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Serga [27]
1 year ago
11

How many molecules are in 2A1203? *

Chemistry
1 answer:
Triss [41]1 year ago
5 0

Answer:

6

Explanation:

2A is a molecule Argon bond. so, it's 1 + 2+ 3

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The equilibrium constant in terms of pressure. Kp for the following process is 0.179 at 50 °C. It increases to 0.669 at 86 °C. C
Vikentia [17]

Answer:

a) ΔHvap=35.3395 kJ/mol

b) Tb=98.62 °C

Explanation:

Given the reaction:

C₇H₁₆ (l) ⇔ C₇H₁₆ (g)

Kp=P(C₇H₁₆) since the concentration ratio for a pure liquid is equal to 1.

When

T₁=50°C=323.15K ⇒P₁=0.179

T₂=86°C=359.15K ⇒P₂=0.669

The Clasius-Clapeyron equation is:

ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})

ln(\frac{0.669}{0.179}) =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{359.15K}-\frac{1}{323.15K})

1.3184 =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (-3.10186*10^{-4}K{^-1})

ΔHvap=35339.5 J/mol=35.3395 KJ/mol

Normal boiling point ⇒ P=1 atm

Hence, we find the normal boiling point where:

T₁=323.15K

P₁=0.179 atm

P₂=1 atm

ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})

ln(\frac{1atm}{0.179atm}) =-\frac{35339.5 J/mol}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{T_2}-\frac{1}{323.15K})

1.7203=-4250.34 (\frac{1}{T_2}-\frac{1}{323.15K})

T₂=371.77 K= 98.62 °C

5 0
3 years ago
Rolls of foil are 302 mm wide and 0.018 mm thick. (The density of foil is 2.7 g/cm3 .)
Kruka [31]

Answer:

The length of foil will be 8107.81 cm or 81.7081 m.

Explanation:

Given data:

Width of roll of foil = 302 mm

Height or thickness = 0.018 mm

Density of foil = 2.7 g/cm³

Mass of foil = 1.19 Kg

Length of foil = ?

Solution:

d = m/ v

v = length (l) × width (w) × height (h)

First of we will convert the Kg into gram and mm into cm.

one Kg = 1000 g

1.19 × 1000 = 1190 g

one cm = 10 mm

302 / 10 = 30.2 cm

0.018 / 10 =  0.0018 cm

Now we will put the values in formula:

d = m/ l× h× w

l = m / d × h× w

l = 1190 g / 2.7 g/cm³× 30.2 cm × 0.0018 cm

l = 1190 g/ 0.146772 g/cm

l = 8107.81 cm or 81.7081 m

3 0
3 years ago
What is the percent composition by mass of sulfur in the compound MgSO4 (gram formula mass = 120 g/mol)?
slavikrds [6]

26.7% is the percent composition by mass of sulfur in a compound named magnesium sulfate. Explanation: Molar mass of compound = 120 g/mol.

7 0
2 years ago
The monovalent salt concentration (the predominant solute in the blood cell) for a sample of red blood cells is 0.13 moles/liter
11111nata11111 [884]

Answer:

The osmotic pressure of cell is 648.3 KPa

Explanation:

As we know the osmotic pressure is equal to

\pi = icRT

Where

i is the Van Hoff factor

c is the concentration of solution

R is the ideal gas constant

and T is the temperature.

Substituting the given values, we get -

\pi = 2 * 0.13 * 0.08206 * 300\\

\pi = 648.3 KPa

4 0
3 years ago
Two objects are brought into contact Object 1 has mass 0.76 kg, specific heat capacity 0.87) g'c and initial temperature 52.2 'C
taurus [48]

Answer:

T_F=77.4\°C

Explanation:

Hello there!

In this case, according to the given information, it turns out possible to set up the following energy equation for both objects 1 and 2:

Q_1=-Q_2

In terms of mass, specific heat and temperature change is:

m_1C_1(T_F-T_1)=-m_2C_2(T_F-T_2)

Now, solve for the final temperature, as follows:

T_F=\frac{m_1C_1T_1+m_2C_2T_2}{m_1C_1+m_2C_2}

Then, plug in the masses, specific heat and temperatures to obtain:

T_F=\frac{760g*0.87\frac{J}{g\°C} *52.2\°C+70.7g*3.071\frac{J}{g\°C}*154\°C}{760g*0.87\frac{J}{g\°C} +70.7g*3.071\frac{J}{g\°C}} \\\\T_F=77.4\°C

Yet, the values do not seem to have been given correctly in the problem, so it'll be convenient for you to recheck them.

Regards!

4 0
3 years ago
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