We are given that 1 drop of 0.25 M HCl is added to 3 mL of water, and we need to find the concentration of H+ ions and the pH of the solution is 2.39
First, let's determine the volume of the HCl solution in the mixture. Since 1 drop of acid is equal to 50 microliters, we have 50 microliters = 0.05 mL
Now, let's find the total volume of the mixture (HCl + water):
0.05 mL (HCl) + 3 mL (water) = 3.05 mL
Next, we need to calculate the moles of H+ ions from the HCl solution. We know that the concentration of the HCl solution is 0.25 M, so:
moles of H+ = (0.25 mol/L) × (0.05 L/1000) = 0.0000125 mol
To find the concentration of H+ ions in the mixture, we divide the moles of H+ by the total volume of the mixture:
[H+] = (0.0000125 mol) / (3.05 L/1000) = 0.004098 mol/L
Now we can calculate the pH of the solution using the formula:
pH = -log10[H+]
pH = -log10(0.004098) ≈ 2.39
The pH of the solution is approximately 2.39 after adding 1 drop of 0.25 M HCl to 3 mL of water.
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Please show explanation: If 1 drop of acid is equal to 50 microliter. Calculate the concentration of H+ ion and the pH of the solution when 1 drop of 0.25 M HCl is added to 3 mL water?
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if 10 grams of aluminum reacts with 4 grams of oxygen, what is the expected grams of product?
Expected grams of aluminum oxide product from the given masses of reactants are 18.93 g.
What is aluminum?Aluminum is chemical element with symbol Al and atomic number is 13.
4Al + 3O₂ → 2Al₂O₃
10 g Al × 1 mol Al / 26.98 g Al = 0.371 mol Al
4 g O₂ × 1 mol O₂ / 32.00 g O₂ = 0.125 mol O₂
We determine the limiting reactant by comparing the mole ratios of aluminum and oxygen in the balanced equation and reactant that produces smaller amount of product is limiting reactant. In this case, aluminum is the limiting reactant because it produces only 0.1855 moles of aluminum oxide, which is less than the 0.25 moles of aluminum oxide produced by the oxygen:
0.371 mol Al × 2 mol Al₂O₃ / 4 mol Al = 0.1855 mol Al₂O₃
0.125 mol O₂ × 2 mol Al₂O₃ / 3 mol O2 = 0.2083 mol Al₂O₃
0.1855 mol Al₂O₃ × 101.96 g/mol = 18.93 g Al₂O₃
Therefore, expected grams of aluminum oxide product from the given masses of reactants are 18.93 g.
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What is the concentration (in molality) of an aqueous solution of NaCl made by adding
4.56 g of NaCl to enough water to give 20.0 mL of solution. Assume the density of the
solution is 1.03 g/mL
Answer:
data given
mass of NaCl 4.56
dissolved volume 20ml(0.02l)
density of solution 1.03g/ml
Required molality
Explanation:
molarity=m/mr×v
where
m is mass
mr molar mass
v is volume
now,
molarity=4.56/58.5×0.02
molarity =3.9
: .molarity is 3.9mol/dm^3
According to molal concentration, the concentration (in molality) of an aqueous solution of NaCl is 0.0047 mole/kg.
What is molal concentration?Molal concentration is defined as a measure by which concentration of chemical substances present in a solution are determined. It is defined in particular reference to solute concentration in a solution . Most commonly used unit for molal concentration is moles/kg.
The molal concentration depends on change in volume of the solution which is mainly due to thermal expansion. Molal concentration is calculated by the formula, molal concentration=mass/ molar mass ×1/mass of solvent in kg.
In terms of moles, it's formula is given as molal concentration= number of moles /mass of solvent in kg.
Substitution in formula gives the answer but first mass of solution is determined which is density×volume= 1.03×20=20.6 g , mass of solvent= 20.6-4.56=16.05, thus molal concentration=4.56/58.5×1/16.05=0.0047 moles/kg.
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tollens's test shows the presence of aldehydes . a positive tollens's test appears as a silver precipitate . a negative tollens's test appears as
Tollens's test shows the presence of aldehydes . a positive Tollens's test appears as a silver precipitate . a negative Tollens's test appears as presence of ketone.
