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2 years ago

If you start with 64g of a radioactive element how many half-lives would occur before 8g remain?

Chemistry

1 answer:

Nonamiya [84]2 years ago

3 0

Answer:

3 half-lives

Explanation:

The half-life is the time that it takes to a radioactive element to decay to half of its initial amount.

Let's suppose we start with 64 g of the radioactive element.

After 1 half-life, the mass of the element will be 32 g.

After 2 half-lives, the mass of the element will be 16 g.

After 3 half-lives, the mass of the element will be 8 g.

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A gas stream containing n-hexane in nitrogen with a relative saturation of 0.58 (as a fraction, multiply by 100% if you prefer %

kondor19780726 [428]

This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.

In this case, it is recommended to write the enthalpy for each substance as follows:

$H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)$

Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and $y_{C-6}$ and $y_{N_2}$ are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

$H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol$

It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:

$Q=-40036+(-672.21)=-40708.21J$

Finally we convert this result to kJ:

$Q=-40708.21J*\frac{1kJ}{1000J}\\\\Q=-40.7kJ$

Learn more:

6 0

1 year ago

Write the formula for potassium nitride? <br> (show work)<br> CAN SOMEONE PLEASE HELP?

ohaa [14]

Answer:

It should b KNO3

Explanation:

one Potassium (K) and three Nitrite (NO3)

8 0

2 years ago

The density of aluminum is 2.70 g/cm'. If the volume of a piece of aluminum is 90.4 cm', what is

weeeeeb [17]

Answer:

33.4814814815

Explanation:

90.4/2.70

6 0

2 years ago

Some versions of the periodic table show hydrogen at the top of Group 1A(1) and at the top of Group 7A(17). What properties of h

lawyer [7]

Answer:Hydrogen is placed such because it exhibits some similar characteristics of both group1 and group VII elements.

Explanation:

The reason why hydrogen is similar to group 1 metals:

#It has same valence electron and inorder achieve octet state it can lose that electron and forms H+ ion

#It acts as a good reducing agent similar to group1 metals

#It can also halides

Similarity to halogens:

#hydrogen can also gain one electron to gain noble gas configuration. It can combine with other non metals to form molecules with covalent bonding.

#It exists as diatomin molecule,H2

#Have the same electronegativity nature

#its reaction with other metal

8 0

2 years ago

The expected o-c-h angle in this molecule is degrees. the expected hybridization at the central carbon is

WINSTONCH [101]

Each neutral carbon atom contains four valence electrons and may form up to four electron domains. Possible hybridizations include

$sp^{3}$ , four electron domains, as in ethane $\text{C}_2\text{H}_6$
$sp^{2}$ , three electron domains, as in ethene $\text{C}_2\text{H}_4$
$sp$ , two electron domains, as in ethyne $\text{C}_2\text{H}_2$

Molecules of each of the three hybridization demonstrate spatial configurations that would maximizes the separation between the electron domains.

Carbon atoms with a $sp^{3}$ hybridization would demonstrate a tetrahedral configuration with a bond angle of approximately $109.5\textdegree{}$
Carbon atoms with a $sp^{2}$ hybridization would demonstrate a triangular planar configuration with a bond angle of $120\textdegree{}$
Carbon atoms with a $sp$ hybridization would demonstrate a linear configuration with a bond angle of $180\textdegree{}$

Bond angles are characteristic of the spatial configuration of electron domains and identifies the hybridization of the central carbon atom.

Note that each hydrogen atom contains only one valence electron and would form only single bonds. It takes two valence electrons for oxygen atoms to achieve an octet such that each oxygen form only two bonds at a single time. Therefore given the fact that the carbon is bonded to both hydrogen and oxygen, only the following hybridizations are possible

$sp^{3}$ in which the oxygen atom forms a carbon-oxygen double bond with the central carbon atom;
$sp^{2}$ in which the oxygen atom forms a single bond with the central carbon atom and with a second atom.

3 0

2 years ago