Answer: An electric current
Explanation:
As there is magnetic effect of electric current i.e when there is an electric current passing through any conductor then is produces a magnetic field around it depending on its shape. so due to magnetic field and a current there will be a magnetic force applied.
A buffer solution is prepared by adding NaH2PO4 to a solution of H3PO4. What happens if KOH is added?
should the visible spectrum (380nm to 770 nm) from li show more or fewer lines than the visible spectrum of hydrogen. assuming that all possible transitions occur?
Hydrogen should show more lines in the visible spectrum compared to lithium if all possible transitions occur.
The number of spectral lines in the visible spectrum of an element depends on the energy levels and electron transitions within that element. In general, the number of lines in an element's spectrum is related to the complexity of its atomic structure.
Lithium (Li) has a simpler atomic structure compared to hydrogen. Hydrogen has more energy levels and a greater number of possible electron transitions, resulting in a more complex spectrum with a larger number of spectral lines. Therefore, assuming all possible transitions occur, the visible spectrum of hydrogen would generally show more lines than the visible spectrum of lithium.
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is the smallest non-metal other than noble gases.
Answer: T is the smallest non-metal other than noble gases.
Explanation:
Hello can someone help me with this and explain Why?
Explanation:
v iv are the filter religon
which band would not appear in the product that was in the starting material in the addition of br2 to 2-pentene? group of answer choices 1680-1620 3400-3600 3000-2850 3100-3000 2260-2200 1800-1650
The C=C stretching vibration at 1680-1620 cm-1, where the double bond in 2-pentene is changed to a single bond in 1,2-dibromopentane, would not show the product that was in the beginning material.
What vibrational frequency does C-C have?The C triple C vibrations are therefore noticed at substantially higher frequencies in the range of 2300 to 2050 cm1, whilst a C-C stretching vibration occurs between 1300-800 cm1 and a C=C stretching vibration occurs between 1700 and 1500 cm1.
In what parts of IR do C-C bonds appear?As C-C bonds are typically nonpolar, they rarely manifest as peaks in the IR spectra. Since they are not extremely polar, C-H bonds do not produce prominent peaks in the IR spectra.
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Where does primary succession take place?
Answer:
Primary succession happens when a new patch of land is created or exposed for the first time.
Explanation:
Choose two reasons that the iodination EAS reaction can be described as "green." Select one or more: Use of renewable energy Use of a benign solvent Use of a less hazardous oxidant Use of a catalyst
Two reasons why the iodination electrophilic aromatic substitution (EAS) reaction can be described as "green" are the Use of a benign solvent and Use of a less hazardous oxidant. Correct answers are option : 2 & 3.
The reaction uses a solvent, such as acetic acid, which is relatively non-toxic, biodegradable, and readily available, making it an environmentally friendly choice compared to more toxic and harmful solvents. Overall, the use of benign solvents and less hazardous oxidants reduces the environmental impact of the reaction and makes it more sustainable, earning it the label of a "green" reaction. Option 2 & 3 are correct.
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--The complete Question is, Choose two reasons that the iodination EAS reaction can be described as "green." Select one or more:
1. Use of renewable energy
2. Use of a benign solvent
3. Use of a less hazardous oxidant
4. Use of a catalyst ---
Draw the repeat unit of the addition polymer that can be formed from Pent-4-enoic acid.
The repeat unit of the addition polymer that can be formed from Pent-4-enoic acid is shown below:
H H
| |
H₂- C = C-C(CH₂)₂COOH
| |
H H
How to draw a repeat unit?Since polymer molecules are much larger than most other molecules, the concept of a repeat unit is used when drawing a displayed formula.
When creating one, change the monomer's double bond to a single bond in the repeat unit, and add a bond to each end of the repeat unit. At the end, put the letter n in subscript after the brackets (n represents a very large number of the repeating unit)
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if you add 0.5312 g of benzoic acid to a 100 ml volumetric flask and then add enough water to bring the total volume to 100.00 ml, how many ml of 0.2514 m naoh solution will be required to completely neutralize the benzoic acid (pka
In the 100 ml volumetric flask, 17.3 mL of a 0.2514 M NaOH solution will be needed to fully neutralise the benzoic acid.
