In the nuclear transmutation represented by 168o(p, \alpha) 137n, the emitted particle is ________.

Final answer:

Recrystallization of an anhydride in water or alcohol is not effective because anhydrides will react with these solvents to form carboxylic acids or esters, respectively, thereby altering the compound instead of purifying it.

Explanation:

Recrystallizing an anhydride from water or an alcohol is often not ideal due to the reactive nature of anhydrides. Anhydrides undergo hydrolysis in the presence of water to form carboxylic acids, a process which is energetically favorable and further driven by the stabilization through hydrogen-bonding interactions among water molecules and the carboxylic acids. Moreover, anhydrides react with alcohols to yield esters. This means that instead of recrystallizing the anhydride, you are likely to convert it to other products, which defeats the purpose of recrystallization since you want to purify the anhydride, not change its chemical structure.

The amount of limestone required to neutralize the lake is approximately [tex]\( 2.653 \times 10^4 \, \text{kg} \).[/tex]

To determine how much limestone ([tex]CaCO_3[/tex]) is required to neutralize a lake acidified by sulfuric acid ([tex]H_2SO_4[/tex]), we need to follow these steps:

1. Calculate the total mass of [tex]H_2SO_4[/tex] in the lake.

2. Determine the moles of [tex]H_2SO_4[/tex] present.

3. Use stoichiometry to find the moles of [tex]CaCO}_3[/tex] required to neutralize the [tex]H_2SO_4[/tex]

4. Convert the moles of [tex]CaCO}_3[/tex] to mass in kilograms.

Step 1: Calculate the Total Mass of [tex]H_2SO_4[/tex] in the Lake

The lake volume is [tex]\( 5.2 \times 10^9 \) liters.[/tex]

The concentration of [tex]H_2SO_4[/tex] is [tex]\( 5.0 \times 10^{-3} \) g/L.[/tex]

Total mass of [tex]H_2SO_4[/tex]

[tex]\[ \text{Total mass of H}_2\text{SO}_4 = \text{Concentration} \times \text{Volume} \][/tex]

[tex]\[ \text{Total mass of H}_2\text{SO}_4 = 5.0 \times 10^{-3} \, \text{g/L} \times 5.2 \times 10^9 \, \text{L} \][/tex]

[tex]\[ \text{Total mass of H}_2\text{SO}_4 = 2.6 \times 10^7 \, \text{g} \][/tex]

Step 2: Determine the Moles of [tex]H_2SO_4[/tex]

Molar mass of [tex]H_2SO_4[/tex]

[tex]\[ \text{H}_2\text{SO}_4 = 2 \times 1 + 32 + 4 \times 16 = 98 \, \text{g/mol} \][/tex]

Moles of [tex]H_2SO_4[/tex]

[tex]\[ \text{Moles of H}_2\text{SO}_4 = \frac{\text{Total mass of H}_2\text{SO}_4}{\text{Molar mass of H}_2\text{SO}_4} \][/tex]

[tex]\[ \text{Moles of H}_2\text{SO}_4 = \frac{2.6 \times 10^7 \, \text{g}}{98 \, \text{g/mol}} \][/tex]

[tex]\[ \text{Moles of H}_2\text{SO}_4 = 2.653 \times 10^5 \, \text{mol} \][/tex]

Step 3: Use Stoichiometry to Find the Moles of [tex]CaCO}_3[/tex] Required

The neutralization reaction between [tex]H_2SO_4[/tex] and [tex]CaCO}_3[/tex] is:

[tex]\[ \text{H}_2\text{SO}_4 + \text{CaCO}_3 \rightarrow \text{CaSO}_4 + \text{CO}_2 + \text{H}_2\text{O} \][/tex]

From the balanced equation, 1 mole of [tex]H_2SO_4[/tex] reacts with 1 mole of [tex]CaCO}_3[/tex]

Therefore, moles of [tex]CaCO}_3[/tex] required:

[tex]\[ \text{Moles of CaCO}_3 = \text{Moles of H}_2\text{SO}_4 \][/tex]

[tex]\[ \text{Moles of CaCO}_3 = 2.653 \times 10^5 \, \text{mol} \][/tex]

