Write the function in the simplest form

Q No. 10: Write the function in the simplest form:

Write the function in the simplest form:tan-1{x/(a2-x2)}

Q No. 9: Write the function in the simplest form:

Write the function in the simplest form tan-1

Q No. 8: Write the function in the simplest form:

Write the function in the simplest form

Q No. 7: Write the function in the simplest form:

Write the function in the simplest form

Q No. 6: Write the function in the simplest form:

Write the function in the simplest form

Q No. 5: Write the function in the simplest form:

prove that: 2tan-1(1/2)+tan-1(1/7)=tan-1(31/17)

prove that tan-1(2/11)+tan-1(7/24)=tan-1(1/2)

prove that 3cos-1x=cos-1(4x3-3x)

Prove that 3sin-1x=sin-1(3x-4x3)

 

An electric oven of 2 kW is operated in a domestic electric circuit (220 V) that has a current rating of 5 A

Q. No 2: An electric oven of 2 kW is operated in a domestic electric circuit (220 V) that has a current rating of 5 A. What result do you expect? Explain.

Ans: In this problem we have

Power of the oven (P) = 2 kW = 2000 W

Potential Difference (V) = 220 V

We have P = VI

⇒I=P/V

    = 2000/220

    = 9.09A

Here the current drawn by the electric oven is 9.09 A, which exceeds the safe limit of the circuit that is 5A. Therefore electric fuse will melt and break the circuit.

Name two safety measures commonly used in electric circuits and appliances

Q. No 1: Name two safety measures commonly used in electric circuits and appliances.

Ans: Two safety measures commonly used in electric circuits and appliances are

(i) Electric Fuse: An electric fuse is connected in series it protects the circuit from overloading and prevents it from short circuiting.

(ii) Proper earthing of all electric circuit in which any leakage of current in an electric appliance is transferred to the ground and people using the appliance do not get the shock.

What precaution should be taken to avoid the overloading of domestic electric circuits

Q. No 3: What precaution should be taken to avoid the overloading of domestic electric circuits?

Ans: Following precautions should be taken to avoid the overloading of domestic circuits:

(a) Do not use too many appliances at the same time.

(b) Use the appliances within the safe limit of electric circuit.

(c) Do not connect too many appliances in a single socket.

(d) Fuse should be connected in series in the circuit to protect overloading and short circuiting. 

The decomposition of hydrocarbon follows the equation k = (4.5 × 1011 s−1) e−28000 K/T Calculate Ea.

Question 4.26: The decomposition of hydrocarbon follows the equation  k = (4.5 × 1011 s⁻¹) e^{−28000 K/T} Calculate E_a.
Answer
Given that
k = (4.5 × 10¹¹ s⁻¹) e^{−28000 K/T}
Arrhenius equation is given by,

Plug the values of R we get
E_a= 8.314 J K⁻¹ mol⁻¹ × 28000 K = 232792 J mol⁻¹
Divide by 1000 to convert in KJ we get
= 232.792 kJ mol⁻¹
Answer
Activation energy E_a = 232.792 kJ mol⁻¹

Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?

Question 4.25: Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t_1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?
Answer
Given that
t_1/2 = 3.00 hours

Use formula of rate constant and plug the values, we get


Answer
Fraction of sample of sucrose that remains after 8 hours = 0.158

Consider a certain reaction A → Products with k = 2.0 × 10−2 s−1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L−

Question 4.24: Consider a certain reaction A → Products with k = 2.0 × 10⁻² s⁻¹. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L⁻¹.
Answer
Given that
Rate constant, k = 2.0 × 10⁻² s⁻¹
Unit of rate constant is s^-1 so it is a first order reaction
Time period, t = 100 s
Initial concentration, [A]_o = 1.0 moL⁻¹
Use the formula of first order reaction we get

Plug the values in this formula we get
Use formula log x/ y = log x - log y, and plug log 1 = 0 we get

 solve it  and take antilog both side, we get
[A]= 0.135 mol L⁻¹ (approx)
Answer
Remaining concentration of A = 0.135 mol L⁻¹

Question 4.23: The rate constant for the decomposition of hydrocarbons is 2.418 × 10−5 s−1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.

Question 4.23: The rate constant for the decomposition of hydrocarbons is 2.418 × 10⁻⁵ s⁻¹ at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.
Answer
Given that
Activation energy , E_a = 179.9 kJ mol⁻¹ = 179.9 × 10³ J mol⁻¹
Use the formula of Arrhenius equation,

Plug the values in this formula we get
Log A = (0.3835 − 5) + 17.2082 = 12.5917
Take antilog both sides we get
A = 3.9 × 10¹² s⁻¹ (approx)