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3 years ago

5

Nitrogen and hydrogen gases react to form ammonia gas via the following reaction:

1 answer:

Luda [366]3 years ago

7 0

1mol N2 + 3mol H2 → 2mol NH3
<span>Because you are dealing with gases, at the same temperature and pressure you can say: </span>
<span>1volume N2 + 3 volumes H2 → 2 volumes NH3 </span>
<span>OR: 4 volumes reactants = 2 volumes products. </span>
<span>You have 1.9+5.7 = 7.6 volumes reactants, which must give 7.6/2 = 3.8 volumes of NH3 </span>
<span>Answer: 3.8 litres of NH3 will be formed</span>

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All the elements in one group have the same number of valence electrons.

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Substance X is extremely acidic. Substance Y is extremely basic. What would be formed if substance X and Y were mixed together?

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The product is

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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.

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3 years ago

10) How many grams are there in 1.00 x 10^24 molecules of BC13?

lana66690 [7]

194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

Explanation:

In order to convert the given number of molecules of BCl₃ to grams, first we have to convert the molecules to moles.

It is known that 1 moles of any element has 6.022×10²³ molecules.

Then 1 molecule will have $\frac{1}{6.022*10^{23} }$ moles.

So $1*10^{24} molecules = \frac{1*10^{24} }{6.022*10^{23} } =1.66 moles$

Thus, 1.66 moles are included in BCl₃.

Then in order to convert it from moles to grams, we have to multiply it with the molecular mass of the compound.

As it is known as 1 mole contains molecular mass of the compound.

As the molecular mass of BCl₃ will be

$Molecular mass of BCl_{3}= Mass of Boron + (3*Mass of chlorine)$

Mass of boron is 10.811 g and the mass of chlorine is 35.453 g.

Molar mass of BCl₃ = 10.811+(3×35.453)=117.17 g.

$grams of BCl_{3}=Molar mass of BCl_{3}*Number of moles$

$grams of BCl_{3}=117.17*1.66=194.5 g$

So, 194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

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4 years ago

1. When _____ <br> are added or removed from an atom, _____<br> is formed

Kobotan [32]

Answer:

1.) protons or electrons 2.) ion

Explanation:

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3 years ago