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meriva
3 years ago
8

Write the equilibrium expression of each chemical equation. 2H2S(g) 2H2(g) + S2(g)

Chemistry
1 answer:
hodyreva [135]3 years ago
6 0

Answer:

<u>[H2]2[S2][H2S]2Kc=[H2]2[S2][H2S]2</u>

Explanation:

2H2S(g)⇋2H2(g)+S2(g)2H2S(g)⇋2H2(g)+S2(g)

The equilibrium constant expression in terms of concentrations is:

Kc=<u>[H2]2[S2][H2S]2Kc=[H2]2[S2][H2S]2</u><u>.</u>

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Sulfur undergoes combustion to yield sulfur trioxide by the following reaction equation:
12345 [234]

Answer:

Therefore, the amount of heat produced by the reaction of 42.8 g S = <u>(-5.2965 × 10²) kJ = (-5.2965 × 10⁵) J</u>

Explanation:

Given reaction: 2S + 3O₂ → 2 SO₃

Given: The enthalpy of reaction: ΔH = - 792 kJ

Given mass of S: w₂ = 42.8 g, Molar mass of S: m = 32 g/mol

In the given reaction, the number of moles of S reacting: n = 2

As, Number of moles: n = \frac{mass\: (w_{1})}{molar\: mass\: (m)}

∴  mass of S in 2 moles of S: w_{1} = n \times m = 2\: mol \times 32\: g/mol = 64\: g

<em>Given reaction</em>: 2S + 3O₂ → 2 SO₃

<em>In this reaction, the limiting reagent is S</em>

⇒ 2 moles S produces (- 792 kJ) heat.

or, 64 g of S produces (- 792 kJ) heat.

∴ 42.8 g of S produces (x) amount of heat

⇒ <u><em>The amount of heat produced by 42.8 g S:</em></u>

x = \frac{(- 792\: kJ) \times 42.8\: g}{64\: g} = (-529.65)\: kJ

\Rightarrow x = (-5.2965 \times 10^{2})\: kJ = (-5.2965 \times 10^{5})\: J

(\because 1 kJ = 10^{3} J)

<u>Therefore, the amount of heat produced by the reaction of 42.8 g S = (-5.2965 × 10²) kJ = (-5.2965 × 10⁵) J</u>

8 0
3 years ago
1 A 3.80 g sample of bronze was dissolved in sulphuric acid. The copper in alloy reacted with
s344n2d4d5 [400]

Answer:

yeyeye

Explanation:

hi and bye  no problem you will come and die

3 0
3 years ago
Name the type of ion formed by the atom when it loses three electrons. Show by illustration
Nimfa-mama [501]
The atoms of elements can gain or lose electrons and become ions. Ions are charged particles that have gained or lost electrons. The atoms of elements can gain or lose electrons to form monatomic ions (made from a single atom of an element).
5 0
2 years ago
Read 2 more answers
How many Earths will we need to have enough resources if the whole world lived like Americans?
Oxana [17]

Answer:

2749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`12345678998546897786544785786445082796468578967109878378610978378517629865378632749274826186482737216657574758757863352`1`1234567899854689778654478578644508279646857896710987837861097837851762986537863 earths

Explanation:

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4 0
3 years ago
Read 2 more answers
The average lung capacity of a human is 6.0L.
Darya [45]

Answer:

(a) 0.25 mol

(b) 0.11 mol

(c) 8.77 mol

Explanation:

(a)

We use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 1.00 atm

V = Volume of the gas = 6.0 L

T = Temperature of the gas = 298 K

R = Gas constant = 0.0821\text{ L.atm }mol^{-1}K^{-1}

n = number of moles = ?

Putting values in above equation, we get:

1.00 atm\times 6.0L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1.00\times 6.0}{0.0821\times 298}=0.25mol

(b)

We use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 0.296 atm

V = Volume of the gas = 6.0 L

T = Temperature of the gas = 200 K

R = Gas constant = 0.0821\text{ L.atm }mol^{-1}K^{-1}

n = number of moles = ?

Putting values in above equation, we get:

0.296 atm\times 6.0L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 200K\\\\n=\frac{0.296\times 6.0}{0.0821\times 200}=0.11mol

(c)

We use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 30 atm

V = Volume of the gas = 6.0 L

T = Temperature of the gas = 250 K

R = Gas constant = 0.0821\text{ L.atm }mol^{-1}K^{-1}

n = number of moles = ?

Putting values in above equation, we get:

30 atm\times 6.0L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 250K\\\\n=\frac{30\times 6.0}{0.0821\times 250}=8.77mol

8 0
3 years ago
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