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Irina-Kira [14]
3 years ago
13

The solubility of nitrogen gas at 25°C and a nitrogen pressure of 522 mmHg is 4.7 × 10 –4 mol/L. What is the value of the Henry'

s Law constant in mol/L·atm?
Chemistry
1 answer:
Anestetic [448]3 years ago
3 0

Answer:

6.8 × 10⁻⁴ mol/L.atm

Explanation:

Step 1: Given data

Solubility of nitrogen gas at 25°C (S): 4.7 × 10⁻⁴ mol/L

Partial pressure of nitrogen gas (P): 522 mmHg

Step 2: Convert the partial pressure of nitrogen to atm

We will use the relationship 1 atm = 760 mmHg.

522mmHg \times \frac{1atm}{760mmHg} = 0.687atm

Step 3: Calculate the value of the Henry's Law constant (k)

We will use Henry's law.

S = k \times P\\k = \frac{S}{P} = \frac{4.7 \times 10^{-4}mol/L }{0.687atm} = 6.8 \times 10^{-4} mol/L.atm

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Explanation:

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What is the empirical formula of a compound that contains 25.92 percent nitrogen and 74.07 percent oxygen?
pychu [463]

<span><span>N2</span><span>O5</span></span>

Explanation! 

When given %, assume you have 100 g of the substance. Find moles, divide by lowest count. In this case you'll end up with

<span><span>25.92 g N<span>14.01 g N/mol N</span></span>=1.850 mol N</span>

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The ratio between these is <span>2.502 mol O/mol N</span>, which corresponds closely with <span><span>N2</span><span>O5</span></span>.

4 0
3 years ago
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A concentrated binary solution containing mostly species 2 (but x2 ≠ 1) is in equilib- rium with a vapor phase containing both s
marusya05 [52]

Answer:

x1= 4.5 × 10^-3, y1= 0.9

Explanation:

A binary solution having two species is in equilibrium in a vapor phase comprising of species 1 and 2

Take the basis as the pressure of the 2 phase system is 1 bar. The assumption are as follows:

1. The vapor phase is ideal at pressure of 1 bar

2. Henry's law apply to dilute solution only.

3. Raoult's law apply to concentrated solution only.

Where,

Henry's constant for species 1 H= 200bar

Saturation vapor pressure of species 2, P2sat= 0.10bar

Temperature = 25°C= 298.15k

Apply Henry's law for species 1

y1P= H1x1...... equation 1

y1= mole fraction of species 1 in vapor phase.

P= Total pressure of the system

x1= mole fraction of species 1 in liquid phase.

Apply Raoult's law for species 2

y2P= P2satx2...... equation 2

From the 2 equations above

P=H1x1 + P2satx2

200bar= H1

0.10= P2sat

1 bar= P

Hence,

P=H1x1 + (1 - x1) P2sat

1bar= 200bar × x1 + (1 - x1) 0.10bar

x1= 4.5 × 10^-3

The mole fraction of species 1 in liquid phase is 4.5 × 10^-3

To get y, substitute x1=4.5 × 10^-3 in equation 1

y × 1 bar = 200bar × 4.5 × 10^-3

y1= 0.9

The mole fraction of species 1 in vapor phase is 0.9

4 0
3 years ago
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What is the volume of 13.21 g of ethane gas (C2H6) at STP?
yuradex [85]

The volume at STP : 9.856 L

<h3>Further explanation</h3>

Given

Mass of ethane : 13.21 g

Required

The volume at STP

Solution

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.

mol ethane(C2H6) :

= mass : molar mass

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Volume at STP :

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Answer:

The answer is 11 electron and 11 protons respectively

Explanation:

Since the sodium is in it neutral state, number of electron is the same as the number of proton, which is 11.

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