1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
avanturin [10]
3 years ago
10

4. Rank the listed elements from largest electronegativity at the top, to smallest electronegativity at

Chemistry
1 answer:
Burka [1]3 years ago
7 0
1. Chlorine
2. Sulfur
3. Selenium
You might be interested in
A. 207 kJ<br> B. 4730 kJ<br> O C. 9460 kJ<br> O D. 414 kJ
slavikrds [6]

Answer:

C = 9460 Kj

Explanation:

Given data:

Mass of copper = 2kg

Latent heat of vaporization = 4730 Kj/Kg

Energy required to vaporize 2kg copper = ?

Solution:

Equation

Q= mLvap

by putting values,

Q= 2kg ×  4730 Kj/Kg

Q = 9460 Kj

3 0
3 years ago
A 50.0 N force is applied to an object travels with a constant velocity of 20m/s, calculate
weqwewe [10]

Answer:

trdewasd7rf8t6gy7hu8jik

Explanation:

6 0
3 years ago
Which of the following statements are FALSE of models? (select all that applies)
ira [324]

Answer:

A B and C

Explanation:

Hopefully this helps

7 0
3 years ago
A solution of an unknown nonvolatile nonelectrolyte was prepared by dissolving 0.250 g of the substance in 40.0 g of ccl4. the b
Finger [1]
Answer : The molar mass of the solute will be 87.90 g/mol.

Explanation : We know the formula for elevation in boiling point, which is

Δt = iK_{b}m

given that, Δt = 0.357, K_{b} = 5.02 and mass of CCl _{4} = 40,

on substituting the value we get,
0.357 = (1) X (5.02) X (x/ 0.044), on solving we get x = 2.844 X10^{-3}. 
Now, 0.250/ 2.844 X10^{-3} = 87.90 g/mol. which is the weight of unknown component.
3 0
3 years ago
Consider the following equilibrium: 2SO^2(g) + O2(9) = 2 SO3^(g)
saul85 [17]

Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if \rm O_2 is removed from the mixture. The reason is that collisions between \rm SO_2 molecules and \rm O_2\! molecules would become less frequent.

The reaction would not be at equilibrium for a while after \rm O_2 was taken out of the mixture.

Explanation:

<h3>Equilibrium</h3>

Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of \rm SO_3\, (g) to the system, the backward reaction would have broken down the same amount of \rm SO_3\, (g)\!. So is the case for \rm SO_2\, (g) and \rm O_2\, (g).

Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.

<h3>Collision Theory</h3>

In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that \rm SO_2\, (g) and \rm O_2\, (g) molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield \rm SO_3\, (g), only a fraction of the reactions will be fruitful.

Assume that \rm O_2\, (g) molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer \!\rm O_2\, (g) molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between \rm SO_2\, (g) and \rm O_2\, (g)\! molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

<h3>Equilibrium after Change</h3>

The backward reaction rate is likely going to stay the same right after \rm O_2\, (g) was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

Therefore, this reaction would not be at equilibrium immediately after the change.

As more and more \rm SO_3\, (g) gets converted to \rm SO_2\, (g) and \rm O_2\, (g), the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.

5 0
3 years ago
Other questions:
  • Throughput is a term that refers to the amount of ?
    8·1 answer
  • What would happen to Earth’s tides if the distance between Earth and the Moon increased?
    12·2 answers
  • Science problem please help
    8·2 answers
  • Carry out the following calculation, paying special attention to the significant figures (where 4/3 is exact), rounding, and uni
    10·1 answer
  • The rate constant for the first-order decomposition of n2o is 3.40 s-1. what is the half-life of the decomposition?
    12·2 answers
  • Which energy source provides most of the world’s energy consumption? solar hydroelectric fossil fuels natural gas
    6·2 answers
  • 3) Which pair of formulas represents the empirical formula and molecular
    14·2 answers
  • Answer the questions below:
    6·2 answers
  • How Chemistry and Earth Science are Connected?Can you tell me completely?​
    15·1 answer
  • A marine biologist and her team capture 40 sea turtles and mark them before releasing them. The following year, they capture 60
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!