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SSSSS [86.1K]
3 years ago
13

Rahul and Manav each were given a mixture of iron filings and sulphur powder. Rahul heated the mixture strongly and a new substa

nce was formed. Write three points of difference between the two.Required to answer.​
Chemistry
1 answer:
AlexFokin [52]3 years ago
3 0

Answer:

sure

Explanation:

The substance formed after heating the mixture of that of Rahul is caleed a compound. Whereas, Manav's mixture still remains in its current stae that is a heterogeneous mixture.

The compound formed is in black in color whereas the mixture is a mix of brownish-red and yellow.

The compound is a homogeneous mixture whereas the mixture is a heterogenous mixture because of its uneven distribution.

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Which of the following processes is endothermic? Group of answer choices water droplets condensing on a soda can on a hot summer
fgiga [73]

Answer: an ice pack getting cold (due to ammonium nitrate dissolving in water inside the pack)

Explanation:

Endothermic reactions: These are the reactions in which energy is required in the reaction. The energy of the products are more than the energy of the reactants.

Exothermic reactions: These are the reactions in which energy is released in the reaction. The energy of the products is less than the energy of the reactants.

a) water droplets condensing on a soda can on a hot summer day : As the gaseous molecules are changing into liquid phase, energy is released and is exothermic

b) an ice pack getting cold (due to ammonium nitrate dissolving in water inside the pack) : As the temperature is decreasing , energy is absorbed by the surroundings and thus is endothermic

c) thermite reaction between iron(III) oxide and aluminium (spectacular flames are observed)  : As flames are produced, the energy is released and thus is exothermic

d) freezing water to make ice cubes  : As the liquid molecules are changing into solid phase, energy is released and is exothermic

4 0
3 years ago
A power plant is driven by the combustion of a complex fossil fuel having the formula C11H7S. Assume the air supply is composed
AlekseyPX

(a) 4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 20.68N_2;

(b) 4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2;

(c) 23 900 kg air; (d) air:fuel = 10.2; (e) air:fuel = 12.2:1

(a) <em>Balanced equation including N_2 from air</em>  

The balanced equation <em>ignoring</em> N_2 from air is  

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2  

Moles of N_2 =55 mol O_2 × (3.76 mol N_2/1 mol O_2) = 206.8 mol N_2  

<em>Including</em> N_2 from air, the balanced equation is  

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 206.8N_2  

(b) <em>Balanced equation for 120 % stoichiometric combustion</em>  

Moles of O_2 = 55 mol O_2 × 1.20 = 66.00 mol O_2  

Excess moles O_2 = (66.00 – 55) mol O_2 = 11.00 mol O_2  

Moles of N_2 = 66.00 mol O_2 × (3.76 mol N_2/1 mol O_2) = 248.2 mol N_2  

The balanced equation is

4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2

(c) <em>Minimum mass of air</em>  

Moles of O_2 required = 1700 kg C_11 H_7S

× (1 kmol C_11 H_7S/185.24 kg C_11 H_7S) × (55 kmol O_2/4 kmol C_11 H_7S)

= 126.2 kmol O_2  

Mass of O_2 = 126.2 kmol O_2 × (32.00 kg O_2/1 kmol O_2) = 4038 kg O_2  

Mass of N_2 required = 126.2 kmol O_2 × (3.76 kmol N_2/1 kmol O_2)

× (28.01 kg N_2/1 kmol N_2) = 13 285 kg N_2  

Mass of air = Mass of N_2 + mass of O_2 = (4038 + 13 285) kg = 17 300 kg air  

(d) <em>Air:fuel mass ratio for 100 % combustion</em>  

Air:fuel = 17 300 kg/1700 kg = <em>10.2 :1 </em>

(e) <em>Air:fuel mass ratio for 120 % combustion </em>

Mass of air = 17 300 kg × 1.20 = 20 760 kg air  

Air:fuel = 20 760 kg/1700 kg = 12.2 :1  

6 0
3 years ago
You should have noticed that the Styrofoam ball rolled slightly towards the rod and then appeared to be lifted above the pan. Wh
Kaylis [27]

Answer:

static electricity

Explanation:

8 0
3 years ago
Read 2 more answers
What is true about the valence shell of the noble gasses?
lana [24]

Answer:

The valence shell of noble gases has all 8 valence electrons which means it is full.

Explanation:

8 0
3 years ago
In the nuclear transmutation, 168O (?, α)137N, what is the bombarding particle? In the nuclear transmutation, O (?, )N, what is
xeze [42]

Answer:

The bombarding particle is a Proton

Explanation:

A Nuclear transmutation reaction occurs when radioactive element decay, usually converting them from one element/isotope into another element. Transmutation is the process which causes decay, generally, alpha or beta.

¹⁶₈O(P,alpha) ¹³₇N, can be written as

¹⁶₈O + x goes to ¹³₇N + ⁴₂He

Where x can be anything, balancing the equation in order to give us the correct amount of proton number and nucleus number

16 + x = 13 + 4

x = 17 – 16 = 1, Hence we can say that x = ¹₁P

<u>¹⁶₈O + ¹₁P goes to ¹³₇N + ⁴₂He</u>

Here we can clearly see the bombarding particle is ¹₁P (proton). The ejected particle being ⁴₂He which is also known as an alpha particle

5 0
4 years ago
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