How many kilojoules of heat are required to heat 1.37kg of water from 21.3 to 89.5

Response: The moles in 369 grams of calcium hydroxide are 4.98 moles

Reasoning: Given,

Mass of calcium hydroxide = 369 g

Molar mass of calcium hydroxide = 74.093 g/mole

Formula used:

Now substituting the provided values into this formula, you will find the moles of calcium hydroxide.

Thus, the number of moles in 369 grams of calcium hydroxide is, 4.98 moles

The balanced chemical equation for the neutralization of HCl with is:

Given weight of = 5g

Moles of =

Volume of HCl solution =

Assuming the density of the solution is 1.0 g/mL

Mass of HCl solution = 50 g

Overall mass of the solution = 50 g + 5 g = 55 g

To find the heat of neutralization, we calculate:

Q = m C ΔT

where m equals the mass of the solution = 55 g

C represents the specific heat capacity of the solution = 4.184

ΔT signifies the temperature change = 6.8 K = (6.8 - 273) C = -266.2

The enthalpy of neutralization per mole of

=

Answer:

The amount of calcium sulfate that precipitates is 6.14 grams.

Explanation:

Step 1: Provided data

We are mixing 500.0 mL of 0.10 M Ca^2+ with 500.0 mL of 0.10 M SO4^2−

The Ksp for CaSO4 is 2.40×10^−5.

Step 2: Determine moles of Ca^2+

Moles of Ca^2+ = Molarity of Ca^2+ * Volume

Moles of Ca^2+ = 0.10 * 0.500 L

Moles of Ca^2+ = 0.05 moles

Step 3: Determine moles of SO4^2-

Moles of SO4^2- = 0.10 * 0.500 L

Moles SO4^2- = 0.05 moles

Step 4: Compute total volume

Total volume = 500.0 mL + 500.0 mL = 1000 mL = 1L

Step 5: Compute Q

Q = [Ca2+] [SO42-]

[Ca2+] = 0.050 M and [SO42-]

Qsp = (0.050)(0.050) = 0.0025 >> Ksp

This indicates that precipitation will take place.

Step 6: Calculate molar solubility

Ksp = 2.40 * 10^-5 = [Ca2+][SO42-] =(x)(x)

2.40 * 10^-5 = x²

x = √(2.40 * 10^-5)

x = 0.0049 M (molar solubility)

Step 7: Determine total dissolved CaSO4

Total CaSO4 dissolved = 0.0049 M * 1 L * 136.14 g/mol = 0.667 g

Step 8: Calculate initial mass of CaSO4

Initial moles of CaSO4 = 0.050

Initial mass of CaSO4 = 0.050 * 136.14 g/mol

Initial mass of CaSO4 = 6.807 grams

Step 9: Calculate precipitate mass

6.807 - 0.667 = 6.14 grams.

The mass of calcium sulfate that will emerge as a precipitate is 6.14 grams.

Answer:

The correct choice is: option A.

Justification:

To address this inquiry, we need to evaluate the total number of electrons each orbital can accommodate.

  Orbital                         Number of electrons

   s                                   2

   p                                  6

   d                                 10

   f                                  14

Provided options:

A. 1s² 2s² 2p⁶ 3s²                 This configuration is valid as it aligns with the permitted number of electrons in each orbital and follows the correct sequence.

B. 1s² 2s² 2p⁶ 3s² 3d⁴          This configuration is not accurate because

                                         3d⁴ should follow 3p.

C. 1s² 2s² 2d¹⁰ 2p³                This is incorrect since 2d¹⁰ is not a valid orbital.

D. 1s² 2s^s 2p³ 2d¹⁰            This option contains two errors; s as an exponent does not exist, and 2d¹⁰ is also an invalid description.

Response:

K = 6.5 × 10⁻⁶

Detailed Explanation:

C₅H₆O₃ ⇄ C₂H₆ + 3CO

Apply PV=nRT to determine the initial pressure of C₅H₆O₃

P (2.50) = (0.0493) (0.08206) (473)

P = 0.78atm

C₅H₆O₃ ⇄ C₂H₆ + 3CO

0.78atm      0           0

0.78 - x        x           3x

1.63atm = 0.78 - x + x + 3x

P(total) = 0.288atm

C₅H₆O₃ = 0.78 - 0.288

             = 0.489atm

C₂H₆ = 0.288atm

CO = 0.846atm

     = 0.379

   = 6.5 × 10⁻⁶