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.
Answer:
sulfur
Explanation:
In oxygen family sulfur has yellow color and also having stinky smell. Thus given statements are about sulfur.
It is present in oxygen family.
It has six valance electrons.
Its atomic number is 16.
Its atomic weight is 32 amu.
The electronic configuration of sulfur is given below,
S₁₆ = 1s² 2s² 2p⁶ 3s² 3p⁴
We can see the valance shell is third shell and it have six electrons thus sulfur have six valance electrons. (3s² 3p⁴ )
Sulfur is used in vulcanisation process.
It is used in bleach and also as a preservative for many food.
it is used to making gun powder.
Answer
The empirical formula is CrO₂Cl₂
Explanation:
Empirical formula is the simplest whole number ratio of an atom present in a compound.
The compound contain, Chromium=33.6%
Chlorine=45.8%
Oxygen=20.6%
And the molar mass of Chromium(Cr)=51.996 g mol.
Chlorine containing molar mass (Cl)= 35.45 g mol.
Oxygen containing molar mass (O)=15.999 g mol.
Step-1
Then,we will get,
Cr=
mol
Cl= 
mol.
O=
mol.
Step-2
Divide the mole value with the smallest number of mole, we will get,
Cr=
Cl=
O=
Then, the empirical formula of the compound is CrO₂Cl₂ (Chromyl chloride)
Explanation:
B. HCIO these is the answer
Answer:
4.6 × 10²³ molecules:
Step-by-step solution
You will need a balanced equation with masses, moles, and molar masses, so let's gather the information in one place:
M_r: 22.99
2Na + 2H₂O ⟶ 2NaOH + H₂
m/g: 35
1. Calculate the <em>moles of Na
</em>
Moles of Na = 35 g Na × (1 mol Na/22.99 g Na)
Moles of Na = 1.52 mol Na
2. Calculate the <em>moles of H₂
</em>
Moles of H₂ = 1.52 mol Na × (1 mol H₂/2 mol Na)
Moles of H₂= 0.761 mol H₂
3. Calculate the molecules of H₂
6.022 × 10²³ molecules H₂ = 1 mol H₂
Molecules of H₂ = 0.761 × (6.022 × 10²³
/1)
Molecules of H₂ = 4.6 × 10²³ molecules H₂
The reaction forms 4.6 × 10²³ molecules of H₂.
