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Sergeu [11.5K]
3 years ago
9

Why is helium trapped in the earth's crust? If it's its trapped in the earth's crust how does it get into the air that we breath

e?
Chemistry
1 answer:
lesya [120]3 years ago
7 0
 <span>A small proportion of helium in the crust is helium that was trapped in the Earth when the Earth formed and has not yet escaped. Most helium on Earth, however, forms as a result of alpha decay of uranium and thorium. The emitted alpha particles, once they grab a couple of stray electrons, become helium atoms and can accumulate in gas reservoirs along with things such as methane.</span>
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Types of Organisms:
maxonik [38]

Answer:

Part A

The green plant is a producer

The grasshopper is a herbivore

The Rat is a Omnivore

The snake is a Carnivore

The eagle is a Carnivore

The mushroom is a Decomposer

Part B

The arrow between each organism indicates the direction of energy flow which shows the direction in which the energy transfer takes place from one organism to the next, with the general source of energy for all animals being the Sun.

Part C

A sample food chain can be presented as follows;

Sun, Green plants → worms → Frogs → Snakes → Coyotes → Bears → Tigers → Mushroom → Plants

Explanation:

6 0
3 years ago
Which is a nonpolar molecule?
Colt1911 [192]
The answer is C- sulfur hexachlorine (SF6)

<span>S<span>F6 is the only molecule here that is non-polar. That's due to having the</span></span><span> fluorine atoms arranged in a way that, in pairs, they lie opposite to each other. Also, these pairs are perpendicular to each other on  three different axis.</span>

7 0
3 years ago
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On may 10th it's gonna be my birthday and my mum don't know what to get me what do u think i would like my bio: creative,artisti
pickupchik [31]

Answer:

It’s past your b day... sorry

Explanation:

8 0
2 years ago
Determine the final temperature of sample with a specific heat of 1.1 J/g°C and a mass of 385 g if it starts out at a temperatur
Assoli18 [71]

Answer:

T2 =21.52°C

Explanation:

Given data:

Specific heat capacity of sample = 1.1 J/g.°C

Mass of sample = 385 g

Initial temperature = 19.5°C

Heat absorbed = 885 J

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature - initial temperature

885J = 385 g× 1.1 J/g.°C×(T2 - 19.5°C )

885 J = 423.5 J/°C× (T2 - 19.5°C )

885 J / 423.5 J/°C = (T2 - 19.5°C )

2.02°C = (T2 - 19.5°C )

T2 = 2.02°C + 19.5°C

T2 =21.52°C

8 0
3 years ago
Which must be the same when comparing 1 mol of oxygen gas, O2, with 1 mol of carbon monoxide gas, CO?
Julli [10]

Answer:

The number of molecules.

Explanation:

7 0
3 years ago
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