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EastWind
1 month ago
10

If 75.0% of the isotopes of an element have a mass of 35.0 amu and 25.0% of the isotopes have a mass of 37.0amu what is the atom

ic mass of the element
A) 35.0amu
B)35.5amu
C)36.0amu
D)37.0amu
Chemistry
2 answers:
castortr0y [3K]1 month ago
8 0
 <span>Calculating the average atomic mass involves a method similar to that of a weighted average. Follow this equation:

(mass1)(percentage1) + (mass2)(percentage2) = average atomic mass

(35.0)(0.75) + (37.0)(0.25) = average atomic mass

Ensure the percentages are represented in decimal format for accurate results.

While another response may be accurate, it lacks some clarity. Two potential answers can be dismissed immediately: 35.0 amu and 37.0 amu cannot represent the average. If the isotopes were mixed equally at 50%, then 36.0 amu would be the correct response; however, we have a 75/25 mix. Thus, only one answer remains viable.</span>
Alekssandra [3K]1 month ago
6 0

Answer: It's B if you find the lengthy explanation unclear.

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Explain why CaCl2 is likely to have properties similar to those of CaBr2
Alekssandra [3086]

Answer:

Both CaCl2 and CaBr2 consist of elements (bromine and chlorine) from the same group (group 7).

Explanation:

In the periodic table, elements are arranged into groups based on their valence electron count in the outermost shell. Elements in the same group, which possess a similar number of valence electrons, typically exhibit similar chemical behaviors.

Chlorine and Bromine in CaCl2 and CaBr2 belong to group 7, known as HALOGENS, characterized by having 7 valence electrons in their outer shell.

The similarity in properties between CaCl2 and CaBr2 arises because both contain Chlorine and Bromine, leading to analogous reactions and behaviors when interacting with other compounds.

5 0
2 months ago
A sample of solid naphthalene is introduced into an evacuated flask. Use the data below to calculate the equilibrium vapor press
Tems11 [2777]

Answer: The vapor pressure of naphthalene within the flask remains at 2.906\times 10^{-4} atm.

Explanation:

The transformation from solid naphthalene to its gaseous form follows the equilibrium reaction:

C_{10}H_8(s)\rightleftharpoons C_{10}H_8(g)

  • The formula employed to determine the enthalpy change for the reaction is:  

\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]

The formula for calculating the enthalpy change regarding the aforementioned reaction is:

\Delta H^o_{rxn}=(1\times \Delta H^o_f_{(C_{10}H_8(g))})-(1\times \Delta H^o_f_{(C_{10}H_8(s))})

The provided information includes:

\Delta H^o_f_{(C_{10}H_8(s))}=78.5kJ/mol\\\Delta H^o_f_{(C_{10}H_8(g))}=150.6kJ/mol

Substituting the values into the previous equation produces:

\Delta H^o_{rxn}=(1\times 150.6)-(1\times 78.5)=72.1kJ/mol

  • The formula utilized to compute Gibbs free energy change is of a reaction:

\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f(product)]-\sum [n\times \Delta G^o_f(reactant)]

The equation for the enthalpy change for the reaction is:

\Delta G^o_{rxn}=(1\times \Delta G^o_f_{(C_{10}H_8(g))})-(1\times \Delta G^o_f_{(C_{10}H_8(s))})

The given factors include:

\Delta G^o_f_{(C_{10}H_8(s))}=201.6kJ/mol\\\Delta G^o_f_{(C_{10}H_8(g))}=224.1kJ/mol

By inserting values from the above equation, we arrive at:

\Delta G^o_{rxn}=(1\times 224.1)-(1\times 201.6)=22.5kJ/mol

  • For the calculation of K_1 (at 25°C) regarding the provided value of Gibbs free energy, the following relationship is applied:

\Delta G^o=-RT\ln K_1

where,

\Delta G^o = Gibbs free energy = 22.5 kJ/mol = 22500 J/mol  (Conversion factor: 1kJ = 1000J)

R = Gas constant = 8.314J/K mol

T = temperature = 25^oC=[273+25]K=298K

K_1 = equilibrium constant at 25°C =?

Inserting values into the above equation yields:

22500J/mol=-(8.314J/Kmol)\times 298K\times \ln K_1\\\\K_1=1.14\times 10^{-4}

  • To determine the equilibrium constant at 35°C, we refer to the equation proposed by Arrhenius, which states:

\ln(\frac{K_2}{K_1})=\frac{\Delta H}{T}(\frac{1}{T_1}-\frac{1}{T_2})

where,

K_2 = Equilibrium constant at 35°C =?

K_1 = Equilibrium constant at 25°C = 1.14\times 10^{-4}

\Delta H = Enthalpy change of the reaction = 72.1 kJ/mol = 72100 J

R = Gas constant = 8.314J/K mol

T_1 = Initial temperature = 25^oC=[273+25]K=298K

T_2 = Final temperature = 35^oC=[273+35]K=308K

By plugging values into the equation above, we obtain:

\ln(\frac{K_2}{1.14\times 10^{-4}})=\frac{72100J/mol}{8.314J/K.mol}(\frac{1}{298}-\frac{1}{308})\\\\K_2=2.906\times 10^{-4}

  • In order to calculate the partial pressure of naphthalene at 35°C, we utilize the equation for K_p, which is:

K_p=\frac{p_{C_{10}H_8(g)}}{p_{C_{10}H_8(g)}}=p_{C_{10}H_8(g)

The partial pressure of the solid phase is considered to be 1 at equilibrium.

Therefore, the value for K_2 will equal K_p

p_{C_{10}H_8}=2.906\times 10^{-4}

Consequently, the partial pressure of naphthalene at 35°C is 2.906\times 10^{-4} atm.

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22 days ago
How is the periodic law demonstrated in halogens
Anarel [2989]
Halogens are characterized as a group of 7 on the periodic table. Each of these elements has 7 valence electrons, needing just one additional electron to fill their outer shell, resulting in their high reactivity. Moreover, the periodic law indicates that when the elements are arranged by increasing atomic number, similar patterns in their properties emerge. Thus, halogens exhibit comparable behaviors during chemical reactions with one another, which is indeed accurate.
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1 month ago
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Select the option that correctly completes this statement: If the reaction takes place at a constant volume in a thermally insul
KiRa [2933]

Answer:

The temperature of the gas rises.

Explanation:

This is classified as an ISOCHORIC process where the volume remains unchanged. There is no work done by the system.

The gas only receives internal energy from the heat transferred to it from the surroundings.

In this situation, the pressure also increases.

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