Step 4: Convert the Moles of [tex]CaCO}_3[/tex] to Mass in Kilograms

Molar mass of [tex]CaCO}_3[/tex]

[tex]\[ \text{CaCO}_3 = 40 + 12 + 3 \times 16 = 100 \, \text{g/mol} \][/tex]

Mass of [tex]CaCO}_3[/tex]

[tex]\[ \text{Mass of CaCO}_3 = \text{Moles of CaCO}_3 \times \text{Molar mass of CaCO}_3 \][/tex]

[tex]\[ \text{Mass of CaCO}_3 = 2.653 \times 10^5 \, \text{mol} \times 100 \, \text{g/mol} \][/tex]

[tex]\[ \text{Mass of CaCO}_3 = 2.653 \times 10^7 \, \text{g} \][/tex]

Convert grams to kilograms:

[tex]\[ \text{Mass of CaCO}_3 = 2.653 \times 10^7 \, \text{g} \times \frac{1 \, \text{kg}}{1000 \, \text{g}} \][/tex]

[tex]\[ \text{Mass of CaCO}_3 = 2.653 \times 10^4 \, \text{kg} \][/tex]

The complete Question is

Lakes that have been acidified by acid rain can be neutralized by the addition of limestone (CaCO3). How much limestone in kg would be required to completely neutralize a 5.2 x 10^9 L lake containing 5.0 x 10-3 g of H2SO4 per liter?

Answer: Option (d) is the correct answer.

Explanation:

Species which contain same number of protons but different number of neutrons are known as isotope.  

For example, [tex]^{1}_{1}H[/tex] and [tex]^{3}_{1}H[/tex] are isotopes.

More is the mass of particle colliding with the isotope more will be the change in mass of an isotope due to emission of a heavier particle.

As alpha ([tex]^{4}_{2}He[/tex]) particle is heavier then a neutron, positron and gamma particles.

For example, [tex]^{14}_{7}N + ^{4}_{2}He \rightarrow ^{17}_{8}O[/tex]

Therefore, we can conclude that alpha particle changes the mass of the isotope the most.

The balanced nuclear equation for beta decay of sodium-26 is represented as Na²⁶₁₁ → Mg²⁶₁₂ + β₋₁⁰ + energy.

Beta decay is one of the type of radioactive nuclear decay reaction in which emission of a beta particle takes place from the atomic nuclear.

Beta particle define by the symbol β₋₁⁰ i.e. this particles has no mass but having a negative charge on it. In this reaction atomic mass of the parent atom is equal to the new formed daughter atom. But in this reaction number of neutron is reduced by 1 and number of proton is increased by one of parent atom. So, balanced nuclear equation for β− decay of sodium−26 is represnted as:

Na²⁶₁₁ → Mg²⁶₁₂ + β₋₁⁰ + energy

Hence balanced nuclear equation for the beta decay is represented above.

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Carbonyl compounds that do not undergo an aldol reaction include those without alpha-hydrogens like formaldehyde and benzaldehyde, as well as esters, acids, and amides.

These compounds either lack necessary alpha-hydrogens or have structures that are unfavorable for aldol reactions.

In an aldol reaction, carbonyl compounds like aldehydes and ketones with alpha-hydrogens can react in the presence of a base such as hydroxide (−OH) in water. However, not all carbonyl compounds undergo this reaction.

Carbonyl compounds that do not undergo an aldol reaction include those without alpha-hydrogens, such as formaldehyde and benzaldehyde. Additionally, carbonyl compounds like esters, acids, and amides typically do not participate in aldol reactions due to their structural properties.

Examples of Carbonyl Compounds Not Undergoing Aldol Reaction:

These compounds either lack the necessary alpha-hydrogens or have structures that are unfavorable for aldol reactions.

Final answer:

Butanoic acid can be converted to butanal via a reduction process using a reducing agent such as lithium aluminum hydride, followed by hydrolysis.

