Whats the number of moles of O2(g) needed to completely react with 8 moles of CO(g).

For the reaction A+B+C→D+E, the initial reaction rate was measured for various initial concentrations of reactants. The followin

erastovalidia [21]

Answer : The initial rate for a reaction will be $3.4\times 10^{-3}Ms^{-1}$

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:

$6.0\times 10^{-5}=k(0.20)^a(0.20)^b(0.20)^c$ ....(1)

Expression for rate law for second observation:

$1.8\times 10^{-4}=k(0.20)^a(0.20)^b(0.60)^c$ ....(2)

Expression for rate law for third observation:

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

Expression for rate law for fourth observation:

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

Dividing 1 from 2, we get:

$\frac{1.8\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1$

Dividing 1 from 3, we get:

$\frac{2.4\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2$

Dividing 3 from 4, we get:

$\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0$

Thus, the rate law becomes:

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

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$6.0\times 10^{-5}=k(0.20)^2(0.20)^0(0.20)^1$

$k=7.5\times 10^{-3}M^{-2}s^{-1}$

Now we have to calculate the initial rate for a reaction that starts with 0.75 M of reagent A and 0.90 M of reagents B and C.

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

$\text{Rate}=(7.5\times 10^{-3})\times (0.75)^2(0.90)^0(0.90)^1$

$\text{Rate}=3.4\times 10^{-3}Ms^{-1}$

Therefore, the initial rate for a reaction will be $3.4\times 10^{-3}Ms^{-1}$

8 0

3 years ago

Routs<br> Which of the following is a physical change?

kotykmax [81]

Answer:

Explanation:

A physical change is a change in which the physical properties of matter are altered. These are properties are the forms and state.

Most physical changes are easily reversible and are pure state changes.

They do not lead to the production of new compounds.

They involve no mass change and requires little to no energy.

Examples are melting, boiling, freezing, sublimation e.t.c

5 0

3 years ago

To calculate the enthalpy change for the reaction, 2CO (g) + O2 (g) Imported Asset 2 CO2 (g), you can use ΔHf0 values for each r

svetoff [14.1K]

Answer:

ΔH0reaction = [ΔHf0 CO2(g)] - [ΔHf0 CO(g) + ΔHf0 O2(g)]

Explanation:

Chemical equation:

CO + O₂ → CO₂

Balanced chemical equation:

2CO + O₂ → 2CO₂

The standard enthalpy for the formation of CO = -110.5 kj/mol

The standard enthalpy for the formation of O₂ = 0 kj/mol

The standard enthalpy for the formation of CO₂ = -393.5 kj/mol

Now we will put the values in equation:

ΔH0reaction = [ΔHf0 CO2(g)] - [ΔHf0 CO(g) + ΔHf0 O2(g)]

ΔH0reaction = [-393.5 kj/mol] - [-110.5 kj/mol + 0]

ΔH0reaction = [-393.5 kj/mol] - [-110.5 kj/mol]

ΔH0reaction = -283 kj/mol

7 0

3 years ago

What is the name of Pb(NO3)2? Explain how you determined the bond type and the steps you used to determine the naming convention

Elena L [17]

Answer: Lead(II) nitrate but idk the rest

Explanation:

5 0

3 years ago

The density of mercury is 13.5g/mL. What is the volume of this liquid if the sample weighs 12.5 pounds?

Mama L [17]

Answer : The volume of liquid is 420 mL.

Explanation :

Density : The mass per unit volume of a substance is known as density.

Formula used:

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$

As we are given:

Density of mercury = 13.5 g/mL

Mass = 12.5 pounds

First we have to convert mass of sample from pound to gram.

Conversion used:

As, 1 pound = 453.6 g

So, 12.5 pounds = 453.6 × 12.5 g = 5670 g

Now we have to calculate the volume of liquid.

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$

Now putting all the given values in this formula, we get:

$13.5g/mL=\frac{5670g}{\text{Volume}}$

Volume = 420 mL

Therefore, the volume of liquid is 420 mL.

7 0

3 years ago