The first step in calculating the volume of NaOH required to neutralize the benzoic acid is to determine how many moles of benzoic acid are present in the 100 ml volumetric flask.
We can do this by dividing the mass of benzoic acid by its molar mass:
0.5312 g / 122.12 g/mol = 0.004346 mol benzoic acid
Since benzoic acid is a weak acid, we can use the Henderson-Hasselbalch equation to calculate the pH of its solution:
pH = pKa + log([A-⁻]/[HA])
The pKa of benzoic acid is 4.20. At the equivalence point of the titration, [A-] = [HA], so we can simplify the equation to:
pH = pKa + log(1) = pKa = 4.20
This means that the benzoic acid will be fully ionized at pH 4.20, and the volume of NaOH required to neutralize it can be calculated by using the balanced equation:
C₆H₅COOH + NaOH → NaC₆H₅COO + H₂O
The stoichiometric ratio of benzoic acid to NaOH is 1:1, so the moles of NaOH required to neutralize the benzoic acid is also 0.004346 mol.
To calculate the volume of 0.2514 M NaOH required to provide this number of moles, we can use the following equation:
moles = concentration x volume
0.004346 mol = 0.2514 mol/L x volume
volume = 0.0173 L = 17.3 mL
Therefore, 17.3 mL of 0.2514 M NaOH solution will be required to completely neutralize the benzoic acid in the 100 ml volumetric flask.
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how to distinguish between aqueous potassium bromide and aqueous potassium iodide TEST AND RESULT
It is possible to conduct a test to distinguish between potassium chloride and potassium iodide using a silver nitrate solution. Silver nitrate solution and ammonia solution are used in the testing for halide ions.
What happens when potassium iodide and aqueous bromine interact?When bromine-water is introduced to a potassium iodide solution, hydrobromic acid is produced as a byproduct of the oxidation to iodate, which is indicated by a sharp rise in conductivity and a fall in pH.
What is the iodide and bromide ion confirmatory test?The Layer's test is conducted using "carbon disulphide" and "dilute hydrochloric acid." This produces an orange layer when bromide ions are present, and a violet layer when iodide ions are present.
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How many moles of oxygen atoms are in 132. 2 g of MgSO4?
132.2 g of MgSO₄ contains 4.392 moles of oxygen ions.
To determine the number of moles of oxygen atoms in 132.2 g of MgSO₄, we need to first calculate the number of moles of MgSO₄, and then use its chemical formula to determine the number of oxygen atoms present.
The molar mass of MgSO₄ can be calculated by adding the atomic masses of its constituent elements, which are 24.31 g/mol for Mg, 32.06 g/mol for S, and 4x16.00 g/mol for O, respectively. Therefore, the molar mass of MgSO₄ is:
molar mass of MgSO₄ = 24.31 + 32.06 + 4(16.00) = 120.37 g/mol
Next, we can calculate the number of moles of MgSO₄ in 132.2 g as follows:
moles of MgSO₄ = mass of MgSO₄ / molar mass of MgSO₄
moles of MgSO₄ = 132.2 g / 120.37 g/mol
moles of MgSO₄ = 1.098 mol
Finally, we can use the chemical formula of MgSO₄ to determine the number of moles of oxygen atoms present in 132.2 g of MgSO4. The formula of MgSO₄ indicates that there are four oxygen atoms per molecule of MgSO₄. Therefore, the number of moles of oxygen atoms in 132.2 g of MgSO₄ is:
moles of oxygen atoms = moles of MgSO₄ x 4
moles of oxygen atoms = 1.098 mol x 4
moles of oxygen atoms = 4.392 mol
Therefore, there are 4.392 moles of oxygen atoms in 132.2 g of MgSO₄
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a student obtained a ph curve for titration between sr(oh)2 and 25.00ml of 0.10m h3po4 solution. from the ph curve, student found the volume of sr(oh)2 at the end point is 20.00 ml. calculate the concentration of sr(oh)2.
