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2 years ago

What type of reaction occurs without the addition of heat?

Chemistry

1 answer:

maria [59]2 years ago

5 0

The type of reaction that occurs without the addition of heat is called spontaneous reaction. That is option D.

<h3>What is spontaneous reaction?</h3>

Spontaneous reaction is defined as the reaction where by new substances are formed naturally without the addition of extra energy in the form of heat.

Therefore, the type of reaction that occurs without the addition of heat is called spontaneous reaction

Learn more about heat here:

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Given the initial rate data for the reaction A + B –––> C, determine the rate expression for the reaction.

umka21 [38]

Answer:

B) Δ[C]/Δt = 3,60x10⁻² M⁻¹s⁻¹ [A] [B]

Explanation:

For the reaction A + B → C

The formula for rate of reaction is:

Δ[C]/Δt = k [A] [B]

As you have [A], [B] and Δ[C]/Δt information you can multiply [A] times [B] and take this value as X and Δ[C]/Δt as Y. The slope of this lineal regression will be k.

Thus, you must obtain:

y = 3,60x10⁻² X

Thus, rate of reaction is:

B) Δ[C]/Δt = 3,60x10⁻² M⁻¹s⁻¹ [A] [B]

I hope it helps!

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3 years ago

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When a neutrally charged atom loses an electron to another atom, the result is the creation of

kvasek [131]

When a neutrally charged atom loses an electron to another atom, the result is the creation of two ions (cation and anion).

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3 years ago

The vaporization of 1 mole of liquid water (system) at 100.9 C, 1.00 atm, is endothermic.

yan [13]

Answer:

a) The work done is 10.0777 kJ

b) The water's change in internal energy is -122.1973 kJ

Explanation:

Given data:

1 mol of liquid water

T₁ = temperature = 100.9°C

P = pressure = 1 atm

Endothermic reaction

T₂ = temperature = 100°C

1 mol of water vapor

VL = volume of liquid water = 18.8 mL = 0.0188 L

VG = volume of water vapor = 30.62 L

3.25 moles of liquid water vaporizes

Q = heat added to the system = -40.7 kJ

Questions: a) Calculate the work done on or by the system, W = ?

b) Calculate the water's change in internal energy, ΔU = ?

Heat for 3.25 moles:

$Q_{1} =3.25*(-40.7)=-132.275kJ$

The work done:

$W=-nP*delta(V)=-3.25*101.33*(0.0188-30.62)=10077.6637J=10.0777kJ$

The change in internal energy:

$delt(U)=W+Q_{1} =10.0777-132.275=-122.1973kJ$

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3 years ago

100 points! Formation of an Ion

djyliett [7]

$\huge\underline\pink {Answer}$

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ion

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3 years ago

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