What volume of nitrogen gas is equal too 3.42x10^22 molecules of this substance

Chemistry

1 answer:

Bas_tet [7]3 years ago

8 0

if i am correct the volume of nitrogen gas has to equal to molecules density, making the substance 1.27 liters :)

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How many moles of N2 in 57.1 g of N2?

SpyIntel [72]

We are given –

Mass of $\bf N_2$ is 57.1 g and we are asked to find number of moles present in 57.1 g of $\bf N_2$

$\qquad$ $\pink{\bf\longrightarrow { Molar \:mass \:of \: N_2:-} }$

$\qquad$ $\bf \twoheadrightarrow 14\times 2$

$\qquad$ $\bf \twoheadrightarrow 28$

$\qquad$ ____________________

Now,Let's calculate the number of moles present in 57.1 g of $\bf N_2$

$\qquad$ $\purple{\bf\longrightarrow { No \:of \:moles = \dfrac{Given \:mass}{Molar\: mass}}}$

$\qquad$ $\bf \twoheadrightarrow \dfrac{57.1}{28}$

$\qquad$ $\bf \twoheadrightarrow 2.04\: moles$

__________________________________

7 0

2 years ago

Gyhgguuuyiiuyhjuhjiokjhgyhjiuh

Tasya [4]

Answer:

gyhgguuuyiiuyhjuhjiokjhgyhjiuh

Explanation:

lol

3 0

3 years ago

Read 2 more answers

When adjusted for any changes in ΔHΔH and ΔSΔS with temperature, the standard free energy change ΔG∘TΔGT∘Delta G_{T}^{\circ} at

STALIN [3.7K]

The equilibrium constant is 0.0022.

Explanation:

The values given in the problem is

ΔG° = 1.22 ×10⁵ J/mol

T = 2400 K.

R = 8.314 J mol⁻¹ K⁻¹

The Gibbs free energy should be minimum for a spontaneous reaction and equilibrium state of any reaction is spontaneous reaction. So on simplification, the thermodynamic properties of the equilibrium constant can be obtained as related to Gibbs free energy change at constant temperature.

The relation between Gibbs free energy change with equilibrium constant is ΔG° = -RT ln K

So, here K is the equilibrium constant. Now, substitute all the given values in the corresponding parameters of the above equation.

We get,

$1.22 * 10^{5} = - 8.314* 2400 * ln K$