PLEASE HELP ITS URGENT!! How do I answer this? I will mark brainlist

State the definition of the partial molar Gibbs energy.

balu736 [363]

Explanation :

As we know that the Gibbs free energy is not only function of temperature and pressure but also amount of each substance in the system.

$G=G(T,P,n_1,n_2)$

where,

$n_1\text{ and }n_2$ is the amount of component 1 and 2 in the system.

Partial molar Gibbs free energy : The partial derivative of Gibbs free energy with respect to amount of component (i) of a mixture when other variable $(T,P,n_j)$ are kept constant are known as partial molar Gibbs free energy of $i^{th}$ component.

For a substance in a mixture, the chemical potential $(\mu)$ is defined as the partial molar Gibbs free energy.

The expression will be:

$\bar{G_i}=\mu_i=\frac{\partial G}{\partial n_i}_{(T,P,n_j)}$

where,

T = temperature

P = pressure

$n_i\text{ and }n_j$ is the amount of component 'i' and 'j' in the system.

4 0

3 years ago

What is the symbol for the carbon isotope with seven neutrons?

Tresset [83]

Carbon -13 has 7 neutrons and carbon -12 has six neutrons. Carbon -12 is the most common isotope of Carbon. Carbon -14 is radioactive and vary rare. The symbols for the isotopes of Carbon atoms shown here indicate they each have six protons but mass numbers of 14, 13, and 12. Hope this helps. :)

8 0

3 years ago

Which atom would it be most difficult to remove an electron from

Norma-Jean [14]

I would be difficult to remove an electron from a Noble or Inert Gas (also known as the group 8 or 0 elements). This is because they all have filled outermost shells and as such the outermost shell would be held tightly to the nucleus and as such make it difficult to remove. Examples Helium, Neon, Argon, Xenon, Krypton and Radon

6 0

3 years ago

1. 7.000 mmhg = ______ KPa?<br> 2. 10.00 kPa = ________atm?<br> 3. 15.00 kPa______mmHg?

Minchanka [31]

1 kPa = 7.5 mmHg so 7.0 mmHg / 7.5 mmHg x 1 kPa = .93 kPa

101.3 kPa = 1 atm so 10 kPa / 101.3 kPa x 1 atm = .0987 atm

1 kPa = 7.5 mmHg so 15 kPa x 7.5 mmHg / 1 kPa = 112.5 mmHg

7 0

3 years ago

How many moles of gas X are present if the gas has a volume of 2dm³ at room temperature and pressure? Give your answer to 2 deci

bezimeni [28]

Answer:

Approximately $0.08\; \rm mol$ , assuming that this gas is an ideal gas.

Explanation:

Look up the standard room temperature and pressure: $25\; \rm ^{\circ}C$ and $P = 101.325 \; \rm kPa$ .

The question states that the volume of this gas is $V = 2\; \rm dm^{3}$ .

Convert the unit of all three measures to standard units:

$\begin{aligned} T &= 25\; \rm ^{\circ}C \\ &= (25 + 273.15)\; \rm K \\ &= 293.15\; \rm K\end{aligned}$ .

$\begin{aligned}P &= 101.325\; \rm kPa \\ &= 101.325 \; \rm kPa \times \frac{10^{3}\; \rm Pa}{1\; \rm kPa} \\ &= 1.01325 \times 10^{5}\; \rm Pa\end{aligned}$ .

$\begin{aligned}V &= 2\; \rm dm^{3} \\ &= 2 \; \rm dm^{3} \times \frac{1\; \rm m^{3}}{10^{3}\; \rm dm^{3}} \\ &= 2 \times 10^{-3}\; \rm m^{3}\end{aligned}$ .

Look up the ideal gas constant in the corresponding units: $R \approx 8.31\; \rm m^{3}\cdot Pa \cdot mol^{-1} \cdot K^{-1}$ .

Let $n$ denote the number of moles of this gas in that $V = 2\; \rm dm^{3}$ . By the ideal gas law, if this gas is an ideal gas, then the following equation would hold:

$P \cdot V = n \cdot R \cdot T$ .

Rearrange this equation and solve for $n$ :

$\begin{aligned}n &= \frac{P \cdot V}{R \cdot T} \\ &\approx \frac{1.01325 \times 10^{5}\; {\rm Pa} \times 2 \times 10^{-3}\; {\rm m^{3}}}{8.31 \; {\rm m^{3} \cdot Pa \cdot mol^{-1} \cdot K^{-1}} \times 293.15\; {\rm K}} \\ &\approx 0.08\; \rm mol\end{aligned}$ .

In other words, there is approximately $2\; \rm mol$ of this gas in that $V = 2\; \rm dm^{3}$ .

6 0

3 years ago