(G)-->Iron and steel industry is called a heavy industry because all the raw material as well as finished goods are heavy and bulky entailing heavy transportation costs. Iron ore, coking coal and limestone are required in the ratio of 4:2:1 approximately. Some quantity of manganese is also required to harden the steel.

(H)-->Sodium is very reactive in nature. When exposed in air, it automatically forms Na2O. When it is put in water it reacts vigorously and starts burning on water. Due to the above reasons Sodium is called an active metal.

(I)-->Down the group, the effective nuclear charge experienced by valence electrons is decreasing because the outermost electrons are far away from the nucleus. Thus, these electrons can be lost easily by the element to form positive ions. Hence, the chemical reactivity of metals increases on going down a group.

(J)-->While moving from top to bottom in a group of the periodic table, the reactivity of non- metals decreases. While moving from top to bottom in a group of non- metals, the atomic size increases with the additional number of shells and the force of attraction between the nucleus and valence shell decreases.

3 0

2 years ago

What amount of heat is required to raise the temperature of 200 g of water by 15°C (the specific heat of water is 1 cal/g°C)

Sergeeva-Olga [200]

Answer:

Heat energy required (Q) = 3,000 J

Explanation:

Find:

Mass of water (M) = 200 g

Change in temperature (ΔT) = 15°C

Specific heat of water (C) = 1 cal/g°C

Find:

Heat energy required (Q) = ?

Computation:

Q = M × ΔT × C

Heat energy required (Q) = Mass of water (M) × Change in temperature (ΔT) × Specific heat of water (C)

Heat energy required (Q) = 200 g × 15°C × 1 cal/g°C

Heat energy required (Q) = 3,000 J

4 0

3 years ago

The electric motor of a model train accelerates the train from rest to 0.685 m/s in 21.5 ms. The total mass of the train is 875

Natali [406]

Answer:

P=9.58 W

Explanation:

According to Newton's second law, and assuming friction force as zero:

$F_m=m.a\\F_m=0.875kg*a$

The acceleration is given by:

$a=\frac{\Delta v}{t}\\a=\frac{0.685m/s}{21.5*10^{-3}s}\\\\a=31.9m/s^2$

So the force exerted by the motor is:

$F_m=0.875kg*31.9m/s^2\\F_m=27.9N$

The work done by the motor is given by:

$W_m=F_m*d\\\\d=\frac{1}{2}*a*t^2\\d=\frac{1}{2}*31.9m/s^2*(21.5*10^{-3}s)^2\\\\d=7.37*10^{-3}m$

$W_m=27.9N*7.37*10^{-3}m\\W_m=0.206J$

And finally, the power is given by:

$P=\frac{W_m}{t}\\P=\frac{0.206J}{21.5*10^{-3}s}\\\\P=9.58W$

8 0

3 years ago

a water-balloon launcher with mass 4 kg fires a 0.5 kg balloon with a velocity of 3 m/s to the east. what is the recoil velocity

kotykmax [81]

I think we will use the law of conservation of linear momentum;
M1V1 = M2V2
M1 = 4 kg (mass of the water balloon launcher)
V1=?
M2= 0.5 kg ( mass of the balloon)
V2 = 3 m/s

Therefore; 4 V1 = 0.5 × 3
4V1= 1.5
V1= 1.5/4
= 0.375 m/s

5 0

3 years ago

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