The coolest and dimmest space objects emit radiation of longer wavelengths than those we can see. A detector for which type of electromagnetic radiation is most suitable to study cool dim stars?

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?

If a person mixes 20.0 ml of a 3.00m sugar solution with 30.0 ml of a 5.69m sugar solution, you will end up with a sugar solution of:

According to the given question, we can see that there is a mixture of 20 milliliters of 3 moles of sugar solution with 30 milliliters of 5.69 moles of another sugar solution, then we need to find the total sugar solution from the mixtures.

As a result of this, we need to first convert the milliliters to moles.

20 mL/1000 mL x 3 moles = 0.06 moles of sugar

30 mL/ 1000 mL x 5.69 moles = 0.1707 moles of sugar

Please note that we are dividing by 1000 mL to convert to liters.

Now, we add up the two values, 0.06 + 0.1707 = 0.2307 moles

Next, we add the total millimeters we have so far which would be

20 mL + 30 mL = 50 mL

Finally, we would convert the 50 mL to liters before we get our final answer.

1000 mL / 50 mL = 20 L

Now, we multiply 20 L by 0.2307 moles = 4.614 M solution

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Answer : The mass of solute is 4.72 grams.

Explanation :

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

[tex]\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]

or,

[tex]\text{Molarity}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]

In this question, the solute is potassium bromide.

Given:

Volume of solution = 25.4 mL

Molarity = 1.56 M

Molar mass of potassium bromide = 119 g/mole

Now put all the given values in this formula, we get:

[tex]1.56M=\frac{\text{Mass of solute}\times 1000}{119g/mole\times 25.4L}[/tex]

[tex]\text{Mass of solute}=4.72g[/tex]

Therefore, the mass of solute is 4.72 grams.

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 [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]

In the compound K2SO4, the oxidation numbers for potassium (K), sulfur (S) and oxygen (O) are +1, +6 and -2, respectively.

In the compound K2SO4, the oxidation numbers for potassium (K), sulfur (S) and oxygen (O) can be determined based on established guidelines for assigning oxidation states.

In every stable (neutral) atom, the oxidation number is always zero. Therefore, for potassium, in its ionic form, it has an oxidation number of +1.

In general, the oxidation number of oxygen in its compounds is -2. The compound K2SO4 contains 4 oxygen atoms. Therefore, the total oxidation state contributed by oxygen is -8.

To ensure that the compound is electrically neutral, the total oxidation number should be zero. Hence, for sulfur, you would calculate its oxidation state as follows: total oxidation state of compound = (+1 x 2 for potassium) + oxidation state of sulfur + (-2 x 4 for oxygen) = 0. Solving this equation gives the oxidation state for sulfur as +6.

So, the oxidation numbers for potassium, sulfur and oxygen in K2SO4 are +1, +6 and -2, respectively.

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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.

The electrolysis of molten sodium iodide (NaI) yields sodium metal and iodine gas. The process involves the migration of ions to respective electrodes, gaining or losing electrons, thus getting reduced or oxidized respectively.

The products obtained from the electrolysis of molten sodium iodide (NaI) are sodium (Na) and iodine (I2). During electrolysis, sodium ions are reduced at the cathode to form sodium metal, shown by the half-equation 2Na+ + 2e- → 2Na. Meanwhile, iodide ions get oxidized at the anode to produce iodine gas, as depicted by 2I- → I2 + 2e-.

In a nutshell, the process involves electrolysis of molten sodium iodide using a Downs cell. Positively charged sodium ions migrate to the negatively charged cathode and gain electrons, reducing them to sodium metal. Conversely, the negatively charged iodide ions migrate to the positively charged anode and lose electrons, getting oxidized to iodine gas.

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The electrolysis of molten sodium iodide (NaI) produces sodium (Na) and iodine (I2). In the process, sodium ions migrate to the cathode and are reduced to sodium metal, while iodide ions migrate to the anode and are oxidized to iodine gas.

The process of electrolyzing molten sodium iodide, or NaI, is similar both in concept and execution to the electrolysis of molten sodium chloride, which is a better-known and more commonly discussed process. Electrolysis of molten sodium iodide will produce sodium (Na) and iodine (I) as products. Here's a description of how it works:

In the set-up, you would have a Downs cell. The Downs cell contains molten sodium iodide. The passage of a direct current through the cell causes the sodium ions to migrate to the negatively charged cathode and pick up electrons, reducing the ions to sodium metal. Iodide ions migrate to the positively charged anode, lose electrons, and undergo oxidation to iodine gas.

