How does the number of molecules in one mole of carbon dioxide compare with the number of molecules in one mole of water?

Predict whether the changes in enthalpy, entropy, and free energy will be positive or negative for the boiling of water, and exp

Sedbober [7]

Answer:

ΔH > 0; ΔS >0; ΔG = 0

Not spontaneous when T < 100 °C;

Equilibrium when T = 100 °C

Spontaneous when T > 100 °C

Step-by-step explanation:

The process is

H₂O(ℓ) ⇌ H₂O(g)

ΔH > 0 (positive), because we must add heat to boil water

ΔS > 0 (positive), because changing from a liquid to a gas increases the disorder

ΔG = 0, because the liquid-vapour equilibrium process is at equilibrium at 100 °C

ΔG = ΔH – TΔS

Both ΔH and ΔS are positive.

If T = 100 °C, ΔG =0. ΔH = TΔS, and the system is at equilibrium.

If T < 100 °C, the ΔH term will predominate, because T has decreased below the equilibrium value.

ΔG > 0. The process is not spontaneous below 100 °C.

If T > 100 °C, the TΔS term will predominate, because T has increased above the equilibrium value.

ΔG < 0. The process is spontaneous above 100 °C.

4 0

3 years ago

Solute and solvent of C12H22O11

nasty-shy [4]

Answer:

sugar is the solute and water is the solvent

8 0

3 years ago

Which of the following is necessarily characterized by a covalent linkage?a. Base-pairing interactionb. Disulfide bondc. Hydroge

mr_godi [17]

Answer:

b. Disulfide bond

Explanation:

Covalent linkage or bonding is a type of bond in which electrons are shared between atoms in a compound or molecule. Each atom contributes to the shared electron.

This is the case of disulfide bonding i.e. S-S, in which two thiol groups (-SH) share electrons with one another. The process of oxidation between two cysteine (amino acid) molecules forms this bond. Hence, Disulfide bond is characterized by a covalent linkage.

6 0

3 years ago

Why is it important for scientists to describe their procedures in full detail?

Licemer1 [7]

So that other scientist can repeat their experiments
i hope i got i right

6 0

3 years ago

En un balneario necesitan calentarse 1 millón de litros de agua anuales, subiendo la temperatura desde 15 ºC a 50 ºC y para ello

MAXImum [283]

Answer:

a) $m_{CH_4}=2630kg$

b) 1657 €

Explanation:

Hola,

a) En este problema, vamos a considerar el millón de litros de agua anuales, ya que con ellos podemos calcular el calor requerido para dicho calentamiento, sabiendo que la densidad del agua es de 1 kg/L:

$Q_{H_2O}=m_{H_2O}Cp(T_2-T_1)=1x10^6LH_2O*\frac{1kgH_2O}{1LH_2O}*4.18\frac{kJ}{kg\°C}(50-15) \°C\\Q_{H_2O}=146.3x10^6kJ$

Luego, usamos la entalpía de combustión del metano para calcular su requerimiento en kilogramos, sabiendo que la energía ganada por el agua, es perdida por el metano:

$Q_{H_2O}=-Q_{CH_4}=-146.3x10^5kJ=m_{CH_4}\Delta _cH_{CH_4}$

$m_{CH_4}= \frac{Q_{CH_4}}{\Delta _cH_{CH_4}} =\frac{-146.3x10^5kJ}{-890kJ/molCH_4} *\frac{16gCH_4}{1molCH_4} \\\\m_{CH_4}=2630112.36g=2630kg$

b) En este caso, consideramos que a condiciones normales de 1 bar y 273 K, 1 metro cúbico de metano cuesta 0,45 €, con esto, calculamos las moles de metano a dichas condiciones:

$n_{CH_4}=\frac{PV}{RT}=\frac{1atm*1000L}{0.082\frac{atm*L}{mol*K}*273K} =44.67mol$

Con ello, los kilogramos de metano que cuestan 0,45 €:

$44.67molCH_4*\frac{16gCH_4}{1molCH_4}*\frac{1kg}{1000g} =0.715kgCH_4$

Luego, aplicamos la regla de tres:

0.715 kg ⇒ 0.45 €

2630 kg ⇒ X

X = (2630 kg x 0.45 €) / 0.715 kg

X = 1657 €

Regards.

3 0

4 years ago