Since the student found the volume of Sr(OH)₂ at the end point is 20.00 ml, then the concentration of Sr(OH)₂ is 0.0417 M.
The balanced chemical equation for the reaction between Sr(OH)₂ and H₃PO₄ is given as:
3Sr(OH)₂ + 2H₃PO₄ → Sr₃(PO₄)₂ + 6H₂O
At the end point of the titration, the number of moles of H₃PO₄ reacted with the number of moles of Sr(OH)₂. This is given as:
Moles of H₃PO₄ = (25.00/1000) x 0.10 = 0.0025 moles
Moles of Sr(OH)₂ = 0.0025/3 = 0.00083333 moles
Given the volume of Sr(OH)₂ at the end point as 20.00 mL = 0.02 Liters, the concentration of Sr(OH)₂ is calculated using the formula:
Concentration = Number of moles / Volume= 0.00083333 / 0.02= 0.0417 M
Therefore, the concentration of Sr(OH)₂ is 0.0417 M.
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a reactant decomposes with a half-life of 103 103 s when its initial concentration is 0.154 0.154 m. when the initial concentration is 0.664 0.664 m, this same reactant decomposes with the same half-life of 103 103 s. what is the order of the reaction? 2 0 1 what is the value and unit of the rate constant for this reaction?
The order of reaction is 1 and the value of the rate constant is 6.74*10^-5s^-1.What is the order of the reaction? To solve this question we have to check the half-life of a chemical reaction at different concentrations.
It is given in the question that when the initial concentration of reactant is 0.154m, it has half-life of 103s. Similarly, when initial concentration is 0.664m, it has half-life of 103s. To find the order of reaction, we have to use the equation for half-life of reaction. The equation for half-life of reaction is given as:\[\frac{t_{1/2} }{2}=\frac{1}{k}\frac{1}{[A]_{0}}\]Where, t1/2 is half-life of the reaction,[A]0 is the initial concentration of the reactant and k is the rate constant.
Putting the values in equation: When [A]0 = 0.154m and t1/2 = 103s, we get:\[\frac{103}{2}= \frac{1}{k} \frac{1}{0.154}\] Multiplying both sides with 0.154k:\[k*\frac{103}{2}*0.154 = 1\]When [A]0 = 0.664m and t1/2 = 103s, we get:\[\frac{103}{2}= \frac{1}{k} \frac{1}{0.664}\]Multiplying both sides with 0.664k:\[k*\frac{103}{2}*0.664 = 1\]Dividing second equation by first equation:\[\frac{k*\frac{103}{2}*0.664}{k*\frac{103}{2}*0.154}= \frac{0.664}{0.154}\] Simplifying the above equation, we get:\[k=6.74*10^{-5}s^{-1}\]Therefore, the order of reaction is 1 and the value of the rate constant is 6.74*10^-5s^-1.
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Which of these gases diffuses the fastest under the same conditions? a. HCl b. H2 c. Cl2 d. He. H2.
The gas that diffuses the fastest under the same conditions is He. The correct answer is Option D.
What is diffusion?Diffusion is the process by which molecules move from high concentrations to low concentrations. The rate at which diffusion occurs is determined by the type of gas or vapor, temperature, and pressure. In a gas, the rate of diffusion is proportional to the mean free path of its molecules, which in turn is proportional to the square root of its absolute temperature.
What is the significance of He gas?Helium (He) is a chemical element with atomic number 2 and symbol He. It is a colorless, odorless, tasteless, non-toxic, inert monatomic gas that heads the noble gas group in the periodic table of elements. Its boiling and melting points are the lowest of any substance, and it exists just as a gas other than a very small liquid state in a closed cell. It's the second-most abundant element in the universe, after hydrogen. Helium is also a valuable gas, and its use in cryogenics and deep-sea diving has expanded in recent years.
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which of the following would be the most soluble in water? question 4 options: a) cacl2 b) c6h6 c) ch3ch2ch3 d) c2h2 e) ch4
The most water-soluble of the aforementioned chemicals is ethylene glycol (HOCH2 CH2 OH). Two of the hydroxy groups in ethylene glycol create hydrogen bonds with one another.
