Answers
Answer:
Explanation:
Chemical species which can behave as both acids and bases are known as amphoteric species.
can behave as both acid and base.
can donate H+:
can accept H+ as well:
HF can only behave as acid, as it can only donate H+.
is a conjugate acid of
.
is a conjugate base.
Final answer:
In chemistry, an amphoteric species can act as both a base and an acid. Among the given options, HCO3- (bicarbonate ion) is amphoteric because it can either donate or accept protons.
Explanation:
Among the provided options, the species that are amphoteric are HCO3- (bicarbonate ion). The term amphoteric refers to substances that can act both as an acid and a base. In other words, they can either donate or accept protons. Let's take HCO3- as an example. This ion can act as a base by accepting H+, forming H2CO3, or it can act as an acid by donating H+, forming CO32-. This dual behavior makes it an amphoteric species.
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Related Questions
To measure the solubility product of lead (II) iodide (PbI2) at 25°C, you constructed a galvanic cell that is similar to what you used in the lab. The cell contains a 0.5 M solution of a lead (II) nitrate in one compartment that connects by a salt bridge to a 1.0 M solution of potassium iodide saturated with PbI2 in the other compartment. Then you inserted two lead electrodes into each half-cell compartment and closed the circuit with wires. What is the expected voltage generated by this concentration cell? Ksp for PbI2 is 1.4 x 10-8. Show all calculations for a credit.
If u trust urself do itStudy these images.4 photos of clouds. 1: Sky covered with large, flat layers of blue, grey clouds. 2: A tall, fluffy cloud shaped like an anvil. 3: Round, puffy clouds in a blue sky. 4: Thin, wispy clouds high in the sky.Which image shows a cumulonimbus cloud?123
A solution of HNO3HNO3 is standardized by reaction with pure sodium carbonate. 2H++Na2CO3⟶2Na++H2O+CO2 2H++Na2CO3⟶2Na++H2O+CO2 A volume of 27.71±0.05 mL27.71±0.05 mL of HNO3HNO3 solution was required for complete reaction with 0.9585±0.0007 g0.9585±0.0007 g of Na2CO3Na2CO3 , (FM 105.988±0.001 g/mol105.988±0.001 g/mol ). Find the molarity of the HNO3HNO3 solution and its absolute uncertainty.
A 6.13 g sample of an unknown salt (MM = 116.82g/mol) is dissolved in 150.00 g water in a coffee cupcalorimeter. Before placing the sample in the water, thetemperature of the salt and water is 23.72°C. After thesalt has completely dissolved, the temperature of thesolution is 28.54°C.If 3.15 x 10J of heat was gained by the solution, whatis the total heat for the dissolution reaction of the 6.13 gof salt?
Answers
Final answer:
The given reaction will shift towards cis-2-butene once placed in equilibrium. This can be determined by calculating the reaction quotient and comparing it with the equilibrium constant.
Explanation:
The reaction could either shift towards the cis-2-butene or trans-2-butene depending on whether the reaction quotient, Q, is lesser or greater than the equilibrium constant, Kp.
Bear in mind that Kp = Ptrans/Pcis. Let's say that Pt is the partial pressure of trans-2-butene and Pc is the partial pressure of cis-2-butene at equilibrium. If we start with 5 atm of each gas, the change in Pc is -x and the change in Pt is +x.
So, Kp = (5+x)/(5-x). We are given that Kp = 3.4. Solving these two equations will show that x is a negative value, which means that the system shifts towards cis-2-butene.
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Final answer:
For the isomerization reaction cis-2-butene ⇌ trans-2-butene, with an initial pressure of 5.00 atm for both gases and a Kp of 3.40, the system will shift towards the product, trans-2-butene, as Kp > Qp (1). This reflects the principle that a chemical system at equilibrium will shift to counteract any change.
Explanation:
In terms of the equilibrium constant (K), for gas-phase reactions, Kp represents equilibrium in terms of partial pressures, while Kc represents it in molar concentrations. For instance, in the isomerization reaction given cis-2-butene ⇌ trans-2-butene, Kp is given as 3.40. To determine the behavior of the system, we need to compare it to reaction quotient (Q). Given that the flask initially contains 5.00 atm of each gas, Qp is 1 (since Qp = partial pressure of trans-2-butene / partial pressure of cis-2-butene). Since Kp > Qp, the reaction will shift towards the products, hence the system will shift towards trans-2-butene. From this, it is clear that the equilibrium constant and reaction quotient play vital roles in determining the direction of shift in a chemical equilibrium.
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Answers
Answer:
Explanation:
Hello!
