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3 years ago

The compound MgF is an example of a(n)

Chemistry

1 answer:

Delvig [45]3 years ago

7 0

MgF is an ionic compound because it's bond is between a metal and a non-metal

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Need help !!!!! ASAP

Ksivusya [100]

<h2>Hello!</h2>

The answer is:

We have that there were produced 0.120 moles of $CO_{2}$

$n=0.120mol$

We are asked to calculate the number of moles of the given gas, also, we are given the volume, the temperature and the pressure of the gas, we can calculate the approximate volume using The Ideal Gas Law.

The Ideal Gas Law is based on Boyle's Law, Gay-Lussac's Law, Charles's Law, and Avogadro's Law, and it's described by the following equation:

$PV=nRT$

Where,

P is the pressure of the gas.

V is the volume of the gas.

n is the number of moles of the gas.

T is the absolute temperature of the gas (Kelvin).

R is the ideal gas constant (to work with pressure in mmHg), which is equal to:

$R=62.363\frac{mmHg.L}{mol.K}$

We must remember that the The Ideal Gas Law equation works with absolute temperatures (K), so, if we are given relative temperatures such as Celsius degrees or Fahrenheit degrees, we need to convert it to Kelvin before we proceed to work with the equation.

We can convert from Celsius degrees to Kelvin using the following formula:

$Temperature(K)=Temperature(C\°) + 273K$

So, we are given the following information:

$Pressure=760mmHg\\Volume=2.965L\\Temperature=25.5C\°=25.5+273K=298.5K$

Now, isolating the number of moles, and substituting the given information, we have:

$PV=nRT$

$n=\frac{PV}{RT}$

$n=\frac{760mmHg*2.965L}{62.363\frac{mmHg.L}{mol.K}*298.5K}$

$n=\frac{760mmHg*2.965L}{62.363\frac{mmHg.L}{mol.K}*298.5K}\\\\n=\frac{2242mmHg.L}{18615.355\frac{mmHg.L}{mol.}}\\\\n=0.120mole$

Hence, we have that there were produced 0.120 moles of $CO_{2}$

$n=0.120mol$

Have a nice day!

7 0

3 years ago

Which of the following is a mixture of substances?

Stels [109]

Answer:

the answer is c

b/c contain more than one element

7 0

3 years ago

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

How many sulfur atoms are there in 3.90 mol of sulfur?

Ilya [14]

<span>In a mole of anything, there are 6.023 x 10^23 units. So, in 3.9 moles of sulfur, there are 3.9 * 6.023 x 10^23 = 23 x 10^23 = 2.3 x 10^24 atoms (keeping only 2 sig figs). Hope I help!!
</span>

6 0

3 years ago

Help fast please it's timed

lys-0071 [83]

Answer: B its most likely

6 0

3 years ago

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