Describe the relationship between predator and prey in a balanced ecosystem. Please help I'll give brainliest.

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Answer 1

Answer:

Answer:

predators are controlling the population of the species who are below them in the food pyramid . Also if the population of the preys decrease it will alternatively reduce the predator population .therefore the predator prey relationship balance an eco system.

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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.

Answers

Answer:

0.2320V

Explanation:

Voltage can be defined as the amount of potential energy available (work to be done) per unit charge, to move charges through a conductor.

Voltage can be generated by means other than rubbing certain types of materials against each other.

Please look at attached file for solution to the problem.

Final answer:

The expected voltage generated by this concentration cell is approximately 0.113 V.

Explanation:

To calculate the voltage generated by the concentration cell, we can use the Nernst equation. The Nernst equation relates the concentration of the ions in the two compartments to the voltage of the cell. The equation is:

E = E° - (RT/nF) ln(Q)

Where:

E is the voltage of the cell
E° is the standard cell potential
R is the gas constant (8.314 J/mol·K)
T is the temperature in Kelvin (25 + 273 = 298 K)
n is the number of moles of electrons transferred (2 in this case)
F is Faraday's constant (96,485 C/mol)
ln(Q) is the natural logarithm of the reaction quotient

The reaction quotient (Q) can be calculated using the concentrations of the lead (II) and iodide ions in each compartment.

Since this is a concentration cell, the standard cell potential (E°) for this system is 0 V. Therefore, the equation simplifies to:

E = - (RT/nF) ln(Q)

Now we can calculate the voltage:

Calculate Q:

The solubility product constant (Ksp) for PbI2 is 1.4 x 10-8. Because PbI2 is in a saturated solution, the concentration of Pb2+ ions and I- ions are both equal to the solubility of PbI2. We can substitute these values into the equation to calculate Q:

Q = [Pb²+] x [I-]

Q = (1.4 x 10-8) x (1.4 x 10-8) = 1.96 x 10-16

Calculate E:

Now we can calculate the voltage using the given values:

For the Nernst equation, we need to convert the temperature to Kelvin:

T = 25°C + 273 = 298 K

Substitute the values into the equation:

E = - (8.314 J/mol·K x 298 K / 2 x 96,485 C/mol) ln(1.96 x 10-16)

E ≈ 0.113 V

Therefore, the expected voltage generated by this concentration cell is approximately 0.113 V.

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Calculate the mass in grams for 0.251 moles of Na2CO3

Answers

Answer:

Explanation:

the molar mass for Na2CO3 is 2*23+12+3*16=106 g/mole

106*0.251=26.606 grames

A chemist designs a galvanic cell that uses these two half-reactions:O2 (g) + 4H+(aq) + 4e− → 2H2O (l) Eo =+1.23V
Zn+2 (aq) + 2e− → Zn(s) Eo=−0.763V

Answer the following questions about this cell.

Write a balanced equation for the half-reaction that happens at the cathode.
Write a balanced equation for the half-reaction that happens at the anode.
Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous as written. Do you have enough information to calculate the cell voltage under standard conditions

Answers

Answer: The reaction is spontaneous and there is not enough information to calculate the cell voltage.

Explanation:

The substance having highest positive $E^o$ reduction potential will always get reduced and will undergo reduction reaction.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

For a:

The half reactions for the cell occurring at cathode follows:

$O_2(g)+4H^+(aq)+4e^-\rightarrow H_2O(l);E^o_(cathode)=+1.23V$

For b:

The half reactions for the cell occurring at anode follows:

$Zn(s)\rightarrow Zn^(2+)+2e^-;E^o_(anode)=-0.763V$ ( × 2)

For c:

The balanced equation for the overall reaction of the cell follows:

$O_2(g)+4H^+(aq)+2Zn(s)\rightarrow H_2O(l)+2Zn^(2+)(aq)$

For the reaction to be spontaneous, the Gibbs free energy of the reaction must come out to be negative.

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_(cell)$

For a reaction to be spontaneous, the standard electrode potential must be positive.

To calculate the $E^o_(cell)$ of the reaction, we use the equation:

$E^o_(cell)=E^o_(cathode)-E^o_(anode)$

Putting values in above equation, we get:

$E^o_(cell)=1.23-(-0.763)=1.993V$

As, the standard electrode potential of the cell is coming out to be positive, the reaction is spontaneous in nature.

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_(cell)=E^o_(cell)-(0.059)/(n)\log ([Zn^(2+)]^2)/([H^(+)]^4* p_(O_2))$

As, the concentrations and partial pressures are not given. So, there is not enough information to calculate the cell voltage.

Hence, the reaction is spontaneous and there is not enough information to calculate the cell voltage.

Which notations represent atoms that have the same number of protons but a different number of neutrons?

Answers

The atoms having the same number of protons but a different number of neutrons have been termed isotopes.

The neutrons and protons are the constituents of the nucleus. The number of protons and electrons is equal in the atom. When two different atoms have the same number of protons the species have been termed isotopes. The atomic number of the species differ that resulting in the different positions in the periodic table.

For more information about the atoms, refer to the link:

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Atoms of the same element, containing the same number of protons, but different numbers of neutrons, are known as isotopes. Isotopes of any given element all contain the same number of protons, so they have the same atomic number (for example, the atomic number of helium is always 2).

