Answer:
B) Δ[C]/Δt = 3.60x10⁻² M⁻¹s⁻¹ [A] [B]
Explanation:
For the reaction A + B → C
The expression for the reaction's rate is:
Δ[C]/Δt = k [A] [B]
Utilizing the values for [A], [B], and Δ[C]/Δt, multiply [A] by [B] to acquire a value of X and consider Δ[C]/Δt as Y. The slope derived from this linear regression will yield k.
Therefore, the results you should have are:
y = 3.60x10⁻² X
Thus, the reaction's rate is:
B) Δ[C]/Δt = 3.60x10⁻² M⁻¹s⁻¹ [A] [B]
I hope this assists you!
Answer: The complete balanced chemical equation is,
Explanation:
A chemical equation represents the reactants on the left and the products on the right, separated by a right-pointing arrow indicating the reaction.
This representation includes the phases of the substances and uses subscripts and superscripts for numbers.
For the reaction given, the balanced chemical equation with phases is:
To tackle this problem, one must first determine the specific heat of water, which is the energy required to raise the temperature of 1 g of water by 1 degree C. The relationship is given by the formula q = c X m X delta T, where q indicates the specific heat of water, m signifies the mass, and delta T denotes the temperature change. The specific heat of water is 4.184 J/(g X degree C). The temperature of the water increased by 20 degrees, therefore: 4.184 x 713 x 20.0 = 59700 J, rounded to 3 significant digits, equals 59.7 kJ. This value indicates the energy required to produce B2O3 from 1 gram of boron. To convert this to kJ/mole, additional calculations are required. The gram atomic mass of Boron is 10.811, so dividing 1 gram of boron by 10.811 results in.0925 moles of boron. Given that 2 moles of boron are needed for the formation of 1 mole of B2O3, dividing the moles of boron by two yields.0925/2 =.0462 moles. Consequently, dividing the energy in KJ by the number of moles provides KJ/mole: 59.7/.0462 = 1290 KJ/mole.
Answer:
To lower the temperature of the solution from 25.0°C to 5.0°C, it is necessary to use 35.2g of NH₄NO₃ for every 100.0g of water.
Explanation:
In order to cool down the solution, we need:
4.184 J/g°C × (5.0°C - 25.0°C) × (100.0g + X) = -Y
8368 J + 83.68 J/gX = Y (1)
Here, x represents the grams of NH₄NO₃ required, and Y represents the energy needed to remove heat.
Furthermore, the energy Y becomes:
Y = 25700 J/mol × 
X
Y = 321 J/g X (2)
Substituting (2) into (1)
8368 J + 83.68 J/g X = 321 J/g X
8363 J = 237.32 J/gX
X = 35.2g
This means 35.2g of NH₄NO₃ must be used for every 100.0g of water to achieve a temperature decrease from 25.0°C to 5.0°C.
I trust this information will be useful!
