Correct Answer: C. Pastoral farming

Correct Answer: C. breakage of bonds of reactants

Explanation

Activation energy must be achieved before reaction begin. This is because in any chemical reaction, chemical bonds in the reactants are broken, and new bonds in the products are formed

Correct Answer: D) 6.67 atm

Explanation

To solve this problem, we need to use the concept of partial pressures and Dalton's Law of Partial Pressures. The law states that in a mixture of non-reacting gases, the total pressure exerted by the mixture is equal to the sum of the partial pressures exerted by each individual gas. Mathematically, it can be represented as:

\(P_{total} = P_{N_2} + P_{O_2} + P_{He}\)

From the question, we know the moles of each gas and the pressure contribution due to oxygen, which is 2.5 atm. We can use the mole fraction of each gas to find the corresponding partial pressures. The mole fraction of a gas is the ratio of the moles of that gas to the total moles of all gases in the mixture. So, first, we need to find the total moles:

\(n_{total} = n_{N_2} + n_{O_2} + n_{He} = 0.25 + 0.15 + 0.4 = 0.8\)

Now we can find the mole fraction of helium:

\(\chi_{He} = \frac{n_{He}}{n_{total}} = \frac{0.4}{0.8} = 0.5\)

We also know the mole fraction of oxygen:

\(\chi_{O_2} = \frac{n_{O_2}}{n_{total}} = \frac{0.15}{0.8} = 0.1875\)

According to Dalton's Law, the ratio of partial pressures is equal to the ratio of mole fractions:

\(\frac{P_{He}}{P_{O_2}} = \frac{\chi_{He}}{\chi_{O_2}}\)

Plugging in the values, we get:

\(\frac{P_{He}}{2.5} = \frac{0.5}{0.1875}\)

Now, we can solve for the partial pressure of helium:

\(P_{He} = 2.5 \times \frac{0.5}{0.1875} = 6.67 \,\text{atm}\)

So, the partial pressure of helium in the mixture is 6.67 atm, which corresponds to Option D.

Correct Answer: D. cost-push inflation

Explanation

Cost-push inflation occurs when overall prices increase (inflation) as a result of increases in the cost of wages and raw materials.

Correct Answer: C. its vapour pressure equals the atmosphere pressure

Correct Answer: B. disciples believed more in Him

Correct Answer: D. Geographical

Explanation

(a) The specific heat capacity of copper is 400 \(JKg^{-1}K^{-1}\) means that 400 J of heat energy is required to raise (or lower) the temperature of 1 kg of a piece of copper by 1 kelvin.

(b) Experiment to determine the specific heat capacity of copper using a copper ball. The piece of copper ball is suspended by a thread and placed in a beaker of boiling water. While the temperature of the copper is being raised to boiling point of water, a copper calorimeter is half filled with water. The initial temperature of the water is measured and recorded and the hot copper ball is lifted by the thread and transferred to the calorimeter when the water has reached boiling point. The calorimeter cover is put on, the mixture is well stirred and the final steady temperature read from the themometer. The copper ball is weighed either before or after the experiment. The result is tabulated as shown below.

Mass of copper ball - m (kg)

Temperature of boiling water - \(\theta\) (K)

Mass of copper calorimeter - \(m_{c}\) (kg)

Mass of calorimeter + water - \(m_{x}\) (kg)

Mass of water \((m_{x} - m_{c})\) - \(m_{w}\) (kg)

Specific heat capacity - C \((Jkg^{-1} K^{-1})\)

Initial temperature of water = \(\theta_{w}\) (K)

Final temperature of water = \(\theta_{f}\) (K)

Using Heat lost by copper ball = Heat gained by calorimeter + Heat gained by water

\(mc(\theta - \theta_{f}) = m_{c} c (\theta_{f} - \theta_{w}) + m_{w} c_{w} (\theta_{f} - \theta_{w})\)

\(c = \frac{m_{w} c_{w} (\theta_{f} - \theta_{w})}{m (\theta - \theta_{f}) - m_{c} (\theta_{f} - \theta_{w})}\)

Precautions: (i) The copper ball should be transferred quickly into the calorimeter without splashing. (ii) Just before transfer, the copper should be shaken slightly to remove adhering hot water.

(iii) Heat lost by copper ball = heat gained by copper calorimeter + Hoat gained by water.

\(m_{c} (\theta - \theta_{f}) = m c (\theta_{f} - \theta_{w}) + m_{w} c_{w} (\theta_{f} - \theta_{w})\)

\(20 \times 0.4 \times (200 - \theta_{f}) = 60 \times 0.4 \times (\theta_{f} - 30) + 50 \times 4.2 \times (\theta_{f} - 30)\)

\(8(200 - \theta_{f}) = 24(\theta_{f} - 30) + 210 (\theta_{f} - 30)\)

\(\theta_{f} = \frac{8620}{242} = 35.6K\)

Correct Answer: A. 20 years

Explanation

(a) \(f(x) = \int (4x - x^{2}) \mathrm {d} x\)

\(f(x) = 2x^{2} - \frac{x^{3}}{3} + c\)

\(f(3) = 2(3^{2}) - \frac{3^{3}}{3} + c\)

\(21 = 18 - 9 + c\)

\(c = 21 - 9 = 12\)

\(f(x) = 2x^{2} - \frac{x^{3}}{3} + 12\)

(b) \(T_{n} = a + (n - 1) d\) (terms of an A.P)

\(T_{2} = a + d ; T_{4} = a + 3d ; T_{8} = a + 7d\)

These are consecutive terms of a G.P

\(i.e. \frac{T_{4}}{T_{2}} = \frac{T_{8}}{T_{4}}\)

\((T_{4})^{2} = T_{2} \times T_{8}\)

\((a + 3d)^{2} = (a + d)(a + 7d)\)

\(a^{2} + 6ad + 9d^{2} = a^{2} + 8ad + 7d^{2}\)

\(6ad + 9d^{2} = 8ad + 7d^{2} \implies 2ad = 2d^{2}\)

\(a = d\)

\(T_{3} + T_{5} = 20\)

\(a + 2d + a + 4d = 20\)

\(a + 2a + a + 4a = 20 \implies 8a = 20\)

\(a = d = \frac{5}{2}\)

(i) First four terms of A.P = \(\frac{5}{2}, 5, \frac{15}{2}, 10\)

(ii) \(S_{n} = \frac{n}{2} (2a + (n - 1)d)\)

\(S_{10} = \frac{10}{2} (2(\frac{5}{2}) + (10 - 1)(\frac{5}{2}))\)

= \(5(5 + \frac{45}{2})\)

= \(\frac{275}{2} = 137.5\)