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3 years ago

HELP

Chemistry

1 answer:

Vesnalui [34]3 years ago

3 0

Answer:

145.8g

Explanation:

Given parameters:

Number of moles of magnesium hydroxide = 2.5mol

Unknown:

Mass of Mg(OH)₂ = ?

Solution:

To solve this problem we use the expression below;

Mass of Mg(OH)₂ = number of moles x molar mass

Molar mass of Mg(OH)₂ = 24.3 + 2(16 + 1) = 58.3g/mol

Mass of Mg(OH)₂ = 2.5 x 58.3 = 145.8g

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The coefficient of thermal expansion α = (1/V)(∂V/∂T)p. Using the equation of state, compute the value of α for an ideal gas. Th

andreyandreev [35.5K]

Answer:

The coefficient of thermal expansion α is

$\alpha = \frac{1}{T}$

The coefficient of compressibility

$\beta = \frac{1}{P}$

Now considering $(\frac{ \delta P }{\delta T} )V$

From equation (1) we have that

$\frac{ \delta P}{\delta T} = \frac{n R }{V}$

From ideal equation

$nR = \frac{PV}{T}$

$\frac{\delta P}{\delta T} = \frac{PV}{TV}$

=> $\frac{\delta P}{\delta T} = \frac{P}{T}$

=> $\frac{\delta P}{\delta T} = \frac{\alpha }{\beta}$

Explanation:

From the question we are told that

The coefficient of thermal expansion is $\alpha = \frac{1}{V} * (\frac{\delta V}{ \delta P}) P$

The coefficient of compressibility is $\beta = - (\frac{1}{V} ) * (\frac{\delta V}{ \delta P} ) T$

Generally the ideal gas is mathematically represented as

$PV = nRT$

=> $V = \frac{nRT}{P} --- (1)$

differentiating both side with respect to T at constant P

$\frac{\delta V}{\delta T } = \frac{ n R }{P}$

substituting the equation above into $\alpha$

$\alpha = \frac{1}{V} * ( \frac{ n R }{P}) P$

$\alpha = \frac{nR}{PV}$

Recall from ideal gas equation $T = \frac{PV}{nR}$

$\alpha = \frac{1}{T}$

Now differentiate equation (1) above with respect to P at constant T

$\frac{\delta V}{ \delta P} = -\frac{nRT}{P^2}$

substituting the above equation into equation of $\beta$

$\beta = - (\frac{1}{V} ) * (-\frac{nRT}{P^2} ) T$

$\beta =\frac{ (\frac{n RT}{PV} )}{P}$

Recall from ideal gas equation that

$\frac{PV}{nRT} = 1$

$\beta = \frac{1}{P}$

Now considering $(\frac{ \delta P }{\delta T} )V$

From equation (1) we have that

$\frac{ \delta P}{\delta T} = \frac{n R }{V}$

From ideal equation

$nR = \frac{PV}{T}$

$\frac{\delta P}{\delta T} = \frac{PV}{TV}$

=> $\frac{\delta P}{\delta T} = \frac{P}{T}$

=> $\frac{\delta P}{\delta T} = \frac{\alpha }{\beta}$

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Answer:

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Explanation:

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