The overall cell reaction would be: 2NaI -> 2Na + I2

This is a simple explanation of the process and should suffice for your understanding. Do keep in mind that actual industrial processes could have additional steps and complexities not addressed here.

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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 is: the % ionization of hypochlorous acid is 0.14.

Balanced chemical reaction (dissociation) of an aqueous solution of hypochlorous acid:

HClO(aq) ⇄ H⁺(aq) + ClO⁻(aq).

Ka = [H⁺] · [ClO⁻] / [HClO].

[H⁺] is equilibrium concentration of hydrogen cations or protons.

[ClO⁻] is equilibrium concentration of hypochlorite anions.

[HClO] is equilibrium concentration  of hypochlorous acid.

Ka is the acid dissociation constant.

Ka(HClO) = 3.0·10⁻⁸.

c(HClO) = 0.015 M.

Ka(HClO) = α² · c(HClO).

α = √(3.0·10⁻⁸ ÷ 0.015).

α = 0.0014 · 100% = 0.14%.

The percent ionization of the solution is 0.14%.

        HClO(aq) ⇄      H^+(aq)   +    ClO^-(aq)

I        0.015                 0                    0

C       -x                       +x                   +x

E   0.015 - x                  x                      x

Ka = [ H^+] [ClO^-]/[HClO]

3.0 x 10^-8= x^2/ 0.015 - x  

3.0 x 10^-8(0.015 - x) = x^2

4.5 x 10^-10 - 3.0 x 10^-8x = x^2

x^2 + 3.0 x 10^-8x - 4.5 x 10^-10 = 0

x = 0.000021 M

Percent ionization = 0.14%

Hence, the percent ionization of the solution is 0.14%.

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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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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.

Answer:

L Group

Explanation:

Proteins are made up of amino acids, these amino acids determine the protein's functional groups, in the image I annexed you can observe that amino acids have an amino group (NH2), an acid group(COOH) and a R group, a lateral chain that varies depending on the amino acid, in the protein these amino acids are linked through a peptide bond leaving one side of the chain with an acid group and the other side with an amide chain, keeping the R groups as well.

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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 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 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.

Answer: The mass of NaOH for given number of moles is 31.4 grams.

Explanation:

To calculate the mass for given number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

We are given:

Molar mass of NaOH = 40 g/mol

Moles of NaOH = 0.785 moles

Putting values in above equation, we get:

[tex]0.785mol=\frac{\text{Mass of NaOH}}{40g/mol}\\\\\text{Mass of NaOH}=(0.785mol\times 40g/mol)=31.4g[/tex]

Hence, the mass of NaOH for given number of moles is 31.4 grams.

Final answer:

To determine the rate law, the rate-determining step must be identified, which is the third step in the given reaction mechanism. The rate law correlates with the concentration of reactants in this slow step, resulting in a rate law of rate [tex]= k[A]^2[C][/tex] correct answer to the question is a. rate = [tex]k[A]^2[C].[/tex]

Explanation:

To determine the rate law for a reaction with a given mechanism, it is essential to identify the rate-determining step, as the rate law is based on this slowest step. In the provided mechanism, the slow step is the third one, where B2 reacts with C to form G. The overall reaction is 4 A + C + H → D, and the steps are:

2 A → B (fast)

2 B → B2 (fast)

B2 + C → G (slow)

G + H → D (fast)

Because step 3 is the slow, rate-determining step, the rate law will be based on the concentrations of the reactants in this step. Since B is formed from A in a fast step and B2 is formed from 2 B, the rate of formation of B2 is dependent on the concentration of A. However, as B2 forms immediately before the slow step, we look at the concentration of A instead of B2 when writing the rate law.

With the stoichiometry of 2 A forming B and then 2 B forming B2, we can note that the concentration of B (and hence B2) is proportional to [tex][A]^2[/tex]efore, for the rate-determining step, the rate law is:

[tex]rate = k[B2][C] → rate = k[A]^2[C][/tex]

So the correct rate law for the overall reaction is:

[tex]rate = k[A]^2[C][/tex]

[tex]rate = k[A]^2[C][/tex]

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 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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