calculate the amount of heat energy required to heat up 23.2 grams of ice from -21° C to 56° C ** please show your work**

Answer:

0.41 M

Explanation:

A -> B

rate constant (k) = 0.039L/mol s

t = 23

Final concentration, [A]  = 0.30M

Initial concentration, [A]o = x

1 / [A]  = kt + 1 / [A]o

1 / [A]o = 1 / [A] - kt

1 / [A]o  = 1 / 0.30   - 0.039 (23)

1 / [A]  = 3.33 -  0.897 = 2.433

[A] = 0.41 M

Answer:

The second experiment (reversible path) does more work

Explanation:

Step 1:

A piston confines 0.200 mol Ne(g) in 1.20L at 25 degree °C

(a) The gas is allowed to expand through an additional 1.20 L against a constant of 1.00atm

Irreversible path: w =-Pex*ΔV

⇒ with Pex = 1.00 atm

⇒ with ΔV = 1.20 L

W = -(1.00 atm) * 1.20 L

W = -1.20L*atm *101.325 J /1 L*atm = -121.59 J

(b) The gas is allowed to expand reversibly and isothermally to the same final volume.

W = -nRTln(Vfinal/Vinitial)

⇒ with n = the number of moles = 0.200

⇒ with R = gas constant = 8.3145 J/K*mol

⇒ with T = 298 Kelvin

⇒ with Vfinal/Vinitial  = 2.40/1.20 = 2

W = -(0.200mol) * 8.3145 J/K*mol *298K *ln(2.4/1.2)

W = -343.5 J

The second experiment (reversible path) does more work

Answer:

pH after the addition of 10 ml NaOH = 4.81

pH after the addition of 20.1 ml NaOH = 8.76

pH after the addition of 25 ml NaOH = 8.78

Explanation:

(1)

Moles of butanoic acid initially present = 0.1 x 20 = 2 m moles  = 2 x 10⁻³ moles,

Moles of NaOH added = 10 x 0.1 = 1 x 10⁻³ moles

                          CH₃CH₂CH₂COOH + NaOH ⇄ CH₃CH₂CH₂COONa + H₂O

Initial conc.            2 x 10⁻³                 1 x 10⁻³           0            

Equilibrium             1 x 10⁻³                   0                  1 x 10⁻³

Final volume = 20 + 10 = 30 ml = 0.03 lit

So final concentration of Acid =

Final concentration of conjugate base [CH₃CH₂CH₂COONa]

Since a buffer solution is formed which contains the weak butanoic acid and conjugate base of that acid .

Using Henderson Hasselbalch equation to find the pH

Final answer:

During titration of butanoic acid with NaOH, we can calculate the pH at various points using the Henderson-Hasselbalch equation for buffer scenarios. After 10.00mL of NaOH, the pH will be 4.74. After 20.10 mL, the pH will be 8.27, and after 25.00 mL, the pH will be 12.30.

Explanation:

This involves calculating the pH at various stages during a titration procedure. Here the titration involves a weak acid, butanoic acid, with a strong base, NaOH. We can simplify the reaction as follows: CH₃CH₂CH₂COOH + OH- --> CH₃CH₂CH₂COO- + H₂O.

(a) After 10.00 mL of NaOH is added, the system isn't at equivalence. Here, the reaction hasn't fully completed and a buffer solution is present. Using the Henderson-Hasselbalch equation, we can find the pH: pH = pKa + log([base]/[acid]). After calculating, we can find pH = 4.74.

(b) After 20.10 mL NaOH is added, the system reaches past equivalence. The pH can be determined by finding pOH using the remaining OH- concentration and then subtracting from 14. After the calculation, the pH = 8.27.

(c) For 25.00 mL of NaOH, the system is beyond equivalence and extra OH- ions increase pH. The pH calculation is like previous step and the result will be pH = 12.30.

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The molarity is an important method which is used to calculate the concentration of a solution. The molarity of a solution that contains 0.50 g of NaCl dissolved in 100mL of solution is 0.085 M.

What is molarity?

The molarity of a solution is defined as the number of moles of the solute present per litre of the solution. It is an most important method to calculate the concentration of a binary solution. It is represented as 'M'.

The equation used to calculate the molarity is:

Molarity = Number of moles of the solute / Volume of the solution in litres

1L = 1000 mL

100 mL = 0.1 L

Number of moles (n) = Given mass / Molar mass

n = 0.50 / 58.44 = 0.008

Molarity = 0.0085 /  0.1  = 0.085 M

Thus the molarity of the solution is 0.085 M.

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Reactants take 504.87 yr to reach 12.5% of their original value in first-order decomposition reaction.

Equation for the first-order decomposition reaction:-

....(1)

Here,   is the final concentration, t is the time,   is the initial concentration, and k is the rate constant.

Given:-

k=

Substitute the above value in equation (1) as follows:-

So, 504.87 yr does it take for the reactant to reach 12.5% of its original value.

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Final answer:

The time required for a reactant to reach 12.5% of its original value in a first-order reaction is approximately 1482 years, obtained by applying the formula for the half-life of a first-order reaction and multiplying by 3.

Explanation:

In a first-order reaction, the half-life of the reaction, which is the time it takes for half of the reactant to be consumed, is independent of the concentration of the reactant. Also, for a first-order reaction, it would take approximately 3 half-lives for the reactant to be reduced to 12.5% of its original value. The Integrated Rate Law for a First-Order Reaction can be applied to determine the time it will take.

Given the rate constant (k) is 0.00140 yr¯¹, we will use the formula for the half-life of a first-order reaction: t₁/₂ = 0.693 / k. After calculating the half-life (t₁/₂), multiply it by 3 to determine the time for the reactant concentration to reach 12.5% of its original value. Hence, it would take approximately 1482 years to reach 12.5% of the original value when rounded to the correct number of significant figures.

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Answer:

1. B

2. A

3. C

4. C

5. D

Explanation:

Soil is regarded as the solid unconsolidated material of the earth crust. Soil is of three different types namely: Sandy soil, clay soil and loamy soil. These three different soil types possess different properties that distinguish them. Some of them are:

- CLAY soil is characterized as having the finest particles and can hold greater amount of water i.e. have a high water holding capacity.

- LOAMY SOIL is the best soil type for planting agricultural crops because it has the highest concentration of nutrients that suited for plant growth.

- loamy, Sandy and clay differ in how we feel when touched i.e. texture, and colour.

- SANDY soils are the kind of soils that are found in Sea shores and beaches.

- Soil is important to living things as it forms part of the earth where animals live, provides the necessary nutrients needed by plants, serves as a place where people live.