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2 years ago

Compare and contrast the carbon cycle of 1500 and of 2021

Chemistry

1 answer:

Delvig [45]2 years ago

7 0

Answer:

Carbon moves from the atmosphere to the land, ocean, and life through biological, chemical, geological and physical processes in a cycle called the carbon cycle. Because some carbon gases are greenhouse gases, changes in the carbon cycle that put more carbon in the atmosphere also warm Earth’s climate.

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Given the reaction: HNO2 (aq) <-> H+ (aq) + NO2- (aq); write the ionization constant for the reaction.

Nuetrik [128]

Answer:

see explanation

Explanation:

HNO₂ ⇄ H⁺ + NO₂⁻

Ka = [H⁺][NO₂⁻]/[HNO₂]

4 0

2 years ago

What is the electron configuration for N (nitrogen)?

ollegr [7]

Answer:

$1s^22s^22p^3$

Explanation:

Nitrogen has the atomic number = 7

So, No. of electrons = 7

Electronic Configuration:

$1s^22s^22p^3$

Remember that:

s sub shell holds upto 2 electrons while p sub shell upto 6

7 0

3 years ago

Ramelteon is a hypnotic agent used in the treatment of insomnia: Considering the labeled drawing above, which of the highlighted

9966 [12]

Answer:

1 and 2

Trigonal planar

Tetrahedral

Trigonal Planar

Linear

Bent ( v- shape)

Explanation:

The highlighted atoms has a delocalized lone pair - 1 and 2

All sp2-hybridized carbon atoms have geometry- Trigonal planar

All sp3-hybridized carbon atoms have geometry - Tetrahedral

The nitrogen atom has geometry - Trigonal Planar

The oxygen atom of the C=O bond has geometry - Linear

and the other oxygen atom has geometry - Bent ( v- shape)

P.S - The correct question is -

7 0

2 years ago

If a 125 gms of radioactive element has a half life of 60 min how many half lives will it go through to become 3.90625 gms

Sonja [21]

Answer:

Explanation:

Let the number of half lives be x

Solve this equation to find the value of x:

125*(1/2)ˣ = 3.90625
(0.5)ˣ = 3.90625 / 125
(0.5)ˣ = 0.03125
log (0.5)ˣ = log 0.03125
x = log 0.03125 / log 0.5
x = 5

4 0

2 years ago

How many milliliters of 0.0050 N KOH are required to neutralize 41 mL of 0.0050 M H2SO4?

Kipish [7]

Answer:

V KOH = 41 mL

Explanation:

for neutralization:

( V×C )acid = ( V×C )base

∴ C H2SO4 = 0.0050 M = 0.0050 mol/L

∴ V H2SO4 = 41 mL = 0.041 L

∴ C KOH = 0.0050 N = 0.0050 eq-g/L

∴ E KOH = 1 eq-g/mol

⇒ C KOH = (0.0050 eq-g/L)×(mol KOH/1 eq-g) = 0.0050 mol/L

⇒ V KOH = ( V×C ) acid / C KOH

⇒ V KOH = (0.041 L)(0.0050 mol/L) / (0.0050 mol/L)

⇒ V KOH = 0.041 L

4 0

2 years ago