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3 years ago

Potassium chlorate (KClO3) decomposes to form potassium chloride (KCl) and pure oxygen (O2) gas as shown below.

1 answer:

7 0

The balanced chemical reaction is:

2KClO3 ----> 2KCl + 3O2

We are given the amount of oxygen gas to be produced from potassium chlorate. This will be the starting point for the calculations.

28.0 L O2 ( 1 mol / 22.4 L) (2 mol KClO3 / 3 mol O2) ( 122.55 g / 1 mol) = 102.13 g KClO3 is needed

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For the vaporization of a liquid at a given pressure: A.ΔG is positive at all temperatures. B.ΔG is negative at all temperatures

MArishka [77]

Answer:

C. ΔG is positive at low temperatures, but negative at high temperatures (and zero at some temperature).

Explanation:

Since we need to give energy in the form of heat to vaporize a liquid, the enthalpy is positive. In a gas, molecules are more separated than in a liquid, therefore the entropy is positive as well.

Considering the Gibbs free energy equation:

ΔG= ΔH - TΔS

+ +

When both the enthalpy and entropy are positive, the reaction proceeds spontaneously (ΔG is negative) at high temperatures. At low temperatures, the reaction is spontaneous in the reverse direction (ΔG is positive).

5 0

3 years ago

1. The heat of fusion for the ice-water phase transition is 335 kJ/kg at 0°C and 1 bar. The density of water is 1000 kg/m3 at th

vodomira [7]

Answer:

Expression for the change of melting temperature with pressure..> T₂ = T₁exp(-(P₂-P₁)/(3.61x10⁹ Pa), Freezing Point = 0°C

Explanation:

Derivation from state postulate

Using the state postulate, take the specific entropy, , for a homogeneous substance to be a function of specific volume and temperature .

ds = (partial s/partial v)(t) dv + (partial s/partial T)(v) dT

During a phase change, the temperature is constant, so

ds = (partial s/partial v)(T) dv

Using the appropriate Maxwell relation gives

ds = (partial P/partial T)(v) dv

s(β) – s(aplαha) = dP/dT (v(β) – v(α))

dP/dT = s(β) – s(α)/v(β) – v(α) = Δs/Δv

Here Δs and Δv are respectively the change in specific entropy and specific volume from the initial phase α to the final phase β.

For a closed system undergoing an internally reversible process, the first law is

du = δq – δw = Tds - Pdv

Using the definition of specific enthalpy, h and the fact that the temperature and pressure are constant, we have

du + Pdv = dh Tds,

ds = dh/T,

Δs = Δh/T = L/T

After substitution of this result into the derivative of the pressure, one finds

dp/dT = L/TΔv

<u>This last equation is the Clapeyron equation.</u>

(dP/dT) = dH/TdV => dP/dlnT = dH/dV

=> dP/dlnT = dH/dV = [H(liquid) - H(solid)]/[V(liquid) - V(solid)]

= [335,000 J/kg]/[1000⁻¹ - 915⁻¹ m³/kg]

= -3.61x10⁹ J/m³ = -3.61x10⁹ Pa

=> P₂ = P₁ - 3.61x10⁹ ln(T₂/T₁) Pa

T₂ = T₁exp(-(P₂-P₁)/(3.61x10⁹ Pa)

b) if the pressure in Denver is 84.6 kPa:

T₂(freezing) = 273.15exp[-(84,600-100,000)/(3.61x10⁹)]

≅ 273.15 = 0°C T₁(freezing) essentially no change

5 0

3 years ago

The combustion of palmitic acid is represented by the chemical equation: C16H32O2(s) + 23O2(g) → 16 CO2(g) + 16 H2O(l) The magni

solniwko [45]

Answer:

The correct option is C

Explanation:

From the question we are told that

The reaction is

$C_{16}H_{32}O_2(g) + 23O_2(g) \to 16 CO_2(g) + 16 H_2O(l)$

Generally $\Delta H = \Delta U + \Delta N* RT$

Here $\Delta H$ is the change in enthalpy

$\Delta U$ is the change in the internal energy

$\Delta N$ is the difference between that number of moles of product and the number of moles of reactant

Looking at the reaction we can discover that the elements that was consumed and the element that was formed is $O_2$ and $CO_2$ and this are both gases so the change would occur in the number of moles