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3 years ago

Compare the occurrence of heavier elements on earth and the universe.

1 answer:

3 0

By definition, a heavy element is a specific kind of element with a larger atomic mass. An element could be considered as "heavy" if its atomic number is greater already than 92. One of the known heavy elements on Earth is Tennessine and Tungsten while in the universe are lithium-5 and berryllium-8.

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5. Why doesn't the air at a frontal boundary mix?

Vsevolod [243]

Answer:

At a front, the two air masses have different densities, based on temperature, and do not easily mix. One air mass is lifted above the other, creating a low pressure zone.

Explanation:

Hope this helps!

4 0

3 years ago

Identify which sets of quantum numbers are valid for an electron. Note that each set is ordered (n,l,ml,ms).

Fynjy0 [20]

Answer:

The answer to your question is:

Explanation:

Quantum numbers have the numbers

n = from 1 to 7

l = s = 0 p = 1 d = 2 f= 3

m = ± l

s = ±1/2

Group of answer choice.

A) 2, -1. 1, -1/2 The option is incorrect "l" could not be negative

B) 2, 1, 0 , 1/2 The option is posible

C) 3, 2, 1, -1 This option is incorrect "s" can not value -1

D) 4, 3, 2, 1/2 This option is correct

E) 4, 3, -4, 1/2 This option is incorrect "m2 can not value -4

F) 1, 1, 0, 1/2 This option is incorrect, the second number is not posible

G) 1, 1, 0, -1/2 This option is incorrect the second number could not be 1.

H) 3, 0, 0, 1/2 This option is incorrect "l" can not value 0

I) 0, 2, 0, 1/2 This option is incorrect, n can not value 0

J) 3, 2, 2, 1/2 This option is correct

K) 1, 2, 0, -1/2 This option is incorrect "l" can not value 2.

3 0

3 years ago

The half-life of cesium-137 is 30 years. Suppose we have a 200-mg sample. (a) Find the mass that remains after t years.

scZoUnD [109]

Given:

Half life(t^ 1/2) :30 years

A0( initial mass of the substance): 200 mg.

Now we know that

A= A0/ [2 ^ (t/√t)]

Where A is the mass that remains after t years.

A0 is the initial mass

t is the time

t^1/2 is the half life

Substituting the given values in the above equation we get

A= [200/ 2^(t/30) ] mg

Thus the mass remaining after t years is [200/ 2^(t/30) ] mg

5 0

3 years ago

As an employee of a summer sports camp, one of your jobs is to prepare the tank of vitamin water for use throughout the day. the

TiliK225 [7]

Given:
Mixture of vitamin water = 75% pure water & 25% concentrated vitamin drink.
To find:
Quantity of water to be added to 16 gallons of concentrated vitamin drink to prepare a tank of vitamin water.
Solution:
Let total amount of mixture be x.
25% of x = 16 gallons (from given information)
When 25% of x is 16 gallons, the remaining 75% of x will be calculated as below:
(16/25)*75 = 48
Answer: 75% of x = 48 gallons. This means 48 gallons pure water is required to be added to mixture to prepare a tank of vitamin water.

3 0

3 years ago

A solution is prepared by dissolving 215 grams of methanol, ch3oh, in 1000. grams of water. what is the freezing point of this s

Leya [2.2K]

Answer : The freezing point of the solution is, 260.503 K

Solution : Given,

Mass of methanol (solute) = 215 g

Mass of water (solvent) = 1000 g = 1 kg (1 kg = 1000 g)

Freezing depression constant = $1.86^oC/m=1.86Kkg/mole$