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2 years ago

What is the total number of ions in one formula unit of (NH4)2CO3?

Chemistry

2 answers:

evablogger [386]2 years ago

7 0

Answer:

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Len [333]2 years ago

4 0

There are 3 ions in one formula unit of (NH4)2CO3.

Ions present in Ammonium carbonate.(NH4)2CO3

(NH4)2CO3 is called Ammonium carbonate. The formula indicates that in one mole of ammonium carbonate, There are

Two moles of ammonium ions,NH₄⁺ with valency of +1.

1 mole of Carbonate ions CO₃²⁻ with valency of -2.

When 1 mole of ammonium carbonate is dissolved in water it gives

(NH₄)₂CO₃ ₊ H₂O---->2NH₄⁺ ₊ CO₃²⁻

Ions present are 2NH₄⁺ and CO₃²⁻

Learn more on (NH4)2CO3 here:brainly.com/question/2272720

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Exposing a 250 mL sample of water at 20.∘C to an atmosphere containing a gaseous solute at 27.59 kPa resulted in the dissolution

ElenaW [278]

Answer : The solubility of this gaseous solute will be, $5.18\times 10^{-5}g/mL$

Explanation :

First we have to calculate the concentration of solute.

$\text{Concentration of solute}=\frac{Mass}{Volume}=\frac{4.51\times 10^{-3}g}{250mL}=1.80\times 10^{-5}g/mL$

Now we have to calculate the Henry's law constant.

Using Henry's law :

$C=k_H\times p$

where,

C = concentration of solute = $1.80\times 10^{-5}g/mL$

p = partial pressure = 27.59 kPa

$k_H$ = Henry's law constant = ?

Now put all the given values in the above formula, we get:

$1.80\times 10^{-5}g/mL=k_H\times (27.59kPa)$

$K_H=6.52\times 10^{-7}g/mL.kPa$

Now we have to calculate the solubility of this gaseous solute when its pressure is 79.39 kPa.

$C_{(79.39kPa)}=k_H\times p$

$C_{(79.39kPa)}=(6.52\times 10^{-7}g/mL.kPa)\times (79.39kPa)$

$C_{(79.39kPa)}=5.18\times 10^{-5}g/mL$

Therefore, the solubility of this gaseous solute will be, $5.18\times 10^{-5}g/mL$

7 0

3 years ago

A 13.5 g sample of an unknown gas occupies 5.10 L at 149.83 kPa and 301 K. What is the molar mass of the gas ?

alisha [4.7K]

Answer:

The molar mass of the gas is 44.19 g/mol

Explanation:

Amount of sample of gas = m = 13.5 g

Volume occupied by the gas = V = 5.10 L

Pressure of the gas = P = 149.83 KPa

1 KPa = 0.00986 atm

P = $149.83 \textrm{ KPa} \times 0.00986 \textrm{ atm/KPa} = 1.48 \textrm{ atm}$

Assuming M g/mol to be the molar mass of the gas

Assuming the gas is behaving as an ideal gas

$\textrm{PV} =\textrm{nRT} \\\textrm{PV} = \displaystyle \frac{m}{M}\textrm{ RT } \\1.48 \textrm{ atm}\times 5.10 \textrm{ L} = \displaystyle \frac{13.5 \textrm{ g}}{M}\times 0.0821 \textrm{ L.atm.mol}^{-1}.K^{-1}\times 301\textrm{K} \\M = 44.19 \textrm{ g/mol}$