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expeople1 [14]
3 years ago
6

what is the difference between aqueous hydrogen fluoride and hydrofluoric acid? do they have different formulas or are they just

the same but different names?
Chemistry
1 answer:
larisa86 [58]3 years ago
6 0

Answer:

Hydrofluoric acid is aqueous hydrogen fluoride (dissolved in water). Hydrogen fluoride will form hydrofluoric acid when it comes in contact with water.

Explanation:

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Using your periodic table, determine the electron configuration of Barium ion, Ba+2
seraphim [82]

Answer:

im pretty sure the electron configuration of the barium ion Ba+2 is [Kr]5s24d105p6

Explanation:

4 0
3 years ago
Write a chemical equation for the dissolution of the AgCl precipitate upon the addition of NH,(aq).
Kobotan [32]

Explanation:

White precipitate of silver chloride get dissolves in excess ammonia to formation of complex between silver ions, chloride ions and ammonia molecules.

The chemical reaction is given as:

AgCl(s)+2NH_3(aq)\rightarrow Ag[(NH_3)_2]^+Cl^-(aq)

When 1 mole of silver chloride is added to 2 mole of an aqueous ammonia it form coordination complex of diaaminesilver(I) chloride.

7 0
3 years ago
Please Need help asap
ruslelena [56]

Answer:

A

Explanation:

The number of protons and neutrons of an element is the same. the electrons are the only thing that can differ.  The atomic number equal the protons and neutrons.

4 0
3 years ago
What is the molarity of a solution prepared by diluting 43.72 ml of 1.005 M aqueous K2CR2O7 to 500 ml
UNO [17]

Answer:

43.72

Explanation:

that is the answer hope u liked it and I did this already along time ago

3 0
3 years ago
Calculate the mass of Octane needed to release 6.20 mol Co2
n200080 [17]
The combustion reaction of octane is as follow,

                           C₈H₁₈  +  25/2 O₂     →     8 CO₂  +  9 H₂O

According to balance equation,

8 moles of CO₂ are released when  =  114.23 g (1 mole) Octane is reacted

So,

      6.20 moles of CO₂ will release when  =  X g of Octane is reacted

Solving for X,
                                     X  =  (114.23 g × 6.20 mol) ÷ 8 mol

                                     X  =  88.52 g of Octane
Result:
           88.52 g of Octane is needed to release 6.20 mol CO₂.
8 0
3 years ago
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