Explanation:

Converting butanoic acid to butanal involves a reduction process, which can be carried out using reducing agents such as lithium aluminum hydride (LiAlH₄) or borane (BH₃). The reaction with LiAlH₄ is usually carried out in dry ether as a solvent, and it can be followed by careful hydrolysis to give the aldehyde, butanal. The overall reaction can be summarized as:

Butanoic acid (C₄H₈O₂) is treated with the reducing agent LiAlH₄.

The acid is reduced to the corresponding aldehyde, butanal (C₄H₈O).

After the reduction, the reaction mixture is hydrolyzed, which involves adding water to the reaction mixture.

This is different from esterification, which is the reaction between an acid and an alcohol, as described in the esterification of ethanol and butanoic acid to form ethyl butanoate and water.

The biochemical production of butanol from butyric acid is similarly not a direct route for synthesizing butanal, but it is an important process for biomass conversion.

Answer:

Concentration of solution in percent by mass is 13.3% CaO

Explanation:

% by mass = [tex]\frac{mass of solute}{mass of solution}\times 100[/tex]

Here solute is CaO and solvent is water.

So, mass of solution = (mass of solute)+(mass of solvent)

                                  = (mass of CaO)+(mass of water)

                                   = 32.5 g + 212 g

                                    = 244.5 g

So, Concentration of solution by percent mass = [tex]\frac{mass of CaO}{mass of solution}\times 100[/tex] = [tex]\frac{32.5}{244.5}\times 100[/tex] = 13.3 %

Using the Clausius-Clapeyron equation, we can derive that the boiling point of water at 880 torr is closest to 90°C. Thus, among the given options, the boiling point of water at 880 torr is most likely to be 92°C.

The subject question is considering the boiling point of water at differing pressures. The standard boiling point of water is 100°C at 1 atmospheric pressure (or 760 torr). However, boiling point changes with changes in atmospheric pressure.

 By using the Clausius-Clapeyron equation, which establishes a relationship between the vapor pressure and temperature of a substance, we can establish that at 880 torr, the boiling point is likely to be close to 90°C, given that the vapor pressure of water is 68kPa at about this temperature. Therefore, among the options given, 92°C is the most likely boiling point of water at 880 torr.

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The [tex]\( K_{sp} \)[/tex] for zinc hydroxide is [tex]\[ 3.8 \times 10^{-17}}\][/tex]

1. Molar mass of  [tex]\( \text{Zn(OH)}_2 \)[/tex]:

  [tex]\[ \text{Zn} = 65.38 \, \text{g/mol}, \, \text{O} = 16.00 \, \text{g/mol}, \, \text{H} = 1.01 \, \text{g/mol} \][/tex]

 [tex]\[ \text{Molar mass} = 65.38 + 2 \times (16.00 + 1.01) = 99.40 \, \text{g/mol} \][/tex]

2. Molar solubility [tex]\( s \)[/tex]:

  [tex]\[ \text{Solubility} = 2.1 \times 10^{-4} \, \text{g/L} \][/tex]

 [tex]\[ s = \frac{2.1 \times 10^{-4}}{99.40} \approx 2.11 \times 10^{-6} \, \text{mol/L} \][/tex]

3. Ksp expression:

 [tex]\[ \text{Zn(OH)}_2 \rightleftharpoons \text{Zn}^{2+} + 2\text{OH}^- \][/tex]

 [tex]\[ K_{sp} = [\text{Zn}^{2+}][\text{OH}^-]^2 = s \cdot (2s)^2 = 4s^3 \][/tex]

4. Calculate [tex]\( K_{sp} \):[/tex]

 [tex]\[ K_{sp} = 4 \times (2.11 \times 10^{-6})^3 = 4 \times 9.39 \times 10^{-18} \approx 3.76 \times 10^{-17} \][/tex]

After 20 minutes, approximately 32 mg of the 128 mg sample of Nitrogen-13 would remain.

The half-life of a radioactive isotope is the time required for half of the atoms in a sample to decay.

In this case, we are given that the half-life of Nitrogen-13 is 10 minutes. This means that after every 10 minutes, half of the sample will decay.

Since 20 minutes have passed, we need to determine how many half-lives have occurred. There have been 2 half-lives because 20 divided by 10 equals 2.

