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jekas [21]
3 years ago
12

What is the barometric pressure exerted in pascals (Pa) of a column of decane (density = 0.7300 g/cm3) 636.2 mm in height?

Chemistry
1 answer:
BartSMP [9]3 years ago
5 0
<h3>Answer:</h3>

4551.37 Pascals

<h3>Explanation:</h3>

The pressure refers to the force exerted by a substance per unit area.

The pressure is liquid is calculated by;

P = height × density × gravitational acceleration

In this case;

Height = 636.2 mm

Density = 0.7300 g/cm³

g = 9.8 N/kg

We need to convert mm to m and g/cm³ to kg/m³

Therefore;

Height = 636.2 mm ÷1000

           = 0.6362 m

Density = 0.73 g/cm³ × 1000 kg/m³

             = 730 kg/m³

Then, we can calculate the pressure;

Pressure = 0.6362 m × 730 kg/m³ × 9.8 N/kg

              = 4551.3748 pascals

               = 4551.37 Pascals

Therefore, the pressure of the column of decane is 4551.37 Pascals

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Assuming complete distribution, what is the molarity of 10 milligrams of lisinopril in 100 liters?
SVEN [57.7K]

Answer:

2.47x10^{-7}   M

Explanation:

Lisoprisil's molecular mass is 405.488g/mol, we'll use this fact to calculate molarity, which units are mol/L, and we proceed to the calculus:

  • First, we'll unify unities, the 10 milligrams of lisinopril we'll transform into grams.

10mg*\frac{1g}{1000mg}=0.01g

  • Now that we have the same unities we'll calculate molarity using the molecular mass, the grams of lisinopril and the liters in which these grams are, let's consider that our final unities have to be mol/L.

\frac{1mol}{405.488g}*\frac{0.01g}{100L}=2.47x10^{-7}   M

I hope you find this information useful and interesting! Good luck!

3 0
3 years ago
For a particular reaction, ΔH∘=20.1 kJ/mol and Δ????∘=45.9 J/(mol⋅K). Assuming these values change very little with temperature,
tia_tia [17]

Answer:

The temperature should be higher than 437.9 Kelvin (or 164.75 °C) to be spontaneous

Explanation:

<u>Step 1:</u> Data given

ΔH∘=20.1 kJ/mol

ΔS is 45.9 J/K

<u>Step 2:</u> When is the reaction spontaneous

Consider temperature and pressure = constant.

The conditions for spontaneous reactions are:

ΔH <0

ΔS  > 0

ΔG <0  The reaction is spontaneous at all temperatures

ΔH <0

ΔS  <0

ΔG <0 The reaction is spontaneous at low temperatures ( ΔH - T*ΔS <0)

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ΔS  >0

ΔG <0 The reaction is spontaneous at high temperatures ( ΔH - T*ΔS <0)

<u>Step 3:</u> Calculate the temperature

ΔG <0 = ΔH - T*ΔS

T*ΔS > ΔH

T > ΔH/ΔS

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T > (20100 J)/(45.9 J/K)

T > 437.9 K

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The temperature should be higher than 437.9 Kelvin (or 164.75 °C) to be spontaneous

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