50.00 mL of 0.10 M HNO 2 (nitrous acid, K a = 4.5 × 10 −4) is titrated with a 0.10 M KOH solution. After 25.00 mL of the KOH solution is added, the pH in the titration flask will be a. 2.17
b. 3.35
c. 2.41
d. 1.48
e. 7.00

Answers

Answer:

b. 3.35

Explanation:

To calculate the pH of a solution containing both acid and its salt (produced as a result of titration) we need to use Henderson’s equation i.e.

pH = pKa + log ([salt]/[acid]) (Eq. 01)

Where

pKa = -log(Ka) (Eq. 02)

[salt] = Molar concentration of salt produced as a result of titration

[acid] = Molar concentration of acid left in the solution after titration

Let’s now calculate the molar concentration of HNO2 and KOH considering following chemical reaction:

HNO2 + KOH ⇆ H2O + KNO2 (Eq. 03)

This shows that 01 mole of HNO2 and 01 mole of KOH are required to produce 01 mole of KNO2 (salt). And if any one of them (HNO2 and KOH) is present in lower amount then that will be considered the limiting reactant and amount of salt produced will be in accordance to that reactant.

Moles of HNO2 in 50 mL of 0.01 M HNO2 solution = 50/1000x0.01 = 0.005 Moles

Moles of KOH in 25 mL of 0.01 M KOH solution = 25/1000x0.01 = 0.0025 Moles

As it can be seen that we have 0.0025 Moles of KOH therefore considering Eq. 03 we can see that 0.0025 Moles of KOH will react with only 0.0025 Moles of HNO2 and will produce 0.0025 Moles of KNO2.

Therefore

Amount of salt produced i.e [salt] = 0.0025 moles (Eq. 04)

Amount of acid left in the solution [acid] = 0.005 - 0.0025 = 0.0025 moles (Eq.05)

Putting the values in (Eq. 01) from (Eq.02), (Eq. 04) and (Eq. 05) we will get the following expression:

pH= -log(4.5x10 -4) + log (0.0025/0.0025)

Solving above we get

pH = 3.35

Final answer:

The pH value in the titration flask after 25.00 mL of the 0.10 M KOH solution is added to 50.00 mL of 0.10 M HNO2 solution is 3.35.

Explanation:

The subject of this question is titration, which is a method used in chemistry to measure the concentration of an unknown solution. Given 50.00 mL of 0.10 M HNO2 (nitrous acid, Ka = 4.5 × 10-4), titrated with 0.10 M KOH (potassium hydroxide), we need to calculate the pH after 25.00 mL of the KOH solution is added.

First, we need to find the moles of the HNO2 and the KOH. Moles equals Molarity times Volume. So, for HNO2, it is 0.10 M * 0.050 L which equals 0.005 moles. For KOH, it is 0.10 M * 0.025 L which equals 0.0025 moles.

Then, subtract the moles of OH- from the moles of HNO2 to determine the concentration of HNO2 left, which is 0.005 moles - 0.0025 moles = 0.0025 moles. Divide this by the total volume of the solution (50.00 mL + 25.00 mL = 75.00 mL or 0.075 L to determine the new concentration of HNO2, 0.0025 moles / 0.075 L = 0.033 M. Then use the given Ka value with the equation [H+] = sqrt(Ka * [HNO2]) to get [H+].

To find acids' pH, we use the formula pH = -log[H+]. Use the calculated [H+] to find the pH.

Upon performing these calculations, the resulting pH value should be approximately 3.35 after 25.00 mL of the KOH solution is added, so the answer is (b) 3.35.

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What volume will 12 g of oxygen gas (O2) occupy at 25 °C and a pressure of 53 kPa?

Answers

ANSWER

The volume of the oxygen gas is 17.5 L

EXPLANATION

Given that;

The mass of oxygen gas is 12 grams

The temperature of the gas is 25 degrees Celcius

The pressure of the gas is 53 kPa

To find the volume of the oxygen gas, follow the steps below

Step 1; Assume the gas behaves like an ideal gas

Therefore, apply the ideal gas equation to find the volume of the gas

$\text{ PV }=\text{ nRT}$

Where

P is the pressure of the gas

V is the volume of the gas

n is number of moles of the gas

R is the universal gas constant

T is the temperature of the gas

Step 2: Find the number of moles of the oxygen gas using the below formula

$\text{ mole }=\text{ }\frac{\text{ mass}}{\text{ molar mass}}$

Recall, that the molar mass of the oxygen gas is 32 g/mol

$\begin{gathered} \text{ mole }=\text{ }\frac{12}{\text{ 32}} \n \text{ mole }=\text{ 0.375 mol} \end{gathered}$

Step 3; Convert the temperature to degree Kelvin

$\begin{gathered} \text{ T }=\text{ t }+\text{ 273.15} \n \text{ t }=\text{ 25}\degree C \n \text{ T }=25\text{ }+\text{ 273.15} \n \text{ T }=\text{ 298.15K} \end{gathered}$

Step 4; Substitute the given data into the formula in step 1

Recall, that R is 8.314 L kPa K^-1 mol^-1

$\begin{gathered} \text{ 53 }*\text{ V }=\text{ 0.375}*\text{ 8.314}*\text{ 298.15} \n \text{ 53V }=\text{ 929.557} \n \text{ Divide both sides by 53} \n \text{ }\frac{\cancel{53}V}{\cancel{53}}\text{ }=\text{ }(929.557)/(53) \n \text{ V }=\text{ }(929.557)/(93) \n \text{ V }=\text{ 17.5 L} \end{gathered}$

Hence, the volume of the oxygen gas is 17.5 L

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