All categories

3 years ago

Write the Rate Law

Chemistry

1 answer:

Ivanshal [37]3 years ago

5 0

As we know that rate law expression is

$r = k { |a| }^{2} { |b| }^{2}$

(a)

$r = k { |o2| }^{1} { |no2| }^{2}$

(b)

$r = k { |cl2| }^{1}$

(c)

$r = k { |chcl3|}^{1} { |cl2| }^{1}$

#$# HOPE YOU UNDERSTAND #$#

#$¥ THANK YOU ¥$#

❤ ☺ ☺ ☺ ☺ ☺ ☺ ❤

You might be interested in

If the density of a diamond is 3.15 g/cm3 then what is the mass of a 25cm diamond?

Yanka [14]

Answer:

78.75g

Explanation:

The density of diamond is the mass of diamond per unit of its volume. The density of a substance is calculated using the formula:

Density (g/cm^3) = mass (g) / volume (cm^3)

According to the question, the density of diamond= 3.15g/cm^3, volume of diamond= 25cm^3, mass = ?

Hence, to calculate the mass, we say;

Mass = Volume × density

Mass = 25 × 3.15

Mass = 78.75

Therefore, the mass of diamond is 78.75grams.

3 0

3 years ago

3. Does entropy increase or decrease in the following processes?

tamaranim1 [39]

Answer:

HOPE IT helps much as you can

5 0

3 years ago

Calculate the standard enthalpy change of formation of CHA given that the standard enthalpies change of combustion of methane, g

Nostrana [21]

Answer : The standard enthalpy of formation of methane is, -74.8 kJ/mole

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of $CH_4$ will be,

$C(s)+2H_2(g)\rightarrow CH_4(g)$ $\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

(1) $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)$ $\Delta H_1=-890kJ/mole$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.4kJ/mole$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_3=-285.7kJ/mole$

Now we will reverse the reaction 1, multiply reaction 3 by 2 then adding all the equations, we get :

(1) $CO_2(g)+2H_2O(l)\rightarrow CH_4(g)+2O_2(g)$ $\Delta H_1=890kJ/mole$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.4kJ/mole$

(3) $2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_3=2\times (-285.7kJ)=-571.4kJ/mole$

The expression for enthalpy of formation of $CH_4$ will be,

$\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(+890kJ/mole)+(-393.4kJ/mole)+(-571.4kJ/mole)$

$\Delta H=-74.8kJ/mole$

Therefore, the standard enthalpy of formation of methane is, -74.8 kJ/mole

3 0

3 years ago

Is nuclear waste and radioactive waste the same thing?

ElenaW [278]

Answer: yes

Explanation: Radioactive (or nuclear) waste is a byproduct from nuclear reactors, fuel processing plants, hospitals and research facilities. Radioactive waste is also generated while decommissioning and dismantling nuclear reactors and other nuclear facilities. There are two broad classifications: high-level or low-level waste.

3 0

3 years ago

Given the balanced ionic equation representing the reaction in an operating voltaic cell: zn(s) + cu2+(aq) → zn2+(aq) + cu(s) th

soldi70 [24.7K]

Answer: from the Zn anode to the Cu cathode

Justification:

1) The reaction given is: Zn(s) + Cu₂⁺ (aq) -> Zn²⁺ (aq) +Cu(s)

2) From that, you can see the Zn(s) is losing electrons, since it is being oxidized (from 0 to 2⁺), while Cu²⁺, is gaining electrons, since it is being reduced (from 2⁺ to 0).

3) Then, you can already tell that electrons go from Zn to Cu.

4) The plate where oxidation occurs is called anode, and the plate where reduction occus is called cathode.

So you get that the electrons flow from the anode (Zn) to the cathode (Cu).

Always oxidation occurs at the anode, and reduction occurs at the cathode.

3 0

4 years ago