What makes water so soluble?Water is a great solvent that can dissolve a wide variety of compounds due to its polarity and capacity to create hydrogen bonds. Because it is more polar than methane, methanol dissolves better in water. Water is polar, whereas methane is non-polar.
What substances dissolve best in water?All salts of sodium, potassium, and ammonium are water soluble. 3. All metals, with the exception of lead, silver, and mercury(I), have chlorides, bromides, and iodides that are soluble in water. In water, HgI2 is insoluble.
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the individual thermodynamic contribution of w (rm)chain a was found to increase the interaction energy of the mkr681h dimer. if so, what must be true for chain a?
The individual thermodynamic contribution of w(rm)chain that was found to be increase the interaction energy of the MKR681H dimer. if so, The ture for the chain A is ΔGsolv < 0. The option A is correct.
The expression is as :
W (RM)int = W (RM)dimer - W (RM)chain A - W (RM)chain B
If the W (RM)chain A will increases the interaction energy for the MKR681H dimer, the W (RM)int, the W (RM)chain A must be the negative quantity, Like that the -W (RM)chain A term in the above equation becomes the positive value.
If the W (RM)chain A < 0, then the one or the both of the terms H intra and the ΔGsolv must be negative. Therefore, the option A is correct.
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This question is incomplete, the complete question is :
The individual thermodynamic contribution of W (RM)chain A was found to increase the interaction energy of the MKR681H dimer. If so, what must be true for chain A?
A. ΔGsolv < 0
B. ΔGsolv = 0
C. W (RM)chain A > 0
D. Hintra < 0
A ball is rolling 0. 25 meters per second. How fast is it moving in units of miles per hour?
Answer:
0.56 mi/hr
Explanation:
hope this helps
A saturated solution of potassium nitrate is prepared at 60°C in 100 g of water. How many grams of
Would it precipitate the potassium nitrate of the solution if it cooled to 30°C?
Certain potassium nitrate solutes precipitate when a saturated potassium nitrate solution is cooled.
What temperature does potassium nitrate KNO3 dissolve at?90 percent sodium nitrate dissolves in 100 grams of liquid at 30 degrees Celsius. 150g of potassium nitrate (KNO) is added to 100g of water, heated until the solute dissolves, and then cooled to 55 to create a supersaturated solution.
What transpires after cooling potassium nitrate?With a temperature drop, sodium nitrate becomes less soluble. As a result, extra potassium nitrate crystallises when a saturation potassium nitrate solution is cooled.
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which two subunits of na -nqr can be separated by gel filtration but not by ion exchange chromatography?
NqrB and NqrC are the two subunits of Na-NQR which can be separated by gel filtration but not by ion exchange chromatography.
The sodium-dependent NADH-quinone oxidoreductase (Na-NQR) is a membrane-bound enzyme complex found in bacteria that participates in the electron transport chain. It consists of six subunits: NqrA, NqrB, NqrC, NqrD, NqrE, and NqrF.
we need to consider the properties of the subunits and the mechanisms of the separation techniques. Gel filtration separates molecules based on their size, while ion exchange chromatography separates molecules based on their charge.
Based on this information, we can infer that the two subunits that can be separated by gel filtration but not by ion exchange chromatography are those that have similar sizes but different charges. Among the six subunits of Na-NQR, NqrB and NqrC are the two subunits that have similar molecular weights (~45 kDa) but different charges.
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The balanced equation below represents the reaction that occurs inside the airbag:
2NaN 3 (s) à 2Na(s) + 3N 2 (g)
How many grams of NaN 3 are needed to produce the number of moles you calculated in problem 1 above?
Approximately 402.27 grams of sodium azide (NaN₃) must decompose to fill a 55.0-gallon airbag.
What is ideal gas law?The ideal gas law is a fundamental equation that describes the behavior of ideal gases under a wide range of conditions. It relates the pressure (P), volume (V), temperature (T), and number of moles of gas (n) of an ideal gas through the equation:
PV = nRT
where R is the gas constant, which has a value of 8.314 J/(mol·K) or 0.0821 L·atm/(mol·K) in SI units.
