For which of the following reactions is ΔHrxn equal to ΔHf of the product? You do not need to look up any values to answer this

Solution:

The gas's new temperature is 604K

Justification:

Assuming standard temperature and pressure, we can determine the gas's temperature using the ideal gas law;

Step 1: Formulate the general gas law equation

P1V1/T1 = P2V2/T2

Step 2: Insert the values, converting as needed to standard units.

P1 = 0.800 atm

V1 = 0.180 L

T1 = 29°C = 273 + 29 = 302K

P2 = 3.20 atm

V2 = 90 mL = 90 * 10^-3 L = 0.09 L

Step 3: Solve for T2

The new gas temperature T2 is calculated as:

T2 = P2V2T1/(P1V1)

T2 = 3.20 * 0.09 * 302 / (0.800 * 0.180)

T2 = 86.976 / 0.144

T2 = 604K

The gas's new temperature is 604K.

Response:

The specific heat of the alloy

Clarification:

Weight of the alloy = 25 gm

Initial temperature = 100°c = 373 K

Weight of the water = 90 gm

Initial temperature of water = 25.32 °c = 298.32 K

Final temperature = 27.18 °c = 300.18 K

Using the energy balance equation,

Heat released by the alloy = Heat absorbed by the water

[[ - ] = ( - )

25 × × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)

This gives us the specific heat of the alloy.

Response:

9.9 ml of 0.200M NH₄OH(aq)

Reasoning:

3NH₄OH(Iaq) + FeCl₃(aq) => NH₄Cl(aq) + Fe(OH)₃(s)

What volume in ml of 0.200M NH₄OH(aq) will fully react with 12ml of 0.550M FeCl₃(aq)?

1 x Molarity of NH₄OH x Volume of NH₄OH Solution(L) = 2 x Molarity of FeCl₃ x Volume of FeCl₃ Solution

1(0.200M)(Volume of NH₄OH Soln) = 3(0.550M)(0.012L)

=> Volume of NH₄OH Soln = 3(0.550M)(0.012L)/1(0.200M) = 0.0099 Liters = 9.9 milliliters

Answer:

The reaction will proceed in the forward direction, resulting in the reduction of NiO to Ni

Explanation:

Step 1: Provided data

Kp = 6.0 * 10²

Pressure of CO = 150 torr

Total pressure remains under 760 torr

Step 2: The balanced reaction equation:

NiO(s) + CO(g) ⇆ Ni(s) + CO2(g)

Step 3: Determine P(CO) and P(CO2)

For this equilibrium, the Kp expression is:

Kp = P(CO2)/P(CO) = 6.0*10^2

Given that NiO and Ni are solids, they do not affect the Kp

⇒ with P(CO) = 150 torr

⇒ P(CO) = 760 - 150 = 610 torr

Step 4: Calculate the reaction quotient

Q = 610/150 = 4.1

Since Q is significantly less than Kp, there are more reactants than products. Some reactants will convert into products, driving the reaction rightward.

Thus, the reaction will proceed forward, leading to the reduction of NiO to Ni.

The amino acids classified under the first group include alanine, aspartate, and glutamate, whereas those in the second group consist of glycine, valine, proline, leucine, isoleucine, methionine, serine, threonine, cysteine, asparagine, glutamine, phenylalanine, tryptophan, tyrosine, lysine, arginine, and histidine.