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2 years ago

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Write the equilibrium expression for N2 (g) + 3 H2 (g) ↔ 2 NH3 (g) + heat I have no idea where to even start, due to the corona

we have to learn by ourselves and my teacher is slacking off. Please Help!

1 answer:

motikmotik2 years ago

5 0

Answer:

The expression of an equilibrium constant will given as:

$K_c=\frac{[NH_3]^2}{[N_2][H_2]^3}$

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{eq}$

K is the constant of a certain reaction when it is in equilibrium

$N_2 (g) + 3 H_2 (g)\rightleftharpoons 2 NH_3 (g) + heat$

The expression of an equilibrium constant will given as:

$K_c=\frac{[NH_3]^2}{[N_2][H_2]^3}$

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Which of the following will have the longest-term impact on organisms in a ecosystem?

bearhunter [10]

Answer:

D. Global warming

Explanation:

Hope this Helps.

8 0

2 years ago

There are two isotopes of an unknown element, X-19 and X-21. The abundance of X-19 is 14.55%. A weighted average uses the percen

alekssr [168]

Answer:

2.765amu is the contribution of the X-19 isotope to the weighted average

Explanation:

The average molar mass is defined as the sum of the molar mass of each isotope times its abundance. For the unknown element X that has 2 isotopes the weighted average is defined as:

X = Mass X-19 * Abundance X-19 + MassX-21 * Abundance X-21

The contribution of the X-19 isotope is its mass (19.00 amu) times its abundance (14.55% = 0.1455). That is:

19.00amu * 0.1455 =

2.765amu is the contribution of the X-19 isotope to the weighted average

8 0

2 years ago

Which of the following gases would diffuse the fastest? <br> CH 4<br> CCl 4 <br> O 2<br> H 2

Sholpan [36]

H2 <span>because the smaller the gas molecule, the faster the diffusion. (the lightest molecule will diffuse the quickest)</span>

8 0

3 years ago

1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

1)

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2)

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3.

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

3 0

3 years ago

C (graphite) is used as a lubricant, whereas C (diamond) is used as an abrasive. Why is this?

zysi [14]

Answer:

Carbon atoms in graphite and diamond are arranged in different ways. Hence, the two allotropes of carbon have different physical properties.

Explanation:

Both graphite and diamond are both made of only carbon atoms. However, their physical properties differ from each other. Hence, they are called allotropes. Think about how these carbon atoms are arranged in each of the allotropes.

<h3>Graphite</h3>

In graphite, each carbon atom is bonded to three other carbon atoms. These carbon atoms will be located in the same plane. A chunk of graphite can contain many of these planes.

Each carbon atom has four valence electrons. Three of these electrons will be used in the bonds. The other electron will be delocalized. These electrons would flow between the sheets of carbon atoms. That keeps the sheets separate and allow them to slide on top of each other.

<h3>Diamond</h3>

In diamond, each carbon atom is bonded to four other carbon atoms. These carbon atoms will form a tetrahedral network.

In graphite, there's a significant separation between two adjacent sheets of carbon atoms. The force between the two sheets is rather weak. When a piece of graphite is between two objects that move over one another, the layers in the graphite would also slide over one another. Since the attraction between two adjacent sheets isn't very strong, there wouldn't be much resistance. Hence the graphite acts as a lubricant.

In contrast, most of the carbon atoms in a piece of diamond would be connected to each other. Unlike the sheets in graphite, in a diamond there are almost no moving parts. Also, the forces between neighboring carbon atoms are very strong. When an external force acts on a chunk of diamond, the carbon atoms would barely move. Hence, the structure appears to be very rigid. That gives diamond its abrasive properties.

4 0

3 years ago