We know that the volume of the airbag is 55.0 gallons, so we need to convert this to liters using a conversion factor.
1 gallon = 3.78541 liters
Therefore, the volume of the airbag in liters is:
55.0 gallons x 3.78541 liters/gallon = 208.20 liters
Next, we need to calculate the number of moles of N₂ gas that would be produced from the decomposition of NaN₃ required to fill the airbag.
From the balanced equation:
2 NaN₃ (s) -> 2 Na (s) + 3 N₂ (g)
We can observe that two moles of NaN₃ result in three moles of N₂. Therefore, the number of moles of N2 produced is:
moles of N2 = (2/3) x moles of NaN₃
To fill the airbag, we need enough N₂ gas to occupy a volume of 208.20 liters. One mole of any gas has a volume of 22.4 litres at standard temperature and pressure (STP).
Therefore, the number of moles of N₂ required is:
moles of N₂ = (208.20/22.4) = 9.29 moles
Now we can use the balanced equation to calculate the number of moles of NaN₃ required:
2 NaN₃ (s) -> 2 Na (s) + 3 N2 (g)
For every 3 moles of N₂ produced, we need 2 moles of NaN3. Therefore, the number of moles of NaN₃ required is:
moles of NaN₃ = (2/3) x moles of N₂ = (2/3) x 9.29 = 6.19 moles
Finally, we can use the molar mass of NaN₃ to calculate the mass required:
mass of NaN₃ = moles of NaN₃ x molar mass of NaN₃
The molar mass of NaN₃ is:
Molar mass of NaN₃ = (1 x 22.99) + (3 x 14.01) = 65.01 g/mol
Therefore, the mass of NaN₃ required to fill the airbag is:
mass of NaN₃ = 6.19 moles x 65.01 g/mol = 402.27 grams
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The molar heat of fusion for Gold is 12.5 kJ/mol. The specific heat capacity of gold is 0.129 J/g oC.
Calculate the amount of energy required to melt 70.0 g of solid Gold and then heat the liquid to 1213 oC? The melting point of gold is 1063 oC.
Explanation:
The heat of fusion given has units of kJ / MOL
so we need to find the number of moles in 70 . 0 g
using periodic table, mole wt for AU =196.97 g/mol
then 70 g is 70 g / 196.97 g/mol = .355 mole
Now to MELT the gold .355 mole * 12.5 kJ/mol = 4.44 kJ ( = 4440 J)
Then to HEAT the liquid to 1213 degrees C from the melting point :
70 g * (1213 - 1063) C * .129 J / (g C) = 1355 J
Then add together 4.44kJ + 1355 J = 4440 J + 1355 J = 5795 J
Why should a standardized NaOH solution be kept in a stoppered bottle with a rubber stopper?
A standardized NaOH (sodium hydroxide) solution should be kept in a stoppered bottle with a rubber stopper to prevent it from reacting with atmospheric carbon dioxide.
Carbon dioxide can dissolve in the solution and react with the NaOH to form sodium carbonate, which can change the concentration of the solution. This reaction can also produce heat and gas, which can cause pressure to build up in the bottle and possibly lead to the stopper popping out.
By using a rubber stopper, the solution is protected from atmospheric carbon dioxide and the stopper is able to release any pressure that may build up inside the bottle. Additionally, the use of a stoppered bottle prevents evaporation and contamination of the solution.
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What are foods that contain chemical changes
meat cooking is a chemical change
Heating a meat changes its colour to brownish due to Maillard reaction
what is civic engagement and what are your tips on water scarcity
when filling a burette for a titration, adjust the burette so that choose... , preferably over a sink. then, choose... to add the titrant into the burette. the titrant should be filled choose... .
When filling a burette for a titration, adjust the burette so that the tip is located slightly below the level of the meniscus, preferably over a sink. Then, use a funnel to add the titrant into the burette. The titrant should be filled above the 0.00 mL line.
