.

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Answer: catabolism

Explanation:

Catabolism refers to the set of metabolic pathways which is necessary for the breaking down of molecules into smaller units. This is then oxidized for the release of energy or can be used to perform other anabolic reactions.

Catabolism is regarded as the opposite direction of anabolism which is simply the building-up of molecules. It should be noted that anabolism and catabolism work together in every living organisms and perform functions such as the production of energy and the repair of cells.

Higher temperatures and lower pressures produce optimum gas behaviour. Because of the gas's intermolecular interactions, potential energy becomes significantly lower than kinetic energy under these circumstances.

What is potential energy?

Potential energy is a form of energy that is stored but is dependent on the relative positions of different system components. Stretching or compressing a spring increases its potential energy. A steel ball has more potential energy when it is raised above the surface of the earth than when it is brought to Earth.

Any object that is raised from rest has energy that can be released at a later time; for this reason, it is referred to as potential energy.

Therefore, Higher temperatures and lower pressures produce optimum gas behaviour.

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After 24.3 days, 3 half-lives of iodine-131 have passed. Therefore, the amount remaining can be found by multiplying the original amount of 75.0 mg by \((1/2)^3\), which equals 9.38 mg. Option B is correct.

The decay of radioactive isotopes can be modeled using the concept of half-life. Half-life is the amount of time it takes for half of the original sample of the isotope to decay.

In this problem, we are given that the half-life of iodine-131 is 8.1 days. This means that after 8.1 days, half of the original sample will remain, and after another 8.1 days, half of that remaining sample will decay, and so on.

We can use this information to find how much of a 75.0 mg sample of iodine-131 will remain after 24.3 days.

First, we need to determine the number of half-lives that have elapsed. To do this, we divide the elapsed time by the half-life:

24.3 days / 8.1 days per half-life = 3 half-lives

So, after 3 half-lives, the amount of iodine-131 remaining can be found by multiplying the original amount (75.0 mg) by \((1/2)^3\) (since 3 half-lives have passed):

Amount remaining = 75.0 mg * \((1/2)^3\)

= 75.0 mg * 0.125

= 9.38 mg

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Complete question:

The half-life of iodine-131 is 8.1 days. how much of a 75.0 mg sample will remain after 24.3 days? group of answer choices

A - 75.0 mg

B - 9.38 mg

C - 4.68 mg

D - 18.8 mg

E - 37.5 mg

4.68 moles are their in 2m gas balloon.

As we know that,

1.5 m balloon contains 3.5 moles of gas

1 m balloon = \(\frac{3.5}{1.5} moles\)

1 m balloon = 2.34 moles

For 2 m balloon,

2 * 2.34 = 4.68 moles of gas are their.

A pure gas may consist of individual atoms (such as a noble gas like neon), elemental molecules made of a single type of atom (such as oxygen), or complex molecules made of several different types of atoms (e.g. carbon dioxide). There are many different pure gases in a gas mixture like air. The great separation of the individual gas particles is what sets gases apart from liquids and solids. A colorless gas becomes invisible to a human observer due to this gap.

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Answer:

Not exactly such but Experimental manipulation describes the process by which researchers purposefully change, alter or influence the independent variables(IVs), which are also called treatment variables or factors in an experimental research design.

The new height of the water is 17.7 mL

The density of a substance is simply defined as mass per unit volume of the substance. Mathematically, it is expressed as:

With the above formula, we can obtain the volume of the copper. This can be obtained as follow:

Mass of copper = 20 g

Density of copper = 8.96 g/mL

Density = mass / volume

8.96 = 20 / volume

Cross multiply

8.96 × Volume = 20

Divide both side by 8.96

Volume = 20 / 8.96

Finally, we shall determine the new height of the water. This can be obtained as follow:

Initial height of water = 15.5 mL

Volume of copper = 2.2 mL

New height of water = 15.5 + 2.2

Therefore, the new height of the water is 17.7 mL

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Answer:

A: sink

Explanation: If an object's density is:

GREATER than \(1gcm^{3}\), it will SINK in water.

LESS than \(1gcm^{3}\), it will FLOAT in water.

hope this helps

132 gm of CO₂ is generated by combustion of 342 gm of C₈H₁₈.

The molar mass of a chemical compound is defined as the ratio between the mass and the amount of that substance in any sample. A material's molar mass is a bulk characteristic rather than a molecular one.

The molar mass of a substance is its mass expressed in grams per mole. G/mol, or grams per mole, is the sign for molar mass. The ratio between the mass of an isotope and the mass of the isotope carbon is known as the isotopic atomic mass, or mass of a single isotope of any particular element. -12

The reaction is :

C₈H₁₈ + 12.5O₂ ----> 8CO₂ + 9H₂O

Molar mass of C₈H₁₈ is = 12*8 + 18*1 = 114 gm

Molar mass of CO₂ is = 12+ 16*2 = 44gm

44gm of CO₂ is generated by 114 gm of C₈H₁₈

1 gm of CO₂ is generated by 114/44 gm of C₈H₁₈

132 gm of CO₂ is generated by combustion of 114*132/44 = 342 gm of C₈H₁₈

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The net present value (NPV) of the RGM-7000 water filtration system for Caspian Sea Drinks is $3.79 million.

To calculate the net present value (NPV), we need to discount the cash flows associated with the RGM-7000 over its 20-year life using the given discount rate of 9.00%. The formula for NPV is:

NPV = (Cash inflows - Cash outflows) / (1 + Discount rate)^t

Cash inflows:

Additional revenues per year = $2.73 million

Cash outflows:

Operating costs per year = $646,303.00

Initial cost of the RGM-7000 = $14.00 million

Tax rate = 24.00%

To calculate the annual net cash flows, we subtract the operating costs from the additional revenues:

Net cash flows per year = Additional revenues - Operating costs

Next, we calculate the present value of the annual net cash flows for each year using the discount rate of 9.00%:

Present value of net cash flows = Net cash flows / (1 + Discount rate)^t

Finally, we sum up the present values of the net cash flows over the 20-year life of the RGM-7000 and subtract the initial cost:

NPV = Sum of present values - Initial cost

By performing the calculations, the net present value (NPV) of the RGM-7000 for Caspian Sea Drinks is determined to be $3.79 million.

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In light of the aforementioned claim Argon (Ar) monoatomic gases is most likely the sample.

Monatomic gases, which differ from diatomic, triatomic, other typically polyatomic gases in that they contain particles (molecules) thought up of only one atom, include substances like helium and sodium vapour. One type of noble gas is a monatomic gas like helium. Radon, neon.

Starting out: The mass of the gas is 1.66.

The computed temperature is 293 degrees, or T = 273 + 20.

In order to determine the gas, we must calculate the molarity:

RT/PV = Molar Mass

M := (1.66 x 8.314 x 293) | (101.3 x 1000 x 0.001) (101.3 x 1000 x 0.001)

M Equals 4043.76 / 101.3 (= 39.92 g/mol

Based solely on molar calculations, this gas must be argon (Ar).

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Answer: C.) 446 g NaCl.

Explanation:

c.) 446 g NaCl. This is because 4.75 moles of NaCl is equal to 446 grams. This is calculated by multiplying the molar mass of NaCl (58.44 g/mol) by 4.75 moles. Hence, the answer is 446 g NaCl.