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oksian1 [2.3K]
2 years ago
11

Element X has two isotopes. If 72.0% of the element has an isotope mass of 84.9 atomic mass units, and 28.0% of the element has

an isotopic mass of 87.0 atomic mass units, the average atomic mass of element X is numerically equal to
Chemistry
1 answer:
bija089 [108]2 years ago
8 0
<h2>Answer:</h2>

Average atomic mass of an element is the sum of the masses of its isotopes each multiplied by its natural abundance

\footnotesize \longrightarrow \:  \rm Average \:  atomic  \: mass =  \dfrac{ \sum\limits \: \% age \: of \: each \: isotope \times Atomic  \: mass }{100} \\

\footnotesize \longrightarrow \:  \rm Average \:  atomic  \: mass =  \dfrac{ 72 \times84.9 + 28 \times 87  }{100} \\

\footnotesize \longrightarrow \:  \rm Average \:  atomic  \: mass =  \dfrac{ 6112.8 + 2436  }{100} \\

\footnotesize \longrightarrow \:  \rm Average \:  atomic  \: mass =  \dfrac{ 8548.8  }{100} \\

\footnotesize \longrightarrow \:  \bf Average \:  atomic  \: mass =  85.488 \: amu  \\

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Be sure to answer all parts. The equilibrium constant (Kp) for the reaction below is 4.40 at 2000. K. H2(g) + CO2(g) ⇌ H2O(g) +
nikklg [1K]

<u>Answer:</u>

<u>For 1:</u> The value of \Delta G for the chemical equation is -24.636 kJ/mol

<u>For 2:</u> The value of \Delta G for the chemical equation is -20.925 kJ/mol

<u>Explanation:</u>

For the given chemical equation:

H_2(g)+CO_2(g)\rightleftharpoons H_2O(g)+CO(g)

  • <u>For 1:</u>

To calculate the \Delta G for given value of equilibrium constant, we use the relation:

\Delta G=-RT\ln K_p      .....(1)

where,

\Delta G = ? kJ/mol

R = Gas constant = 8.314J/K mol

T = temperature = 2000 K

K_p = equilibrium constant in terms of partial pressure = 4.40

Putting values in above equation, we get:

\Delta G=-(8.314J/Kmol)\times 2000K\times \ln (4.40)\\\\\Delta G=-24636.12J/mol

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, -24636.12 J/mol = -24.636 kJ/mol

Hence, the value of \Delta G for the chemical equation is -24.636 kJ/mol

  • <u>For 2:</u>

The expression of K_p for the given chemical equation is:

K_p=\frac{p_{CO}p_{H_2O}}{p_{H_2}p_{CO_2}}

We are given:

p_{CO}=1.18atm\\p_{H_2O}=0.66atm\\p_{CO_2}=0.82atm\\p_{H_2}=0.27atm

Putting values in above equation, we get:

K_p=\frac{1.18\times 0.66}{0.27\times 0.82}\\\\K_p=3.52

Now, calculating the value of \Delta G by using equation 1:

R = Gas constant = 8.314J/K mol

T = temperature = 2000 K

K_p = equilibrium constant in terms of partial pressure = 3.52

Putting values in equation 1, we get:

\Delta G=-(8.314J/Kmol)\times 2000K\times \ln (3.52)\\\\\Delta G=-20925.68J/mol

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, -20925.68 J/mol = -20.925 kJ/mol

Hence, the value of \Delta G for the chemical equation is -20.925 kJ/mol

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What is the nuclear charge of a berrylium atom?
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Suppose now that you wanted to determine the density of a small crystal to confirm that it is phosphorus. From the literature, y
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Answer:

Volume of CHCl_3   = 15.31 mL

Volume of CHBr_3 = 4.69 mL

Explanation:

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the density of the mixture = 1.82 g/mL

From the density of the pure samples

The density of CHCl_3 = 1.492 g/mL

The density of CHBr_3 = 2.890 g/mL

The total volume of the liquid mixture = 20.0 mL

Suppose the volume of  CHCl_3 = P ml

and the volume of CHBr_3 = Q ml

the sum of their volumes should be equal to the total volume of the mixture

P \ ml + Q \ ml = 20 ml  ----- (1)

However, we know that Density = mass/volume

∴ mass = density × volume

The equation can now be expressed as:

\mathtt{(Density \ of  \ CHCl_3 \times Vol. \ of  \ CHCl_3 ) + (Density  \  of \  CHBr_3  \times \ volume \ of \ CHBr_3)} = \mathtt{ (Density  \ of \ mixture \times volume \ of \ the \ mixture)}

1.492 g/mL × P mL + 2.890 g/mL × Q mL = 1.82 g/mL × 20 mL  ---- (2)

From equation (1) ;

let Q = 20 - P

The replace the value of P into equation (2)

1.492 g/mL × P mL + 2.890g/mL × (20 - P) mL = 1.82 g/mL × 20 mL

1.492 P g + 57.8g - 2.890 P g =  36.4g

1.492 P g - 2.890 P g = 36.4g - 57.8g

-1.398 P g = -21.4g

P = -21.4g/-1.398g

P = 15.31 mL

Q = 20 - P

Q = (20 - 15.31) mL

Q = 4.69 mL

∴

Volume of CHCl_3   = 15.31 mL

Volume of CHBr_3 = 4.69 mL

6 0
3 years ago
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