A burette is a laboratory equipment that is used to dispense known volumes of liquid in experimental procedures. It is usually made of glass and has a long, cylindrical shape with a stopcock at the bottom to control the flow of liquid.
Burettes are commonly used in titration experiments to accurately measure the volume of the titrant added to the sample. Adjust the burette so that the tip is located slightly below the level of the meniscus, preferably over a sink. This is to prevent the loss of any of the titrant that may overflow. Using a funnel, carefully add the titrant into the burette. Make sure to pour the titrant slowly to avoid splashing or spilling any of it. The titrant should be filled above the 0.00 mL line.
This allows the initial volume of the titrant to be measured accurately before titration commences. If there are air bubbles present, they should be removed by gently tapping the burette.
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SOMEONE PLEASE HELP ME PLEASE
Tommy can expect to produce 6.10 grams of potassium chloride.
How to calculate number of moles?To determine the amount of potassium chloride produced, we first need to determine the limiting reagent, which is the reactant that is completely consumed in the reaction.
To do this, we can convert the given masses of potassium phosphate and magnesium chloride into moles using their respective molar masses:
Molar mass of K₃PO₄ = 3 x 39.1 g/mol (K) + 1 x 30.97 g/mol (P) + 4 x 16.00 g/mol (O) = 212.27 g/mol
Molar mass of MgCl₂ = 1 x 24.31 g/mol (Mg) + 2 x 35.45 g/mol (Cl) = 95.21 g/mol
Moles of K₃PO₄ = 5.79 g / 212.27 g/mol = 0.0273 mol
Moles of MgCl₂ = 4.92 g / 95.21 g/mol = 0.0517 mol
Now, we need to determine which reactant is the limiting reagent by comparing the mole ratios of the reactants in the balanced chemical equation:
For every 2 moles of K₃PO₄, we need 3 moles of MgCl₂ to react completely.
The mole ratio of K₃PO₄ to MgCl₂ is therefore 2:3.
Since we have more moles of MgCl₂ than required by the mole ratio, MgCl₂ is in excess and K₃PO₄ is the limiting reagent. This means that all of the K₃PO₄ will be used up in the reaction and there will be some MgCl₂ left over.
Using the mole ratio of the balanced equation, we can now calculate the moles of KCl produced:
For every 2 moles of K₃PO₄, we get 6 moles of KCl.
The mole ratio of K₃PO₄ to KCl is therefore 2:6, or 1:3.
Moles of KCl produced = 0.0273 mol K₃PO₄ x (6 mol KCl / 2 mol K₃PO₄) = 0.0819 mol KCl
Finally, we can convert the moles of KCl produced into grams using the molar mass of KCl:
Molar mass of KCl = 1 x 39.1 g/mol (K) + 1 x 35.45 g/mol (Cl) = 74.55 g/mol
Grams of KCl produced = 0.0819 mol KCl x 74.55 g/mol = 6.10 g KCl
Therefore, Tommy can expect to produce 6.10 grams of potassium chloride.
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For each of the following series of compounds, arrange the molecules in order of decreasing basicity (most basic to least basic): NH4 :NH, INH₂
Explanation:
The basicity of a compound depends on its ability to donate a pair of electrons to an acid. The more easily a compound can donate electrons, the stronger the base it is. Based on this, we can order the compounds NH4+, NH, and INH₂ in terms of decreasing basicity as follows:
NH > NH4+ > INH₂
Here's why:
NH is the most basic of the three compounds because it has a lone pair of electrons on the nitrogen atom that is not shared with any other atoms. This makes it a very good electron donor, and therefore a strong base.
NH4+ is the next most basic compound because it is a positively charged ion, meaning it has lost one of its electrons. As a result, it is not as good at donating electrons as NH, but it is still a stronger base than INH₂.
INH₂ is the least basic of the three compounds because it has two electron-withdrawing groups (the Iodine atoms) attached to the nitrogen atom. These groups decrease the electron density around the nitrogen atom, making it less able to donate electrons and therefore a weaker base than NH4+ and NH.
The molar heat of vaporization for liquid water is 40.6 kJ/mole.
