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3 years ago

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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Answer:

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Explanation:

Given data:

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Q = m.c. ΔT

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Answer : The [H] is increasing at the rate of 0.36 mol/L.s

Explanation :

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$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

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$2D(g)+3E(g)+F(g)\rightarrow 2G(g)+H(g)$

The expression for rate of reaction :

$\text{Rate of disappearance of }D=-\frac{1}{2}\frac{d[D]}{dt}$

$\text{Rate of disappearance of }E=-\frac{1}{3}\frac{d[E]}{dt}$

$\text{Rate of disappearance of }F=-\frac{d[F]}{dt}$

$\text{Rate of formation of }G=+\frac{1}{2}\frac{d[G]}{dt}$

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As,

$-\frac{1}{2}\frac{d[D]}{dt}=+\frac{d[H]}{dt}=0.18mol/L.s$

and,

$+\frac{d[H]}{dt}=2\times 0.18mol/L.s$

$+\frac{d[H]}{dt}=0.36mol/L.s$

Thus, the [H] is increasing at the rate of 0.36 mol/L.s

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