Tollens's test is a chemical test used to differentiate between aldehydes and ketones. In this test, a solution called Tollens's reagent, which contains silver nitrate and ammonia, is used to detect the presence of aldehydes. When an aldehyde is present, it undergoes oxidation by reacting with the Tollens's reagent, forming a silver precipitate.
A positive Tollens's test is indicated by the formation of this silver precipitate, which appears as a shiny silver layer on the inside of the test tube. This silver layer is also referred to as a "silver mirror." This reaction occurs because the aldehyde group is oxidized to a carboxylic acid, while the silver ions in the Tollens's reagent are reduced to metallic silver.
On the other hand, a negative Tollens's test means that no aldehyde is present, and thus, no silver precipitate forms. This is typically observed when a ketone is present in the test sample, as ketones do not readily undergo oxidation like aldehydes do. In this case, the test tube remains clear or slightly cloudy, depending on the reaction conditions and the substances being tested.
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Complete question is :-
tollens's test shows the presence of aldehydes . a positive tollens's test appears as a silver precipitate . a negative tollens's test appears as ______.
which of the following is true about the absorption and metabolism of alcohol? alcohol is metabolized by most tissue and organs in the body. the majority of alcohol is absorbed in the stomach. men and women do not metabolize alcohol at significantly different rates. acetaldehyde produced during alcohol metabolism is highly toxic.
The statement "acetaldehyde produced during alcohol metabolism is highly toxic" is true about absorption and metabolism of alcohol. Option 4 is correct.
Acetaldehyde is a byproduct of alcohol metabolism, and it is a toxic substance that can cause various symptoms such as facial flushing, nausea, and headache. Acetaldehyde is rapidly converted to acetate by the enzyme aldehyde dehydrogenase, which is then metabolized further to carbon dioxide and water.
However, if alcohol is consumed at a high rate, the liver may not be able to metabolize all of the acetaldehyde, leading to a buildup of this toxic substance in the body. This can result in more severe symptoms such as vomiting, rapid heartbeat, and difficulty breathing. Therefore, it is important to consume alcohol in moderation and allow enough time for the liver to metabolize the alcohol and its byproducts. Hence Option 4 is correct.
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addictive substances, for which demand is inelastic, are products for which producers can pass higher costs on to consumers.
The statement is correct. Producers of addictive substances, for which demand is inelastic, can pass higher costs on to consumers.
Inelastic demand refers to a situation where changes in price have little effect on the quantity demanded of a product. Addictive substances, such as tobacco or drugs, often have inelastic demand because users are willing to pay high prices for the product regardless of changes in price.
Producers of addictive substances can take advantage of this inelastic demand by increasing prices without seeing a significant decrease in demand. This means that they can pass on any higher costs, such as increased taxes or production costs, to the consumers, who are likely to continue purchasing the product even at a higher price.
This is often seen in the tobacco industry, where governments may increase taxes on cigarettes as a way to discourage smoking, but the tobacco companies can simply pass on the higher costs to consumers who continue to buy the product.
Therefore, it can be concluded that producers of addictive substances, for which demand is inelastic, can pass higher costs on to consumers.
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you have 400 grams (g) of a substance with a half life of 10 years. how much is left after 100 years?
After 100 years, there will be 6.25 grams of the substance remaining.
What is half life?Half-life is the time it takes for half of the radioactive atoms in a sample to decay or for the concentration of a substance to decrease by half.
Amount remaining = initial amount x (1/2)^(number of half-lives)
In this case, half-life of the substance is 10 years, which means that after 10 years, half of the substance will have decayed. After another 10 years (20 years total), half of remaining substance will decay, leaving 1/4 of the original amount. After another 10 years (30 years total), half of that remaining amount will decay, leaving 1/8 of the original amount. This process continues every 10 years.
To find the amount of substance remaining after 100 years, we need to know how many half-lives have occurred in that time: 100 years / 10 years per half-life = 10 half-lives
Amount remaining = 400 g x (1/2)¹⁰= 6.25 g
Therefore, after 100 years, there will be 6.25 grams of the substance remaining.
