Neither dry soil nor pure water conducts electricity. But wet soil will conduct electricity. Explain why this happens?

At standard temperature and pressure, it is established that 1 mole of gas has a volume of 22.4 liters.

According to the periodic table:
the molar mass of oxygen is 16 g
and the molar mass of hydrogen is 1 g
Hence, the molar mass of water vapor is calculated as 2(1) + 16 = 18 g

Thus, 18 g of water occupies 22.4 liters, therefore:
the volume for 32.7 g is (32.7 x 22.4) / 18 = 40.6933 liters

A secondary alkyl halide would be characterized by having a carbon atom connected to two other carbon atoms, with bromine attached to that carbon.

Therefore, bromo-hexane qualifies as a 2-degree or secondary alkyl halide

Answer: Reaction 2 is a spontaneous one.

Explanation:

According to our understanding:

= +ve, meaning the reaction is non-spontaneous

= -ve, indicating the reaction is spontaneous

= 0, stating that the reaction is at equilibrium

For a reaction to be classified as spontaneous, the Gibbs free energy must yield a negative value.

Reaction 1:

Glucose + Pi ⟶ glucose-6-phosphate + H₂O, ΔG = +13.8 kJ/mol

Reaction 2:

ATP + H₂O ⟶ ADP + Pi, ΔG = -30.5 kJ/mol

From this, we can conclude that ΔG being negative indicates that reaction 2 is indeed spontaneous.

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.

Among the listed options, the one indicating a chemical property is B. Water does not ignite, whereas gasoline is capable of burning. This distinction arises because both combustion and flammability are categorized as chemical properties.