What is the molarity of NaCL solution that has 58 grams of NaCL dissolved in 4.0 liters of water

Solid ammonium chloride, NH4Cl, is formed by the reaction of gaseous ammonia, NH3, and hydrogen chloride, HCl. NH3(g)+HCl(g)⟶NH4

Mashutka [201]

9.41 atm is the pressure in atmospheres of the gas remaining in the flask

<h3>What is the pressure in atmospheres?</h3>

The equation NH3(g) + HCl(g) ==> NH4Cl(s) is balanced.

Divide the moles of each reactant by its coefficient in the balanced equation, and the limiting reagent is identified as the one whose value is less. With the issue we now have...

6.44 g NH3 times 1 mol NH3/17 g equals 0.3688 moles of NH3 ( 1 = 0.3688)

HCl: 6.44 g of HCl times one mole of HCl every 36.5 g equals 0.1764 moles ( 1 = 0.1764). CONTROLLING REAGENT

NH4Cl will this reaction produce in grams

0.1764 moles of HCl multiplied by one mole of NH4Cl per mole of HCl results in 9.44 g of NH4Cl (3 sig. figs.)

the gas pressure, measured in atmospheres, that is still in the flask

NH3(g) plus HCl(g) results in NH4Cl (s)

0.3688......0.1764............0..........

Initial

-0.1764....-0.1764........+0.1764...Change

Equilibrium: 0.1924.......0...............+0.1924

There are 0.1924 moles of NH3 and no other gases in the flask. This is at a temperature of 25 °C (+273 = 298 °K) in a volume of 0.5 L. After that, we may determine the pressure by using the ideal gas law (P).

PV = nRT

P = nRT/V = 0.1924 mol, 0.0821 latm/mol, and 298 Kmol / 0.5 L

P = 9.41 atm

9.41 atm is the pressure in atmospheres of the gas remaining in the flask

To learn more about balanced equation refer to:

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7 0

1 year ago

Heart, 5 stars, and Brainiest to first right answer!

mezya [45]

Answer:

Give person above me brainliest

Explanation:

7 0

2 years ago

The element that has a valence configuration of 5s25p6 is ________.

Ivanshal [37]

Its period 5 from 5s25p6, with Xenon(54) as the noble gas. 2+6 = 8 electrons

54+8 = 62, or Sm.

3 0

2 years ago

How does dissolving salt in water allow electric current to flow?

Mumz [18]

Salt is an ionic compound NaCI. So the salt will dissolve and spread apart and the ions in the salt will move freely, letting electricity to flow freely....!

Hope i helped! If you need anything else ask me!! :)

4 0

3 years ago

Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.

Ad libitum [116K]

Explanation:

The given data is as follows.

Concentration = 0.1 $mol/dm^{3}$

= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

= $6.022 \times 10^{25} ions/m^{3}$

T = $30^{o}C$ = (30 + 273) K = 303 K

Formula for electric double layer thickness ( $\lambda_{D}$ ) is as follows.

$\lambda_{D}$ = $\frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$

where, $n^{o}$ = concentration = $6.022 \times 10^{25} ions/m^{3}$

Hence, putting the given values into the above equation as follows.

$\lambda_{D}$ = $\sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$

= $\sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}$

= $9.669 \times 10^{-10}$ m

or, = $9.7 A^{o}$

= 1 nm (approx)

Also, it is known that $\lambda_{D}$ = $\sqrt \frac{1}{n^{o}}$

Hence, we can conclude that addition of 0.1 $mol/dm^{3}$ of KCl in 0.1 $mol/dm^{3}$ of NaBr " $\lambda_{D}$ " will decrease but not significantly.

7 0

3 years ago