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2 years ago

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

Chemistry

2 answers:

lisabon 2012 [21]2 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]2 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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The active ingredient in Milk of Magnesia™ is Mg(OH)2. Magnesium hydroxide is insoluble in water, so the product is a mixture of

Liula [17]

Answer: This chemical reaction is a neutralization reaction between the Milk of Magnesia and the HCl from the stomach. The balanced equation is Mg(OH)2 (s) + 2 HCl (aq) → MgCl2 (aq) + 2 H2O (l)

Explanation:

The reaction between the HCl and the Mg(OH)2 is a neutralizacion reaction , because the HCl is a strong acid and the Mg(OH)2 is a weak base, then both react and the pH of the medium will increase, so the stomach trouble will dissapear.

Mg(OH)2 (s) + 2 HCl (aq) → MgCl2 (aq) + 2 H2O (l)

Magnesium hydroxide is a weak base due to its very limited solubility in water. This property is a great advantage when treating the excess of HCl in the stomach, because the Mg(OH)2 molecule does not dissociated easily until it reacts with the hydrogen ion, H+ of the HCl. So the effect of the Mg(OH)2 will last longer until the annoyance dissapear.

8 0

2 years ago

Determine the number of moles in 4.21 x 10^23 molecules of CaCl2

Paha777 [63]

<h3>Answer:</h3>

0.699 mole CaCl₂

<h3>Explanation:</h3>

To get the number of moles we use the Avogadro's number.

Avogadro's number is 6.022 x 10^23.

But, 1 mole of a compound contains 6.022 x 10^23 molecules

In this case;

we are given 4.21 × 10^23 molecules of CaCl₂

Therefore, to get the number of moles

Moles = Number of molecules ÷ Avogadro's constant

= 4.21 × 10^23 molecules ÷ 6.022 x 10^23 molecules/mole

= 0.699 mole CaCl₂

Hence, the number of moles is 0.699 mole of CaCl₂

7 0

2 years ago

if 203 mL of water is added to 5.00 mL of 4.16 M KCl solution, what is the concentration of the dilute solution

Sever21 [200]

Answer:0.100

Explanation:

5 0

2 years ago

14) The central iodine atom in the ICl4- ion has __________ nonbonded electron pairs and __________ bonded electron pairs in its

masha68 [24]

Answer:

Two non bonded electron pairs and four bonded electron pairs

Explanation:

An image of the compound as obtained from chemlibretext is attached to this answer.

The ion ICl4- ion, is an AX4E2 ion. This implies that there are four bond pairs and two lone pairs of electrons. As expected, the shape of the ion is square planar since the lone pairs are found above and below the plane of the square. This is clear from the image attached.

7 0

1 year ago

Use average bond energies to calculate ΔHrxn for the following hydrogenation reaction: H2C=CH2(g)+H2(g)→H3C−CH3(g)

marissa [1.9K]

Answer:

The $\Delta H_{rxn}$ of the given reaction is -129.6 kJ

Explanation:

The given chemical reaction is as follows.

$H_{2}C=CH_{2}(g)+H_{2}(g)\rightarrow H_{3}C-CH_{3}(g)$

Enthalpy of each reactant and products are as follows.

$\Delta H_{C=C}\,=615.0\,kJ\,mol^{-1}$

$\Delta H _{H-H}\,=435.1\,kJ\,mol^{-1}$

$\Delta H _{C-C}\,=347.3\,kJ\,mol^{-1}$

$\Delta H _{C-H}\,=416.2\,kJ\,mol^{-1}$

In the given chemical reaction involved two C-H bonds in the reactant side and one C-C bond in the product side therefore, the enthalpy of formation will be the negative.

$\Delta H_{rxn}=-\Delta H_{C-C}-2\Delta H_{C-H}+\Delta H_{C=C}+\Delta H_{H-H}$

$=-347.4-2\times416.2+615.0+435.1$

$=-129.6 \,kJ$

Therefore, The $\Delta H_{rxn}$ of the given reaction is -129.6 kJ

4 0

2 years ago