How much energy is required to change 25 g of liquid water to steam if the water is already at 100oC?
Answer:
Answer:
56.4 kJ
Explanation:
First, let's convert the mass of water from grams to moles. We can do this by dividing the mass by the molar mass of water, which is approximately 18 g/mol.
25 g ÷ 18 g/mol ≈ 1.39 mol
So we have 1.39 moles of water that we want to vaporize.
Next, we need to use the molar heat of vaporization to calculate how much energy is required to vaporize one mole of water. The molar heat of vaporization tells us how much energy is needed to vaporize one mole of a substance at a constant temperature and pressure. In this case, the molar heat of vaporization for water is 40.6 kJ/mol.
So, to vaporize 1 mole of water, we need 40.6 kJ of energy.
Finally, we can use this information to calculate how much energy is required to vaporize 1.39 moles of water. We can multiply the energy required to vaporize one mole of water by the number of moles we have:
40.6 kJ/mol × 1.39 mol ≈ 56.4 kJ
Therefore, it would take approximately 56.4 kJ of energy to vaporize 25 g of liquid water at 100°C.
I hope this explanation helps!
Draw Lewis structures for the fulminate ion including possible resonance forms.
Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all hydrogen atoms and nonbonding electrons. Show the formal charges of all atoms in the correct structure.
The fulminate ion has the chemical formula CNO-, and its Lewis structure can be drawn as follows:
Place the carbon atom in the center since it is the least electronegative atom among C, N, and O.
Connect the carbon atom to the nitrogen atom with a triple bond, as nitrogen is more electronegative than carbon.
Connect the nitrogen atom to the oxygen atom with a single bond since oxygen is more electronegative than nitrogen.
Add a lone pair of electrons to the oxygen atom to satisfy its octet.
Place a negative charge on the oxygen atom since it has gained an extra electron.
The Lewis structure with all atoms and bonds is as follows:
markdown
O
||
C ≡ N -
||
H
All atoms except for the nitrogen atom have a formal charge of 0. The nitrogen atom has a formal charge of +1 because it has four valence electrons but only three bonding electrons. The oxygen atom has a formal charge of -1 because it has six valence electrons but seven electrons around it.
The Lewis structure can also be represented by showing the possible resonance forms:
makefile
O O
|| ||
C = N ↔ C ≡ N
|| ||
H H
In this case, the double bond is delocalized between the carbon and nitrogen atoms, and both resonance structures contribute to the overall electronic structure of the fulminate ion.
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describe the processes by which energy stored in nadh molecules is transferred to atp by oxidative phosphorylation
The energy stored in NADH molecules is transferred to ATP by oxidative phosphorylation through a series of processes involving the Electron Transport Chain, the creation of a proton gradient, and ATP synthesis via chemiosmosis.
What are the process involved in Oxidative Phosphorylation?
1. Electron Transport Chain (ETC): The first step in oxidative phosphorylation involves the Electron Transport Chain, which is a series of protein complexes located in the inner mitochondrial membrane. NADH molecules transfer their high-energy electrons to the first complex in the ETC, known as Complex I.
2. Transfer of electrons: As electrons move through the ETC, they are transferred from one complex to another (from Complex I to Complex II, Complex II to Complex III, and Complex III to Complex IV). During this transfer, energy is released, which is used to pump hydrogen ions (protons) across the inner mitochondrial membrane from the matrix into the intermembrane space.
3. Proton gradient: The pumping of protons across the inner mitochondrial membrane creates an electrochemical gradient, which is also known as a proton gradient. This gradient represents a form of potential energy.
4. ATP synthase: As protons flow back into the mitochondrial matrix through a protein complex called ATP synthase, the energy from the proton gradient is harnessed to synthesize ATP from ADP (adenosine diphosphate) and inorganic phosphate (Pi). This process is known as chemiosmosis.
5. Final electron acceptor: The electrons reach the end of the ETC and are transferred to the final electron acceptor, which is molecular oxygen ([tex]O_{2}[/tex]). This transfer results in the formation of water ([tex]H_{2}O[/tex]), as oxygen combines with protons from the matrix.
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