Therefore, after 2 half-lives, one-fourth of the sample will remain (since half of the original sample will decay after each half-life).

To find out how much of a 128 mg sample would remain after 20 minutes, we multiply the original amount by one-fourth:

128 mg x 1/4 = 32 mg

After 20 minutes, approximately 32 mg of the 128 mg sample of Nitrogen-13 would remain.

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The change in entropy in the system is 308.3 J/K.

The change in entropy in the system can be calculated using the formula ∆S = ∆Hvap/T. Here, ∆Hvap is the molar enthalpy of vaporization of water, which is given as 40.67 kJ/mol. T is the temperature in Kelvin, which can be calculated by adding 273.15 to the boiling point of water in Celsius. So, T = 100.0 + 273.15 = 373.15 K. Plugging in these values in the formula, we get:

∆S = (40.67 kJ/mol)/(373.15 K) = 0.1089 kJ/(mol·K)

Now, we need to convert grams of steam to moles of steam. The molar mass of water is 18.015 g/mol. So, 51.1 g of steam is equal to (51.1 g)/(18.015 g/mol) = 2.835 mol. Multipling this with the change in entropy, we get:

∆S = (0.1089 kJ/(mol·K)) · (2.835 mol) = 0.3083 kJ/K

Finally, to convert kJ/K to J/K, we multiply by 1000:

∆S = (0.3083 kJ/K) · (1000 J/1 kJ) = 308.3 J/K

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The Jaeger reading cards are used to test near vision. The correct answer is "near" vision.

The Jaeger reading cards are used to test near vision, which is the ability to see objects clearly at a close distance, typically around 14 to 16 inches away from the eyes. This type of vision is important for activities such as reading, writing, and working on a computer. The Jaeger chart consists of blocks of text in various sizes, with the smallest print at the bottom and the largest at the top. A person being tested reads the text from top to bottom until they can no longer clearly distinguish the letters, which helps determine their near visual acuity.

Answer: The correct answer is A) Use an inclined plane of 8 m

Work :

Work is defined as force applied to an object to move it to some distance . It is calculated as product of force and displacement .

W = F* d

Where : W =work (N-m) F = force (N) d = displacement (m)

When a body is lifted , work can be expressed as :

W = m*g*h ( h = height )

Also force due to gravitation can be given as

F = m* g

where :

g = gravitational acceleration ([tex] 9.8 \frac{m}{s^2} [/tex]

Hence Work can be written as : W = F * h

Given :

Force on load due to gravitation = 100 N Height = 2 m

Force applied = 25 N displacement = ?

Work done to lift the load to 2 m = F * h

= 100 N * 2 m = 200 N-m

Plugging W = 200 N-m in work formula

W = F* d

200 N-m = 25 N * d

Dividing both side by 25 N

[tex] \frac{200N-m}{25 N} = \frac{25 N * d}{25 N} [/tex]

d = 8 m

Hence, to lift the load using 25 N , the inclined plane of 8 m can be used .

The change in the randomness of the system is the entropy change. The entropy change after condensation at the standard boiling point is 84.64 J/K.

When a system undergoes the addition or deletion of the reactant and the products, then the disorder of the system is known as entropy change.

Given,

Enthalpy of vaporization = 38560 J/mol

Boiling point temperature = 351.3 K

[tex]\begin{aligned}\text{Entropy change of vaporization} &= \dfrac{\text{enthalpy of vaporization}}{\text{boiling point temperature}}\\\\&= \dfrac{38560}{351.3}\\\\&=109.76 \;\rm J/K/mol\end{aligned}[/tex]

Here, liquid has less entropy than gas hence the change in entropy is  -109.76 J/K/mol.

Moles of ethanol is calculated as:

[tex]\begin{aligned}\rm moles &= \dfrac{\rm mass}{\rm molar\; mass}\\\\&= \dfrac{42.2}{46}\\\\&= 0.92 \;\rm moles\end{aligned}[/tex]

If 1 mole of ethanol has an entropy change of -109.76 J/K/mol. Then, 0.92 moles will have,

[tex]\dfrac{-109.76 \times 0.92}{1} = 84.64\;\rm J/K[/tex]

Therefore, 84.64 J/K is the entropy change.

