Why is it important to use the net force instead of the applied force when calculating the acceleration?

In the reversible reaction: 2NO2 (g) ⇌ N2O4 (g) the formation of dinitrogen tetroxide releases heat and the formation of nitroge

xenn [34]

Answer: Option (B) is the correct answer.

Explanation:

According to Le Chatelier's principle, any disturbance causes in an equilibrium reaction will shift the equilibrium in a direction that will oppose the change.

For example, $2NO_{2}(g) \rightleftharpoons N_{2}O_{4}(g)$

When we increase the temperature then the reaction will shift in a direction where there will be decrease in temperature.

This, means that the reaction will shift in the backward direction.

Thus, we can conclude that if the reaction is at equilibrium and the temperature increases, the equilibrium will shift so that there is more nitrogen dioxide.

7 0

4 years ago

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How many grams (mass) are in 1.7 moles of Nickel II oxide NiO?

WARRIOR [948]

Answer:

126.99g

Explanation:

NiO has a molar mass of 74.7 g/mol. This means that one mole of nickel (II) oxide has a mass of 74.7 grams. We can use stoichiometry/dimensional analysis to figure out how many grams are in 1.7 moles.

$1.7mol NiO (\frac{74.7grams}{1 mole} ) = 126.99g$

3 0

3 years ago

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Q 9.3: how many triplets would you expect to observe in the 1h nmr spectra for o-chlorotoluene

agasfer [191]

Hello!

You would expect to observe 2 triplets in the ¹H NMR spectra for o-chlorotoluene.

Multiplicity observed in ¹H NMR spectra when the atom couples with a neighbor ¹H atom. The multiplicity is equal to N+1 where N is the number of neighbor atoms.

To observe a triplet, you'll need a molecule with 2 neighbor atoms. In o-chlorotoluene (shown in the figure), only protons C and D have 2 neighbor atoms (B and D; A and C, respectively), so you'll expect to see a 2 triplets.

Have a nice day!

3 0

4 years ago

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At 298 K, what is the Gibbs free energy change (ΔG) for the following reaction?

9966 [12]

Answer:

(a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

Explanation:

Given that,

Temperature = 298 K

Suppose, density of graphite is 2.25 g/cm³ and density of diamond is 3.51 g/cm³.

$\Delta H\ for\ diamond = 1.897 kJ/mol$

$\Delta H\ for\ graphite = 0 kJ/mol$

$\Delta S\ for\ diamond = 2.38 J/(K mol)$

$\Delta S\ for\ graphite = 5.73 J/(K mol)$

(a) We need to calculate the value of $\Delta G$ for diamond

Using formula of Gibbs free energy change

$\Delta G=\Delta H-T\Delta S$

Put the value into the formula

$\Delta G= (1897-0)-298\times(2.38-5.73)$

$\Delta G=2895.3$

$\Delta G=2.895\ kJ$

The Gibbs free energy change is positive.

(b). When it is compressed isothermally from 1 atm to 1000 atm

We need to calculate the change of Gibbs free energy of diamond

Using formula of gibbs free energy

$\Delta S=V\times\Delta P$

$\Delta S=\dfrac{m}{\rho}\times\Delta P$

Put the value into the formula

$\Delta S=\dfrac{12\times10^{-6}}{3.51}\times999\times10130$

$\Delta S=34.59\ J/mole$

(c). Assuming that graphite and diamond are incompressible

We need to calculate the pressure

Using formula of Gibbs free energy

$\beta= \Delta G_{g}+\Delta G+\Delta G_{d}$

$\beta=V(-\Delta P_{g})+\Delta G+V\Delta P_{d}$

$\beta=\Delta P(V_{d}-V_{g})+\Delta G$

Put the value into the formula

$0=\Delta P(\dfrac{12\times10^{-6}}{3.51}-\dfrac{12\times10^{-6}}{2.25})\times10130+2895.3$

$0=-0.0194\Delta P+2895.3$

$\Delta P=\dfrac{2895.3}{0.0194}$

$\Delta P=14924\ atm$

(d). Here, $C_{p}=0$

So, The value of $\Delta H$ and $\Delta S$ at 900 k will be equal at 298 K

We need to calculate the Gibbs free energy of diamond relative to graphite

Using formula of Gibbs free energy

$\Delta G=\Delta H-T\Delta S$

Put the value into the formula

$\Delta G=(1897-0)-900\times(2.38-5.73)$

$\Delta G=4912\ J$

Hence, (a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

7 0

3 years ago

11. 2073 Set D Q.No. 1 Ammonia molecule has got trigonal

Nikolay [14]

Answer:

See explanation

Explanation:

I know that we usually associate the sp3 hybridization to the tetrahedral shape. This is common in molecules such as CH4. So it may sound somewhat strange that NH3 molecule has an sp3 hybridized nitrogen atom and a trigonal pyramidal geometry.

Let us recall that the central nitrogen atom in NH3 has a lone pair of electrons. These lone pairs causes more repulsion than bond pairs. As a result of the presence of this lone pair, the bond angle in the NH3 molecule is distorted away from the expected 109.7 degrees in tetrahedral geometry and the bonding groups are now arranged in a trigonal pyramidal geometry(with bond angle less than 109.7 degrees) to minimize electron pair repulsions.

5 0

3 years ago