In this case, since the phosphoric acid is a triprotic acid, we infer it has three stepwise ionization reactions in which one hydrogen ion is released at each step, considering they are undergone due to the presence of water, thus, we proceed as follows:
Moreover, notice each step has a different acid dissociation constant, which are quantified in the following order:
Ka1 > Ka2 > Ka3
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Answers
A student pipets 5.00 mL of a 5.103 M aqueous NaOH solution into a 250.00 mL volumetric flask and dilutes up to the mark with distilled water. the final molarity of the dilute solution is 0.102 M.
From the question given above, the following data were obtained:
Volume of stock solution (V1) = 5 mL
Molarity of stock solution (M₁) = 5.103 M
Volume of diluted solution (V₂) = 250 mL
Molarity of diluted solution (M₂) =?
The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:
M₁V₁ = M₂V₂
5.103 × 5 = M2 × 250
25.515 = M2 × 250
Divide both side by 250
M2 = 25.515 / 250
M2 = 0.102 M
Thus, the molarity of the diluted solution is 0.102 M.
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Answer:
0.102 M.
Explanation:
From the question given above, the following data were obtained:
Volume of stock solution (V1) = 5 mL
Molarity of stock solution (M1) = 5.103 M
Volume of diluted solution (V2) = 250 mL
Molarity of diluted solution (V2) =?
The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:
M1V1 = M2V2
5.103 × 5 = M2 × 250
25.515 = M2 × 250
Divide both side by 250
M2 = 25.515 / 250
M2 = 0.102 M
Thus, the molarity of the diluted solution is 0.102 M.
released the drugs into the cancer cells?
a) Graphite and diamond because these well known substances have been used
for years to treat cancer
b) Graphite because it is found in pencils and is a good lubricant
c) Fullerene and carbon nanotubes because they have empty spaces inside the
molecules
d) Diamond because it is one of the hardest substances
Answers
Answer:
c) Fullerene and carbon nanotubes because they have empty spaces inside the molecules
Explanation:
Fullerene and carbon nanotubes would be the most desired in order to hold the cancer fighting drugs and to carry them through the body safely.
- These molecules have empty spaces in them.
- The cavities makes it possible for storage.
- As they pass through the body, they can be held perfectly well to their target site of action.
Answers
∴, activity = decay rate / mass = 1580/0.00094
= 1.681 X 10^6 decays per min per gram
Using the half-life formula, we have:
activity of sample / activity of modern carbon = (1/2)^(age / half-life)
∴, Age = half-life x log (base 2) (modern activity / coal activity)
= 5730 x log(base 2)(1.681X10^6/ 15)
= 96115 years.
Answer: Age of the charcoal = 96115 years
Final answer:
Using the radiocarbon dating technique and applying the decay formula, it is calculated that the age of the charcoal from the an ancient campsite is approximately 9,500 years.
Explanation:
The age of the charcoal can be found using the technique of radiocarbon dating, which capitalizes on the process of radioactive decay. The isotope carbon-14 (¹4C) is used in this method as it has a known half-life of 5730 years. The number of decays per minute per gram of carbon in a live organism is known as its activity.
Initially, the activity was given as 15 decays per minute per gram. The present activity of the carbon in the charcoal is provided at 1580 decays per minute for a 0.94 kg or 940 gram sample. Thus, the current activity per gram is 1580/940 equals approximately 1.68 decays per minute per gram.
Given that the half-life of ¹4C is 5730 years, we can apply the formula for calculating the time passed using the rate of decay, which is given as T = (t1/2 / ln(2)) * ln(N0/N), where 'ln' is the natural logarithm, 'N0' is the initial quantity (15 decays/minute per gram), 'N' is the remaining quantity (1.68 decays/minute per gram).
Plugging in the given values, we get T = (5730 / ln(2)) * ln(15/1.68), which gives us approximately 9,500 years. Therefore, the age of the charcoal is around 9,500 years.
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Answers
9.03*10^23 molecules
Final answer:
The number of water molecules in a 1.50 mol block of ice is calculated by multiplying the number of moles of water by Avogadro's number. The result is approximately 9.033 x 10^23 water molecules.
Explanation:
In chemistry, the amount of substance in moles is related to the number of particles (atoms, molecules) through Avogadro's number. Avogadro's number, which is 6.022 x 1023 particles/mol, tells us the number of molecules in one mole of a substance.
To calculate the number of water molecules in 1.50 mol of water, you would multiply the number of moles of water by Avogadro's number:
1.50 mol of water x 6.022 x 1023 water molecules/mol of water = 9.033 x 1023 water molecules
Therefore, there are approximately 9.033 x 1023 water molecules in a 1.50 mol block of ice.
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