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2 years ago

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What is the volume in liters of 5.25 moles of He gas?

1 answer:

serious [3.7K]2 years ago

5 0

Answer:

$\boxed {\boxed {\sf 117.6 \ L \ He}}$

Explanation:

Regardless of the type of gas, 1 mole at standard temperature and pressure (STP) occupies a volume of 22.4 liters. In this case the gas is helium (He).

We can set up a ratio.

$\frac { 22.4 \ L \ He}{ 1 \ mol \ He}$

Multiply by the given number of moles.

$5.25 \ mol \ He *\frac { 22.4 \ L \ He}{ 1 \ mol \ He}$

The moles of helium will cancel.

$5.25 *\frac { 22.4 \ L \ He}{ 1 }$

$5.25 * { 22.4 \ L \ He}$

Multiply.

$117.6 \ L \ He$

5.25 moles of helium gas at STP is 117.6 liters of helium.

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Explanation:

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179.1 g of water is in a Styrofoam calorimeter of negligible heat capacity. The initial T of the water is 16.1oC. After 306.9 g

taurus [48]

Answer:

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Explanation:

Generally the change in temperature of water is evaluated as

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$\Delta T = 27.4 - 16.1$

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$\Delta T_u = T_3 -T_2$

Substituting 27.4°C for $T_2$ and 94.3°C for $T_3$

$\Delta T = 94.3 - 27.4$

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Since there is an increase in temperature then heat is gained by water and this can be evaluated as

$H_w = mc_w \Delta T$

Substituting 179.1 g for m , 4.18 J/g.C for $c_w$ (specific heat of water)

$H_w = 4.18 * 179.1 * 11.3$

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Since there is a decrease in temperature then heat is lost by unknown compound and this can be evaluated as

$H_u = m_uc_u \Delta T_u$

By conservation of energy law

Heat lost = Heat gained

Substituting 306.9 g for $m_u$ , 8459.6J for $H_u$

$8459.6 = 306.9 * c_u * 66.9$

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Answer:

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