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erastova [34]
3 years ago
7

PLEASE HELP! WILL DO BRAINLIEST! What do scientists call all of the compounds that contain carbon and are found in living things

?
organic

inorganic

acidic

nonacidic
Chemistry
2 answers:
Gnesinka [82]3 years ago
8 0

Organic

Hope this helped.

harkovskaia [24]3 years ago
3 0

Answer:

organic compounds contained in all living things

Explanation:

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Why is this reaction considered to be exothermic? a Because energy difference A is greater than energy difference C. b Because e
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Answer:

we would need a bit more info for this

Explanation:

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James exclaims to his chemistry teacher that a new element has been discovered that fits between Nickel and Copper on the period
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What is the mass, in grams, of a sample of 1.20 × 1022 atoms of mercury (Hg)? Show your work or explain the steps that you used
o-na [289]
Atomic mass Hg = 200.59 u.m.a

200.59 g --------------- 6.02x10²³ atoms
( mass Hg ) ----------- 1.20 x10²² atoms

mass Hg = ( 1.20x10²² ) x 200.59 / 6.02x10²³

mass Hg = 2.407x10²⁴ / 6.02x10²³

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3 years ago
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. What mass of ammonium chloride must be added to 250. mL of water to give a solution with pH
Eduardwww [97]

The mass of ammonium chloride that must be added is : ( A ) 4.7 g

<u>Given data :</u>

Volume of water ( V )  = 250 mL = 0.25 L

pH of solution = 4.85

Kb = 1.8 * 10⁻⁵

Kw = 10⁻¹⁴

Given that the dissolution of NH₄Cl gives NH₄⁺⁺ and Cl⁻ ions the equation is written as :

NH₄CI  +  H₂O  ⇄  NH₃ + H₃O⁺

where conc of H₃O⁺

[ H₃O⁺ ] = \sqrt{Ka.C}   and Ka = Kw / Kb

∴ Ka = 5.56 * 10⁻¹⁰

Next step : Determine the concentration of H₃O⁺  in the solution

pH = - log [ H₃O⁺ ] = 4.85

∴ [ H₃O⁺ ] in the solution = 1.14125 * 10⁻⁵

Next step : Determine the concentration of NH₄CI in the solution

C = [ H₃O⁺ ]² / Ka

  = ( 1.14125 * 10⁻⁵ )² /  5.56 * 10⁻¹⁰

  = 0.359 mol / L

Determine the number of moles of NH₄CI in the solution

n = C . V

  = 0.359 mol / L  * 0.25 L =  0.08979 mole

Final step : determine the mass of ammonium chloride that must be added to 250 mL

mass = n * molar mass

         = 0.08979 * 53.5 g/mol

         = 4.80 g  ≈ 4.7 grams

Therefore we can conclude that the mass of ammonium chloride that must be added is 4.7 g

Learn more about ammonium chloride : brainly.com/question/13050932

8 0
2 years ago
In an experiment, a student needs 250.0 mL of a 0.100 M copper (II) chloride solution. A stock solution of 2.00 M copper (II) ch
LekaFEV [45]

Answer : The volume of stock solution needed are, 12.5 mL

Explanation :

Formula used :

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the initial molarity and volume of copper (II) chloride.

M_2\text{ and }V_2 are the final molarity and volume of stock solution of copper (II) chloride.

We are given:

M_1=0.100M\\V_1=250.0mL\\M_2=2.00M\\V_2=?

Putting values in above equation, we get:

0.100M\times 250.0mL=2.00M\times V_2\\\\V_2=12.5mL

Hence, the volume of stock solution needed are, 12.5 mL

8 0
3 years ago
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