What is a fire blanket MOST often used for?

Porcelain, paper, glass, and rubber are all examples of electrical

Allisa [31]

Electrical Insulators

6 0

2 years ago

Given the data calculated in Parts A, B, C, and D, determine the initial rate for a reaction that starts with 0.85 M of reagent

elixir [45]

Answer : The initial rate for a reaction will be $3.8\times 10^{-4}Ms^{-1}$

Explanation :

Rate law : It 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.

The chemical equation will be:

$A+B+C\rightarrow P$

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:

$6.1\times 10^{-5}=k(0.2)^a(0.2)^b(0.2)^c$ ....(1)

Expression for rate law for second observation:

$1.8\times 10^{-4}=k(0.2)^a(0.2)^b(0.6)^c$ ....(2)

Expression for rate law for third observation:

$2.4\times 10^{-4}=k(0.4)^a(0.2)^b(0.2)^c$ ....(3)

Expression for rate law for fourth observation:

$2.4\times 10^{-4}=k(0.4)^a(0.4)^b(0.2)^c$ ....(4)

Dividing 1 from 2, we get:

$\frac{1.8\times 10^{-4}}{6.1\times 10^{-5}}=\frac{k(0.2)^a(0.2)^b(0.6)^c}{k(0.2)^a(0.2)^b(0.2)^c}\\\\3=3^c\\c=1$

Dividing 1 from 3, we get:

$\frac{2.4\times 10^{-4}}{6.1\times 10^{-5}}=\frac{k(0.4)^a(0.2)^b(0.2)^c}{k(0.2)^a(0.2)^b(0.2)^c}\\\\4=2^a\\a=2$

Dividing 3 from 4, we get:

$\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.4)^a(0.4)^b(0.2)^c}{k(0.4)^a(0.2)^b(0.2)^c}\\\\1=2^b\\b=0$

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:

$6.1\times 10^{-5}=k(0.2)^2(0.2)^0(0.2)^1$

$k=7.6\times 10^{-3}M^{-2}s^{-1}$

Now we have to calculate the initial rate for a reaction that starts with 0.85 M of reagent A and 0.70 M of reagents B and C.

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

$\text{Rate}=(7.6\times 10^{-3})\times (0.85)^2(0.70)^0(0.70)^1$

$\text{Rate}=3.8\times 10^{-3}Ms^{-1}$

Therefore, the initial rate for a reaction will be $3.8\times 10^{-3}Ms^{-1}$

6 0

2 years ago

Why Ammonia is nut dried by passing through concentrated H2SO4.

STALIN [3.7K]

Answer:

Because it is absorbed by conc . H . SO , to form ammonium sulphate.

6 0

2 years ago

Which one of the following equations represents the net ionic equation for the reaction between aqueous potassium chloride and a

Nana76 [90]

<h3>Answer:</h3>

Ag⁺(aq) +Cl⁻(aq) → AgCl(s)

<h3>Explanation:</h3>

The questions requires we write the net ionic equation for the reaction between aqueous potassium chloride and aqueous silver nitrate.

<h3>Step 1: Writing a balanced equation for the reaction.</h3>

The balanced equation for the reaction between aqueous potassium chloride and aqueous silver nitrate will be given by;

KCl(aq) + AgNO₃(aq) → KNO₃(aq) +AgCl(s)

AgCl is the precipitate formed by the reaction.

<h3>Step 2: Write the complete ionic equation.</h3>

The complete ionic equation for the reaction is given by showing all the ions involved in the reaction.

K⁺(aq)Cl⁻(aq) + Ag⁺(aq)NO₃⁻(aq) → K⁺(aq)NO₃⁻(aq) +AgCl(s)

Only ionic compounds are split into ions.

<h3>Step 3: Write the net ionic equation for the reaction.</h3>

The net ionic equation for a reactions only the ions that fully participated in the reaction and omits the ions that did not participate in the reaction.
The ions that are not involved directly in the reaction are known as spectator ions and are not included while writing net ionic equation.

Ag⁺(aq) +Cl⁻(aq) → AgCl(s)

4 0

2 years ago

If Zn and H2SO4 undergo a single-displacement reaction, what is the balanced equation?

ankoles [38]

Answer:Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)

O2Zn(s) + H2SO4(aq) → 22nH(aq) + SO4(s)

Explanation:

6 0

2 years ago