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The correct answer is a. 2 compounds have hydrogen bonding among the listed below.

To determine how many of the given compounds have hydrogen bonding, we must identify if they have hydrogen atoms attached to highly electronegative atoms like nitrogen (N), oxygen (O), or fluorine (F).

The compounds listed are:

Analysis

Based on this analysis, 2 out of the 3 compounds can form hydrogen bonds. Therefore, the correct answer is a. 2.

Correct question is: How many compounds, of the ones listed below, have hydrogen bonding?
CH₃(CH₂)₂NH₂  , CH₃(CH₂)₃NH(CH₂)₂CH₃ , (CH₃CH₂)₂N(CH₂)₄CH₃ ?
a. 2
b. 1
c. 0
d. 3

The transfer  of a section of DNA from one organism into the DNA of another organism by scientists is called genetic engineering.

DNA  is a hereditary material  which is present in human beings as well as all other living organisms.  Every cell which is present in an organism's body has DNA  which is the same. Most of the DNA is situated in the cell's nucleus and small amount of it can be found in the cell's mitochondria as well.

Information which is stored in DNA is stored as codes made up of four chemical bases namely, adenine, thymine , cytosine and guanine.Human DNA consists of 3 billion bases .The order of the bases determines information which is required for building and maintaining an organism.

DNA bases are capable of pairing up with each other. Adenine pairs with thymine and guanine pairs up with cytosine .Each base is also attached to a sugar molecule  and a phosphate group. A base, phosphate  sugar are together called as nucleotides.

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The standard emf of a galvanic cell with a Cd and Cr electrode in solutions of their respective 1.0 M nitrates at 25°C is 0.34 V.

To determine the standard emf of a galvanic cell made of a Cd electrode in a 1.0 M Cd(NO₃)₂ solution and a Cr electrode in a 1.0 M Cr(NO₃)₃ solution at 25°C, we must first identify the half-reactions taking place at each electrode and their standard reduction potentials (E°).

The standard half-cell potentials (available in standard reduction potential tables) for Cd2+ and Cr3+ are as follows:

The cathode is where reduction takes place, so the Cr3+ half-reaction will be the reduction (gain of electrons), and the Cd2+ half-reaction will be the oxidation at the anode (loss of electrons).

Next, we calculate the standard cell potential (E°cell) using the formula:

E°cell = E°cathode - E°anode

Since the Cr3+ half-reaction has the more negative standard reduction potential, it will be reversed to represent oxidation when it functions as the anode reaction. This gives us:

E°cell = (-0.40 V) - (-0.74 V) = 0.34 V

The positive standard cell potential indicates that the galvanic cell reaction is spontaneous under standard state conditions.

Final answer:

The daughter nuclide produced when ¹°Sr undergoes beta decay is ¹°Y (yttrium-90), which is represented by the nuclear equation ¹°Sr → ¹°Y + β⁻.

Explanation:

When ¹°Sr undergoes beta decay, it emits an electron (beta particle) and transforms into a different element. The loss of the electron results in an increase of the atomic number by one while the mass number remains the same. Therefore, the new atomic number will be 39 (as strontium has atomic number 38), and the mass number stays at 90. The daughter nuclide produced from this decay is ¹°Y (yttrium-90).

The nuclear equation for this ß-decay process is:
¹°Sr → ¹°Y + β⁻

Answer:

Scientists can come to different conclusions based on the same data

Explanation:

Matter are anything that is made up of atoms. The quantity of matter can be observed only on the basis of mass and volume calculation. Thus,  scientists can come to different conclusions based on the same data is the statement that describes how the scientific community analyzes data.

Matter is a substance that has some mass and can occupy some volume. The matter is mainly used in science. Matter can be solid, liquid or gas.

So as we saw that matter has some mass so mass can be measured in gram only. Mass can also be represented as number of molecules. We also saw that matter occupy some volume and that volume is measured only in liter. Scientists can come to different conclusions based on the same data is the statement that describes how the scientific community analyzes data.

Thus,  scientists can come to different conclusions based on the same data is the statement that describes how the scientific community analyzes data.

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