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Minchanka [31]
3 years ago
13

Enter the ions formed when (NH4)2S dissolves in water.

Chemistry
2 answers:
lisabon 2012 [21]3 years ago
5 0

The ions formed are NH4(+) and S(2-)

The dissolution reaction of (NH4) 2S in water is as follows:


(NH4) 2S ==> 2 NH4 (+) + S (2-).



Ammonium sulfide is the ammonium salt of hydrogen sulfide. It has the formula (NH4) 2S and belongs to the sulfide family.


It is a relatively unstable compound (crystals decomposing at -18 ° C, but exists and is more stable in aqueous solution.) With a pKa exceeding 15, the hydrosulfide ion cannot be significantly deprotonated by ammonia. Thus, such solutions consist mainly of a mixture of ammonia and hydrosulphide of ammonium, it has a smell, close to that of hydrogen sulfide, and its aqueous solutions can be precisely by emitting H2S.

kirill [66]3 years ago
5 0
The ions formed are NH4+ and SO4^2-
When (NH4)2SO4 is dissolved in water, ammonium ion and a sulfate ion are produced due to the reaction between (NH4)2SO4 and water. Ions are produced when simple polar molecules are dissolved in water, the molecules dissociate into ions on dissolving it in water. Ammonium ion formed is positively charged while the sulfate ion is negatively charged. 
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3 years ago
What mass of sodium chloride (NaCl) forms when 7.5 g of sodium carbonate (Na2CO3) reacts with a dilute solution of hydrochloric
Gre4nikov [31]

 The mass of NaCl  formed  is 8.307  grams


<u><em> calculation</em></u>

step 1: write the equation  for reaction

Na₂CO₃  + 2HCl → 2 NaCl  +CO₂ +H₂O

Step 2: find the  moles of Na₂CO₃

moles = mass/molar mass

 The  molar mass of Na₂CO₃  is = (23 x2) + 12 + ( 16 x3) = 106 g/mol

moles  = 7.5 g/106 g/mol =0.071 moles

Step 3: use the  mole ratio to determine the  mole of NaCl

Na₂CO₃:NaCl  is  1:2  therefore the moles of NaCl =0.07  x2 =0.142 moles


Step 4:  calculate mass  of NaCl

mass= moles x molar mass

the molar  mass of NaCl= 23 +35.5 =58.5 g/mol

mass  = 0.142  moles x 58.5 g/mol =8.307  grams

8 0
3 years ago
Read 2 more answers
What is the molar concentration of 35 mL of H2SO4 that neutralizes 25 mL of 0.320M NaOH
stira [4]
V ( H2SO4) = 35 mL / 1000 => 0.035 L

M ( H2SO4) = ?

V ( NaOH ) = 25 mL / 1000 => 0.025 L 

M ( NaOH ) = 0.320 M

number of moles NaOH:

n = M x V

n = 0.025 x 0.320 => 0.008 moles of NaOH

Mole ratio:

<span>2 NaOH + H2SO4 = Na2SO4 + 2 H2O
</span>
2 moles NaOH ---------------------- 1 mole H2SO4
0.008 moles moles NaOH ---------- ??

0.008 x 1 / 2 => 0.004 moles of H2SO4 :

Therefore:

M ( H2SO4) = n / V

M = 0.004 /  0.035

= 0.114 M

hope this helps!



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Answer:

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never [62]

Answer:

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Explanation:

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