Jim, Jill, Robert, and Kim each run paper chromatography on an unknown aqueous mixture. Jim gets a red band and a blue band, Jil

Answer:

25.2 kJ

Explanation:

The full question can be found in the image linked to this response.

It's important to highlight that the heat absorbed by the 2.00 L of water for increasing its temperature from the beginning to the end comes solely from the burning of benzoic acid, as there are no heat transfers to the container or the surroundings.

To find the heat released from benzoic acid combustion, we simply measure the heat needed to warm the water.

Q = mCΔT

To find the mass of the water,

Density = (mass)/(volume)

Mass = Density × volume

Density = 1 g/mL

Volume = 2.00 L = 2000 mL

Mass = 1 × 2000 = 2000 g

C = specific heat of water = 4.2 J/g.°C

ΔT = (final temperature) - (Initial temperature)

Final water temperature = 25°C

Initial water temperature = 22°C

ΔT = 25 - 22 = 3°C

Q = (2000×4.2×3) = 25,200 J = 25.2 kJ

Hope this Helps!!!

Explanation:

To solve this issue, the combined gas law must be applied:

P₁V₁ = P₂V₂ / T₁T₂

The data is as follows: P1 = 0.99 atm, V1 = 2 L, T1 = 273 K, P2 =?, V2 = 4 L, T2 = 137 K.

By isolating P2 in the equation, you find

Answer:

The rate law for the decomposition reaction is:

The unit for the rate constant will be

Explanation:

The rate law can be expressed as:

..[1]

When the drug concentration is tripled, the decomposition rate rises by a factor of nine.

...[2]

[1] ÷ [2]

Solving for x results in:

x = 2.

This indicates a second-order reaction.

The decomposition reaction's rate law is:

The unit for the rate constant will be:

The unit for the rate constant will be .

Answer:

C. connecting an active metal to designate the pipe as the cathode in an electrochemical cell.

Explanation:

Cathodic protection involves a method to manage the accelerated corrosion of a metal surface by designating it as the cathode within an electrochemical cell. This is accomplished by attaching the protected metal to a more sacrificial metal, which acts as the anode.

This method helps to preserve the metal by introducing a highly reactive metal that serves as the anode, supplying free electrons. By adding these free electrons, the active metal gives up its ions, protecting the less reactive steel from corrosion.