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Murrr4er [49]
3 years ago
5

Balance this equation, please.

Chemistry
1 answer:
Digiron [165]3 years ago
7 0

Answer:3Li+YbCl3=Yb +3LiCl

Explanation:as there are three cl in the first one so to balance them we will put 3before LiCl in order to make Cl balanced now there are 3 Li also so put before Li 3 making it also balanced Yb is balanced already

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PLEASE HELP: Which is the state change that occurs when a solid becomes a gas without becoming a liquid first?
nekit [7.7K]
It is called sublimation
4 0
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if the density of a certain spherical atomic nucleus is 1.0x10^14 g cm^-3 and its mass is 2.0x10^-23 g, what is it radius in cm?
bagirrra123 [75]
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6 0
3 years ago
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For the reaction A + B + C → D + E, the initial reaction rate was measured for various initial concentrations of reactants. The
babymother [125]

Answer : The order of reaction with respect to A is, second order reaction.

The order of reaction with respect to B is, zero order reaction.

The order of reaction with respect to C is, first order reaction.

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+B+C\rightarrow D+E

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b[C]^c

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

6.0\times 10^{-4}=k(0.20)^a(0.20)^b(0.20)^c ....(1)

Expression for rate law for second observation:

1.8\times 10^{-3}=k(0.20)^a(0.20)^b(0.60)^c ....(2)

Expression for rate law for third observation:

2.4\times 10^{-3}=k(0.40)^a(0.20)^b(0.20)^c ....(3)

Expression for rate law for fourth observation:

2.4\times 10^{-3}=k(0.40)^a(0.40)^b(0.20)^c ....(4)

Dividing 1 from 2, we get:

\frac{1.8\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1

Dividing 1 from 3, we get:

\frac{2.4\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2

Dividing 3 from 4, we get:

\frac{2.4\times 10^{-3}}{2.4\times 10^{-3}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^0[C]^1

Thus,

The order of reaction with respect to A is, second order reaction.

The order of reaction with respect to B is, zero order reaction.

The order of reaction with respect to C is, first order reaction.

5 0
3 years ago
A reaction vessel is charged with hydrogen iodide, which partially decomposes to molecular hydrogen and iodine: 2HI (g) H2(g) +
leonid [27]

Answer:

The value of the equilibrium constant: K_{p} = 0.25

Explanation:

Given reaction: 2HI (g) ⇌ H₂(g) + I₂(g)

Number of moles of- HI: n₁ = 2 mole; H₂: n₂ = 1 mole; I₂: n₃ = 1 mole

Total number of moles: n = n₁ + n₂ + n₃ = 2 + 1 + 1 = 4 moles

The equilibrium constant for the given reaction is given as:

K_{p} = \frac{pH_{2}\; pI_{2}}{(pHI)^{2}}

Given: Temperature: T = 425 °C = 425 + 273 = 698 K

The partial pressure: pHI = 0.708 atm,

and, pH₂ = pI₂

 

∵ <em>partial pressure of a given gas</em>: pₐ = Χₐ . P

Here, P is the total pressure

Χₐ is the <em>mole fraction</em> of given gas and is given by the equation

\chi_{a} = \frac{number \, of \,moles \,of \,given \,gas (n_{a})}{total \,number \,of \,moles (n)}

Mole fraction for HI: \chi_{1} = \frac {n_{1}}{n} = \frac {2}{4} = 0.5

Mole fraction for H₂: \chi_{2} = \frac {n_{2}}{n} = \frac {1}{4} = 0.25

Mole fraction for I₂: \chi_{3} = \frac {n_{3}}{n} = \frac {1}{4} = 0.25

Thus, Χ₂ = Χ₃ = 0.25

The partial pressure of HI is given by;

pHI = Χ₁ P

0.708 atm = 0.5 × P

⇒ P = 1.416 atm

 

As the partial pressures: pH₂ = pI₂

∴ pH₂ = pI₂ = Χ₂ P = Χ₃ P = 0.25 × 1.416 atm = 0.354 atm

Therefore, the value of Kp can be calculated as:

K_{p} = \frac{pH_{2}\; pI_{2}}{(pHI)^{2}}

K_{p} = \frac{0.354 atm\times 0.354 atm}{(0.708 atm)^{2}} = 0.25

<u>Therefore, the value of the equilibrium constant: </u>K_{p} = 0.25

8 0
3 years ago
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