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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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120 grams of calcium nitrite Ca(NO2)2 is dissolved in a 240 mL solution. What is the molarity of the solution???

expeople1 [14]

Hey there!:

Molar mass Ca(NO2)2 = 132.089 g/mol

Mass of solute = 120 g

Number of moles:

n = mass of solute / molar mass

n = 120 / 132.089

n = 0.0009084 moles of Ca(NO2)2

Volume in liters of solution :

240 mL / 1000 => 0.24 L

Therefore:

Molarity = number of moles / volume of solution

Molarity = 0.0009084 / 0.24

Molarity = 0.003785 M

Hope that helps!

5 0

2 years ago

A 40.0 mL sample of 0.25 M KOH is added to 60.0 mL of 0.15 M Ba(OH)2. What is the molar concentration of OH-(aq) in the resultin

solmaris [256]

Answer:

C) 0.28 M

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Potassium hydroxide will furnish hydroxide ions as:

$KOH\rightarrow K^{+}+OH^-$

Given :

For Potassium hydroxide :

Molarity = 0.25 M

Volume = 40.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 40.0×10⁻³ L

Thus, moles of hydroxide ions furnished by Potassium hydroxide is same as the moles of Potassium hydroxide as shown below:

$Moles =0.25 \times {40.0\times 10^{-3}}\ moles$

Moles of hydroxide ions by Potassium hydroxide = 0.01 moles

Barium hydroxide will furnish hydroxide ions as:

$Ba(OH)_2\rightarrow Ba^{2+}+2OH^-$

Given :

For Barium hydroxide :

Molarity = 0.15 M

Volume = 60.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 60.0×10⁻³ L

Thus, moles of hydroxide ions furnished by Barium hydroxide is twice the moles of Barium hydroxide as shown below:

$Moles =2\times 0.15 \times {60.0\times 10^{-3}}\ moles$

Moles of hydroxide ions by Barium hydroxide = 0.018 moles

Total moles = 0.01 moles + 0.018 moles = 0.028 moles

Total volume = 40.0×10⁻³ L + 60.0×10⁻³ L = 100×10⁻³ L

Concentration of hydroxide ions is:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity_{OH^-}=\frac{0.028 }{100\times 10^{-3}}$

The final concentration of hydroxide ion = 0.28 M

5 0

2 years ago

Read 2 more answers

One of the emission spectral lines for Be3+ has a wavelength of 253.4 nm for an electronic transition that begins in the state w

4vir4ik [10]

Answer:

Explanation:

Relationship between wavenumber and Rydberg constant (R) is as follows:

$wave\ number=R\times Z^2\frac{1}{n_1^2} -\frac{1}{n_1^2}$

Here, Z is atomic number.

R=109677 cm^-1

Wavenumber is related with wavelength as follows:

wavenumber = 1/wavelength

wavelength = 253.4 nm

$wavenumber=\frac{1}{253.4\times 10^{-9}} \\=39463.3\ cm^{-1}$

Z fro Be = 4

$39463.3=109677\times 4^2(\frac{1}{n_1^2} -\frac{1}{5^2})\\39463.3=109677\times 16(\frac{1}{n_1^2} -\frac{1}{5^2})\\n_1=4$

Therefore, the principal quantum number corresponding to the given emission is 4.

6 0

2 years ago

A solution is prepared by condensing 4.00 l of a gas, measured at 27°c and 748 mmhg pressure, into 58.0 g of benzene. calculate

fgiga [73]

First, we are using the ideal gas law to get n the number of moles:

PV = nRT

when P is the pressure = 748 mmHg/760 = 0.984 atm

V is the volume = 4 L

R is ideal gas constant = 0.0821

T is the temperature in Kelvin = 300 K

∴ n = 0.984atm*4L/0.0821*300

= 0.1598 moles

when the concentration = moles * (1000g / mass)

= 0.1598 * (1000g / 58 g )

= 2.755 M

when the freezing point = 5.5 °C

and Kf = - 5.12 °C/m

∴ the freezing point for the solution = 5.5 °C + (Kf*m)

= 5.5 °C - (5.12°C/m * 2.755m)

= -8.6 °C

8 0

2 years ago

eleanor purchased $2568 worth of stock and paid her broker a 0.5% fee. She sold the stock when the stock price increased to 3928

Mrrafil [7]

Answer: $1338.16

Explanation: Total cost of stock= $2568

Total cost of stock including the brokerage $=2568+\frac{0.5}{100}\times {2568}=2580.84$ $

Selling price of stock = $3928

Selling price of stock including trading fee=($3928-$7)=$3919

Net Proceeds = Net selling price of stock - Cost Price of stock

Net Proceeds = ($3919 - $2580.84) = $1338.16

5 0

2 years ago