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a 1.25 g sample of co2 is contained in a 750. ml flask at 22.5 c. what is the pressure of the gas, in atm?
The pressure of gas is 1.05 atm when a 1.25 g sample of CO₂ is contained in a 750ml flask at 22.5°C.
Molecular weight of CO₂ is 1.25g ,Volume of CO₂ is 750ml,Temperature of CO₂ is 22.5°C and the gas constant is 0.08206 L atm/mol K.
Using the ideal gas law equation the pressure is found to be 1.05 atm.
To calculate the pressure of the gas, we can use the ideal gas law equation: [tex]PV=nRT[/tex]
Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
First, we need to convert the volume to liters by dividing by 1000: 750 ml = 0.75 L.
Next, we need to calculate the number of moles of CO₂ present in the flask. We can use the molecular weight of CO₂ to convert from grams to moles:
[tex]1.25 * (1 /44.01 ) = 0.0284 mol[/tex]
Now we can plug in the values into the ideal gas law equation:
[tex]PV=nRT[/tex]
[tex]P * 0.75 L = 0.0284 mol * 0.08206 L*atm/mol*K * (22.5 + 273.15) K[/tex]
Simplifying and solving for P, we get:
[tex]P = (0.0284 * 0.08206 * 295.65) / 0.75 = 1.05 atm[/tex]
Therefore, the pressure of the gas in the flask is 1.05 atm.
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calculate the engery of a photon needed to cause an electron in the 3s orbital to be excited to tthe 3p orbital
The energy of the photon needed to cause an electron in the 3s orbital to be excited to the 3p orbital is approximately 3.04 × [tex]10^{-18}[/tex] J (or about 1.90 eV).
To calculate the energy of a photon needed to cause an electron in the 3s orbital to be excited to the 3p orbital, we need to know the energy difference between these two orbitals.
The energy of an electron in a hydrogenic atom (an atom with one electron) can be calculated using the following formula:
[tex]E = - (Z^2 * Ry) / n^2[/tex]
where Z is the atomic number, Ry is the Rydberg constant (2.18 × [tex]10^{-18}[/tex]J), and n is the principal quantum number.
The energy difference between the 3s and 3p orbitals can be calculated by subtracting the energy of the 3s orbital from the energy of the 3p orbital.
For hydrogen, the energy of the 3s orbital is:
E(3s) = - ([tex]1^2[/tex]* 2.18 × [tex]10^{18}[/tex] J) / [tex]3^2[/tex]
E(3s) = - 0.242 ×[tex]10^{18}[/tex] J
And the energy of the 3p orbital is:
E(3p) = - ([tex]1^2[/tex] * 2.18 × [tex]10^{-18}[/tex] J) / 2^2
E(3p) = - 0.546 × [tex]10^{-18}[/tex] J
The energy difference between the two orbitals is:
ΔE = E(3p) - E(3s)
ΔE = (- 0.546 ×[tex]10^{18}[/tex] J) - (- 0.242 ×[tex]10^{-18}[/tex] J)
ΔE = - 0.304 × [tex]10^{-18}[/tex]J
This energy difference represents the energy required to excite an electron from the 3s orbital to the 3p orbital.
To calculate the energy of the photon needed to provide this energy, we use the formula:
E = hν
where E is the energy of the photon, h is Planck's constant (6.626 × [tex]10^{-34}[/tex]J·s), and ν is the frequency of the photon.
Rearranging this formula to solve for the frequency of the photon, we get:
ν = E / h
Substituting the energy difference we calculated, we get:
ν = (- 0.304 × [tex]10^{18}[/tex] J) / (6.626 × [tex]10^{-34}[/tex] J·s)
ν = - 4.59 × [tex]10^{15}[/tex]Hz
Finally, to calculate the energy of the photon, we use the formula:
E = hν
Substituting the frequency we calculated, we get:
E = (6.626 ×[tex]10^{-34}[/tex] J·s) x (- 4.59 × [tex]10^{15}[/tex] Hz)
E = - 3.04 × [tex]10^{-18}[/tex]J
Therefore, the energy of the photon needed to cause an electron in the 3s orbital to be excited to the 3p orbital is approximately 3.04 × 10^-18 J (or about 1.90 eV).
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