Whats the formula for density

What is the difference in energy between an electron in 1s and an electron in 4s.

Andre45 [30]

Answer:

Explanation:

An electron in 4s is farther away from nucleus and it has higher energy when compared to electron from 1s.

5 0

3 years ago

Pleassse help!!!

11111nata11111 [884]

True

Explanation:

Harry Hess in 1961 resurrected Wegener's continental drift hypothesis and also he mantle convection idea of Arthur Holmes by postulating the theory of sea floor spreading.

After Wegener's postulation of the continental drift hypothesis, not much was done.
It was only around 1931 that Arthur Holmes came up with the idea of a convecting mantle.
After the second world war passed, new instruments were used to investigate and map the ocean floor.
Through this, Arthur Holmes came up with the idea of a spreading sea floor.
This gave a full gleam picture to the idea of a moving earth and the more accepted terminology of "Plate tectonics " came to the fore.

Learn more:

Sea floor spreading brainly.com/question/9912731

Continental drift brainly.com/question/5002949

#learnwithBrainly

8 0

3 years ago

Riley is scootering down the street before they come to a stop at a stop

11111nata11111 [884]

Answer:

just subrracted them and find out whats m

Explanation:

6 0

3 years ago

Which is a way that the elements are groupef on the periodic table

hammer [34]

Your answer is B because elements within the same column have equal numbers of valence electrons which determines the reactivity of an element.

8 0

2 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) Balanced equation including N_2 from air

The balanced equation ignoring 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

Including N_2 from air, the balanced equation is

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

(b) Balanced equation for 120 % stoichiometric combustion

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) Minimum mass of air

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) Air:fuel mass ratio for 100 % combustion

Air:fuel = 17 300 kg/1700 kg = 10.2 :1

(e) Air:fuel mass ratio for 120 % combustion

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

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

6 0

2 years ago