Please balance xe+f2=xef6

A gas sample with a mass of 12.8 g exerts a pressure of 1.2 atm at 15°c and a volume of 3.94 l. what is the molar mass of the ga

Kryger [21]

We can use the ideal gas law equation to find the number of moles in the gas
PV = nRTwhere P - pressure - 1.2 atm x 101 325 Pa/atm = 121 590 Pa
V - volume - 3.94 x 10⁻³ m³
n - number of moles
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 15 °C + 273 = 288 K
substituting the values in the equation
121 590 Pa x 3.94 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 288 K
n = 0.200 mol
molar mass of gas is = mass / number of moles
molar mass = 12.8 g / 0.200 mol = 64 g/mol
molar mass of gas is 64 g/mol

3 0

3 years ago

A reaction begins with 130.5g of substance X and substance Y. How many grams of

Paha777 [63]

Answer:

130.5g

Explanation:

At the of the reaction, the combined mass of X and Y will be 130.5g.

The premise for this conclusion is based on the law of conservation of matter.

This law states that "in a chemical reaction, matter is neither created nor destroyed but changed from one form to another".

In essence, in a chemical reaction, there is no mass loss.

The amount of product in the reaction is expected to be the same as the amount of reactants used in the experiment.
When we start with 130.5g then we should end with 130.5g

7 0

3 years ago

50.0g of N2O4 is introduced into an evacuated 2.00L vessel and allowed to come to equilibrium with its decomposition product,N2O

Lady_Fox [76]

The mass of N2O4 in the final equilibrium mixture is 39.45 grams.

The decomposition reaction of N2O4 to 2NO2 can be expressed as:

$\mathbf{N_2O_4{(g)} \leftrightarrow 2NO_{2(g)}}$

From the parameters given:

The mass of N2O4 = 50.0 g
The molar mass of N2O4 = 92.011 g/mol

The number of mole of N2O4 can be determined as:

$\mathbf{Moles \ of \ N_2O_4 = \dfrac{50.0 g}{92.011 g/mol}}$

$\mathbf{Moles \ of \ N_2O_4 = 0.5434 \ moles}$

The volume of the vessel in which N2O4 was evacuated is = 2.0 L

From stochiometry, the concentration of $\mathbf{[N2O4] = \dfrac{mole \ of \ N_2O_4}{volume \ of \ N_2O_4}}$

$\mathbf{[N2O4] = \dfrac{0.5434}{2}}$

$\mathbf{[N2O4] = 0.2717 \ M}$

The I.C.E table can be computed as:

$\mathbf{N_2O_4{(g)}}$ ↔ $\mathbf{ 2NO_{2(g)}}$

Initial 0.2717 0

Change -x +2x

Equilibrium (0.2717 - x) 2x

The equilibrium constant from the I.C.E table can be expressed as:

$\mathbf{K_c = \dfrac{[NO]^2}{[N_2O_4]}}$

$\mathbf{K_c = \dfrac{(2x)^2}{(0.2717-x)}}$

Recall that; Kc = 0.133

∴

$\mathbf{0.133 = \dfrac{4x^2}{(0.2717-x)}}$

0.0361 - 0.133x = 4x²

4x² + 0.133x - 0.0361 = 0

By solving the above quadratic equation, we have;

x = 0.07978

The Concentration of [NO2] = 2x

[NO2] = 2 (0.07978)
[NO2] = 0.15956 M

The Concentration of [N2O4] = 0.2717 - x

[N2O4] = 0.2717 - 0.07978
[NO2] = 0.19192 M

Again, from the decomposition reaction, we can assert that;

$\mathbf{N_2O_4{(g)} \leftrightarrow 2NO_{2(g)}}$

0.19192 M of N2O4 decompose to produce 0.15956 M of NO2.

However, if 5.0 g of NO2 is injected into the vessel, then the number of moles of NO2 injected becomes;

$\mathbf{= \dfrac{5.0 \ g}{46 \ g/mol}} \\ \\ \\ \mathbf{= 0.10869\ moles}$

The Molarity of NO2 injected now becomes:

$\mathbf{= \dfrac{00.10869 }{2} } \\ \\ \\ \mathbf{= 0.05434 \ M }$

So, the new moles of [NO2] becomes = 0.15956 + 0.05434

= 0.2139 M

The new I.C.E table can be computed as:

$\mathbf{N_2O_4{(g)}}$ ↔ $\mathbf{ 2NO_{2(g)}}$

Initial 0.19192 0.2139 M

Change +x -2x

Equilibrium (0.19192 + x) (0.2139 -2x)

NOTE: The injection of NO2 makes the reaction proceed in the backward direction.

The equilibrium constant from the I.C.E table can be expressed as:

$\mathbf{K_c = \dfrac{[NO]^2}{[N_2O_4]}}$

$\mathbf{K_c = \dfrac{(0.2139 - 2x)^2}{(0.19192+x)}}$

Recall that; Kc = 0.133

$\mathbf{0.133= \dfrac{(0.2139 - 2x)^2}{(0.19192+x)}}$

By solving for x;

x = 0.2246 or x = 0.0225

We need to consider the value of x that is less than the initial concentration of NO2(0.2139 M) which is:

x = 0.0225

Now, the final concentration of [N2O4] = (0.19192 + 0.0225)M

= 0.21442 M

The final number of moles of N2O4 = Molarity(concentration) × volume

The final number of moles of N2O4 = (0.21442 × 2) moles

The final number of moles of N2O4 = 0.42884 moles

The mass of N2O4 in the final equilibrium mixture is:

= final number of moles × molar mass of N2O4

= 0.42884 moles × 92 g/mol

= 39.45 grams

Learn more about the decomposition of N2O4 here:

brainly.com/question/25025725

6 0

3 years ago

Answer pls ASAPPP 50 pointsss!!

amm1812

Answer:

There should only be one chemical reaction, this is because toasting the bread is not only using heat, but you cannot untoast the bread, and also theres a reaction with the amino acids and sugar in bread when it's cooked. I hope this helps you! :)

5 0

3 years ago

Read 2 more answers

Using a mixture of water, sand, and chemicals under high pressure to free tightly held oil and natural gas is known as ____.

uysha [10]

<span>Using a mixture of water, sand, and chemicals under high pressure to free tightly held oil and natural gas is known as Hy</span>draulic fracturing.
<span>
This is a process commonly practiced in the United states where millions of gallons of water, sand, and chemicals are pumped underground. This is used to break apart the rock bed and release the gases that are deep in the ground.</span>

4 0

3 years ago