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4 years ago

Using the information from STP or SATP conditions determine the value of the ideal gas constant.

Chemistry

1 answer:

Dovator [93]4 years ago

4 0

Answer:

0.0821 atm.L/Kmol

Explanation:

At stp, the values temperature, pressure and volume is given below:

Pressure (P) = 1 atm

Temperature (T) = 273 K

Volume (V) = 22.4 L

At stp, 1 mole of a gas occupy 22.4L.

Number of mole (n) = 1 mole

Gas constant (R) =?

The ideal gas equation is given below:

PV = nRT.

With the above equation, the gas constant R can be obtained as follow:

1 atm x 22.4L = 1 mole x R x 273K

Divide both side by (1 mole x 273 K)

R = (1 atm x 22.4L) / (1 mole x 273 K)

R = 0.0821 atm.L/Kmol

Therefore, the gas constant is 0.0821 atm.L/Kmol

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The half-life of carbon 14 is 5,730 years. How much would be left of an original 50-gram sample after 2,292 years

Delvig [45]

Answer:

37.9g of carbon 14 remains

Explanation:

The radioactive decay of an atom follows the equation:

Ln[A] = -kt + Ln[A]₀

<em>Where [A] is amount of isotope after time t, k is decay constant and [A]₀ initial amount of the isotope.</em>

<em />

We can find k from half-life using:

k = ln 2 / Half-Life

k = ln 2 / 5730 years

k = 1.2097x10⁻⁴ yrs⁻¹

Replacing in the equation all values:

Ln[A] = -1.2097x10⁻⁴ yrs⁻¹*2292yrs + Ln[50g]

Ln[A] = 3.635

[A] = 37.9g of carbon 14 remains

<em />

8 0

3 years ago

PLEASE HELP!! Thanks! How much heat (in kJ) is required to warm 13.0 g of ice, initially at -10.0 ∘C, to steam at 111.0 ∘C? The

ZanzabumX [31]

Answer:

Approximately 39.7 kJ.

Assumptions: the specific heat capacity of water is $\rm 4.182\; J \cdot mol^{-1}$ , the melting point of water is $\rm 0\, ^{\circ} C$ , and that the boiling point of water is $\rm 100 \,^{\circ} C$ .

Explanation:

It takes five steps to convert 13.0 grams of $\rm \text{-}10.0\, ^{\circ}C$ ice to steam at $\rm 111.0\,^{\circ}C$ .

Step one: heat the 13.0 gram of ice from $\rm \text{-}10.0\, ^{\circ}C$ to $\rm 0\,^{\circ}C$ . The change in temperature would be $\rm 10.0\,^{\circ}C$ .
Step two: supply the heat of fusion to convert that 13.0 gram of ice to water.
Step three: heat the 13.0 gram of water from $\rm 0\,^{\circ}C$ to $\rm 100\,^{\circ}C$ . The change in temperature would be $\rm 100\,^{\circ}C$ .
Step four: supply the heat of vaporization to convert that 13.0 gram of water to steam.
Step five: heat the 13.0 gram of steam from $\rm 100\,^{\circ}C$ to $\rm 111.0\,^{\circ}C$ . The change in temperature would be $\rm 11.0\,^{\circ}C$ .

<h3>Energy required for step one, three, and five</h3>

The following equation gives the amount of energy $Q$ required to raise the temperature of an object by a $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ .

In this equation,

$c$ is the specific heat of this substance,
$m$ is the mass of the substance, and
$\Delta T$ is the change in the temperature of the object.

Assume that there's no mass loss in this whole process. The value of $m$ would stay the same at $13.0\; \rm g$ .

$\begin{aligned}& &&\text{Energy required for raising temperature} \cr &=&& c(\text{Ice}) \cdot m \cdot \Delta(\text{Ice}) \cr & && + c(\text{Water}) \cdot m \cdot \Delta(\text{Water})\cr & && + c(\text{Steam}) \cdot m \cdot \Delta(\text{Steam}) \cr & = && (2.09 \times 13.0 \times 10) \cr & && + (4.182 \times 13.0 \times 100) \cr & &&+ ( 2.01 \times 13.0 \times 10) \cr & = && 5969.6\;\rm J \cr & = && 5.969\; \rm kJ\end{aligned}$ .

<h3>Energy required for step two and four</h3>

The equations for the energy of fusion and energy of vaporization are quite similar:

$E(\text{Fusion}) = n \cdot \Delta H_\text{Fusion}$ .

$E(\text{Vaporization}) = n \cdot \Delta H_\text{Vaporization}$ .

where $n$ is the number of moles of the substance.

Look up the relative atomic mass of oxygen and hydrogen from a modern periodic table:

H: 1.008,
O: 15.999.

Hence the molar mass of water:

$M(\rm H_2O) = 2\times 1.008 + 15.999 = 18.015\; g \cdot mol^{-1}$ .

Number of moles of $\rm H_2O$ molecules in $\rm 13.0\; g$ :

$\displaystyle n = \frac{m}{M} \approx 0.721621\; \rm mol$ .

$\begin{aligned}& &&\text{Energy required for phase changes} \cr &=&& n \cdot \Delta H_\text{Fusion} \cr & &&+n \cdot \Delta H_\text{Vaporization} \cr & = &&0.721621 \times 6.02 + 0.721621 \times 40.7 \cr & = &&33.7\; \rm kJ \end{aligned}$

<h3>Energy required for all five steps, combined</h3>

$5.969\; \rm kJ + 33.7\; \rm kJ \approx 39.7\; \rm kJ$ .

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3 years ago

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3 years ago

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Chemistry and physics are so closely related that sometimes the fields overlap. Which of the following experiments might both ch

madreJ [45]

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3 years ago

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The enthalpy of reaction changes somewhat with temperature. Suppose we wish to calculate ΔH for a reaction at a temperature T th

Contact [7]

Answer:

-99.8 kJ

Explanation:

We are given the methodology to answer this question, which is basically Kirchhoff law . We just need to find the heats of formation for the reactants and products and perform the calculations.

The standard heat of reaction is

ΔrHº = ∑ ν x ΔfHº products - ∑ ν x ΔfHº reactants

where ν are the stoichiometric coefficients in the balanced equation, and ΔfHº are the heats of formation at their standard states.

Compound ΔfHº (kJmol⁻¹)

SO₂ -296.8

O₂ 0

SO₃ -395.8

The balanced chemical equation is

SO₂(g) + ½O₂(g) → SO₃(g)

Thus

Δr, 298K Hº( kJmol⁻¹ ) = 1 x (-395.8) - 1 x (-296.8) = -99.0 kJmol⁻¹

Now the heat capacity of reaction will be be given in a similar fashion:

Cp rxn = ∑ ν x Cp of products - ∑ ν x Cp of reactants

where ν is as above the stoichiometric coefficient in the balanced chemical equation.

Cprxn ( JK⁻¹mol⁻¹) = 50.7 - ( 39.9 + 1/2 x 29.4 ) = - 3.90

= -3.90 JK⁻¹mol⁻¹

Finally Δr,500 K Hº = Δr, 298K Hº + CprxnΔT

Δr,500 K Hº = - 99 x 10³ J + (-3.90) JK⁻¹ ( 500 - 298 ) K = -99,787.8

= -99,787.8 J x 1 kJ/1000 J = -99.8 kJ

Notice thie difference is relatively small that is why in some problems it is o.k to assume the change in enthalpy is constant over a temperature range, especially if it is a small range of temperatures.

3 0

3 years ago