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3 years ago

A scientist fills a large, tightly sealed balloon with 75,000 mL of helium at STP.

Chemistry

1 answer:

pav-90 [236]3 years ago

3 0

Considering the definition of STP conditions, 3.35 moles of helium are contained within the balloon.

<h3>Definition of STP condition</h3>

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

<h3>Amount of moles of helium within the balloon</h3>

In this case, you know that scientist fills a large, tightly sealed balloon with 75,000 mL of helium at STP.

So, you can apply the following rule of three: if by definition of STP conditions 22.4 liters are occupied by 1 mole of helium, 75 L (75 L= 75000 mL, being 1 L= 1000 L) are occupied by how many moles of helium?

$amount of moles of helium=\frac{75 Lx1 mole}{22.4 L}$

<u><em>amount of moles of helium= 3.35 moles</em></u>

Finally, 3.35 moles of helium are contained within the balloon.

Learn more about STP conditions:

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Calculate the total energy, in kilojoules, that is needed to turn a 46 g block

gogolik [260]

<u>Answer:</u> The amount of heat absorbed is 141.004 kJ.

<u>Explanation:</u>

In order to calculate the amount of heat released while converting given amount of steam (gaseous state) to ice (solid state), few processes are involved:

(1): $H_2O (s) (-25^oC, 248K) \rightleftharpoons H_2O(s) (0^oC,273K)$

(2): $H_2O (s) (0^oC, 273K) \rightleftharpoons H_2O(l) (0^oC,273K)$

(3): $H_2O (l) (0^oC, 273K) \rightleftharpoons H_2O(l) (100^oC,373K)$

(4): $H_2O (l) (100^oC, 373K) \rightleftharpoons H_2O(g) (100^oC,373K)$

Calculating the heat absorbed for the process having the same temperature:

$q=m\times \Delta H_{(f , v)}$ ......(i)

where,

q is the amount of heat absorbed, m is the mass of sample and $\Delta H_{(f , v)}$ is the enthalpy of fusion or vaporization

Calculating the heat released for the process having different temperature:

$q=m\times C_{s,l}\times (T_2-T_1)$ ......(ii)

where,

$C_{s,l}$ = specific heat of solid or liquid

$T_2\text{ and }T_1$ are final and initial temperatures respectively

<u>For process 1:</u>

We are given:

$m=46g\\C=2.108J/g^oC\\T_2=0^oC\\T_1=-25^oC$

Putting values in equation (i), we get:

$q_1=46g\times 2.108J/g^oC\times (0-(-25))\\\\q_1=2424.2J$

<u>For process 2:</u>

We are given:

$m=46g\\\Delta H_{fusion}=334J/g$

Putting values in equation (i), we get:

$q_2=46g\times 334J/g\\\\q_2=15364J$

<u>For process 3:</u>

We are given:

$m=46g\\C=4.186J/g^oC\\T_2=100^oC\\T_1=0^oC$

Putting values in equation (i), we get:

$q_3=46g\times 4.186J/g^oC\times (100-0)\\\\q_3=19255.6J$

<u>For process 4:</u>

We are given:

$m=46g\\\Delta H_{vap}=2260J/g$

Putting values in equation (i), we get:

$q_4=46g\times 2260J/g\\\\q_4=103960J$

Calculating the total amount of heat released:

$Q=q_1+q_2+q_3+q_4$

$Q=[(2424.2)+(15364)+(19255.6)+(103960)]$

$Q=141003.8J=141.004kJ$ (Conversion factor: 1 kJ = 1000J)

Hence, the amount of heat absorbed is 141.004 kJ.

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1) azimuthal quantum number (l) can be l = 0...n-1:

l = 0, 1.

The azimuthal quantum number determines its orbital angular momentum and describes the shape of the orbital.

2) magnetic quantum number (ml) can be ml = -l...+l.

ml = -1, 0, +1.

Magnetic quantum number specify orientation of electrons in magnetic field and number of electron states (orbitals) in subshells.

3) the spin quantum number (ms), is the spin of the electron.

ms = +1/2, -1/2.

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