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marusya05 [52]
3 years ago
11

The atomic number of an atom is equivalent to the number of:

Chemistry
1 answer:
valentinak56 [21]3 years ago
7 0
C the number of protons an neutrons
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assume in a different experiment, you prepare a mixture containing 10.0 M FeSCN2+, 1.0 M H+, 0.1 MFe3+ and 0.1 M HSCN. Is the in
Katarina [22]

Answer:

The mixture is not in equilibrium, the reaction will shift to the left.

Explanation:

<em>Based on the equilibrium:</em>

<em>Fe³⁺+ HSCN ⇄ FeSCN²⁺ + H⁺</em>

<em>kc = 30 = [FeSCN²⁺] [H⁺] / [Fe³⁺] [HSCN]</em>

Where [] are concentrations at equilibrium. The reaction is in equilibrium when  the ratio of concentrations = kc

Q is the same expression than kc but with [] that are not in equilibrium

Replacing:

Q = [10.0M] [1.0M] / [0.1M] [0.1M]

Q = 1000

As Q > kc, the reaction will shift to the left in order to produce Fe³⁺ and HSCN untill Q = Kc

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7 0
3 years ago
If you looked at one of your body's cell under a microscope with 100x magnification, would it be larger, smaller, or the same si
aev [14]
Peoples cells vary . so theres no definite say
7 0
3 years ago
A chemistry teacher carried out several demonstrations, and students recorded their observations. For one of the demonstrations,
PIT_PIT [208]

Answer:

It demonstrates that it a physical change because you can't see nothing happening but something happening

Sry if it doesn't make sense

4 0
3 years ago
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Describe use of H₂S an analytical reagent?<br><br>​
Gemiola [76]

Answer:

The main use for hydrogen sulfide is in the production of sulfuric acid and elemental sulfur. ... H2S is used to prepare the inorganic sulfides you need to make those products. As a reagent and intermediate, hydrogen sulfide is beneficial because it can prepare other types of reduced sulfur compounds.

7 0
3 years ago
if you have a solution that contains 46.85 g of codeine, C18H21NO3, in 125.5 g of ethanol, C2H5OH, what is the mole fraction of
OverLord2011 [107]

Answer:

The mole fraction of codeine is 5.4%

Explanation:

Mole fraction = Mole of solute / Total moles (Mole solute + Mole solvent)

Solute (Codeine)

Molar mass 299.36 g/m

Mass / Molar mass = Mole →  46.85 g /299.36 g/m = 0.156 moles

Solvent (Ethanol)

Molar mass 46.07 g/m

125.5 g / 46.07 g/m = 2.724 moles

Mole fraction (Codeine) 0.156 / (0.156 + 2.724) → 0.054

3 0
3 years ago
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