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3 years ago

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Question 25———-A, YA are two isotopes of element A.

1 answer:

lakkis [162]3 years ago

6 0

Answer:

Option C. 1

Explanation:

Step 1:

Determination of the Neutron of both isotopes. This is illustrated below.

For isotope y xA:

Mass number = y

Atomic number = x

Neutron =..?

Atomic number = proton number = x

Mass number = Proton + Neutron

y = x + Neutron

Rearrange

Neutron = y – x

For isotope (y + 1) xA:

Mass number = y + 1

Atomic number = x

Neutron =.?

Atomic number = proton number = x

Mass number = Proton + Neutron

y + 1 = x + Neutron

Rearrange

Neutron = y + 1 – x

Step 2:

Determination of the difference between the neutron number of both isotopes. This is illustrated below:

For isotope y xA:

Neutron number = y – x

For isotope (y + 1) xA:

Neutron number = y + 1 – x

Difference in neutron number

=> (y + 1 – x) – (y – x)

=> y + 1 – x – y + x

Rearrange

=> y – y + 1 – x + x

=> 1

Therefore, the difference in the neutron number of both isotopes is 1

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The standard emf for the cell using the overall cell reaction below is +2.20 V:

ANEK [815]

Answer:

emf generated by cell is 2.32 V

Explanation:

Oxidation: $2Al-6e^{-}\rightarrow 2Al^{3+}$

Reduction: $3I_{2}+6e^{-}\rightarrow 6I^{-}$

---------------------------------------------------------------------------------

Overall: $2Al+3I_{2}\rightarrow 2Al^{3+}+6I^{-}$

Nernst equation for this cell reaction at $25^{0}\textrm{C}$ -

$E_{cell}=E_{cell}^{0}-\frac{0.059}{n}log{[Al^{3+}]^{2}[I^{-}]^{6}}$

where n is number of electrons exchanged during cell reaction, $E_{cell}^{0}$ is standard cell emf , $E_{cell}$ is cell emf , $[Al^{3+}]$ is concentration of $Al^{3+}$ and $[Cl^{-}]$ is concentration of $Cl^{-}$

Plug in all the given values in the above equation -

$E_{cell}=2.20-\frac{0.059}{6}log[(4.5\times 10^{-3})^{2}\times (0.15)^{6}]V$

So, $E_{cell}=2.32V$

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At about what age does basal metabolic rate begin to gradually decrease?

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2 years ago

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Which substance below exhibits the weakest intermolecular forces?

mestny [16]

Answer:

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Explanation:

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3 years ago

Aclosed system contains an equimolar mixture of n-pentane and isopentane. Suppose the system is initially all liquid at 120°C an

garri49 [273]

Explanation:

The given data is as follows.

T = $120^{o}C$ = (120 + 273.15)K = 393.15 K,

As it is given that it is an equimolar mixture of n-pentane and isopentane.

So, $x_{1}$ = 0.5 and $x_{2}$ = 0.5

According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.

$p^{sat}_{1}$ (393.15 K) = 9.2 bar

$p^{sat}_{1}$ (393.15 K) = 10.5 bar

Hence, we will calculate the partial pressure of each component as follows.

$p_{1} = x_{1} \times p^{sat}_{1}$

= $0.5 \times 9.2 bar$

= 4.6 bar

and, $p_{2} = x_{2} \times p^{sat}_{2}$

= $0.5 \times 10.5 bar$

= 5.25 bar

Therefore, the bubble pressure will be as follows.

P = $p_{1} + p_{2}$

= 4.6 bar + 5.25 bar

= 9.85 bar

Now, we will calculate the vapor composition as follows.

$y_{1} = \frac{p_{1}}{p}$

= $\frac{4.6}{9.85}$

= 0.467

and, $y_{2} = \frac{p_{2}}{p}$

= $\frac{5.25}{9.85}$

= 0.527

Calculate the dew point as follows.

$y_{1}$ = 0.5, $y_{2}$ = 0.5

$\frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}$

$\frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}$

$\frac{1}{P}$ = 0.101966 $bar^{-1}$

P = 9.807

Composition of the liquid phase is $x_{i}$ and its formula is as follows.

$x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{9.2}$

= 0.5329

$x_{z} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{10.5}$

= 0.467

4 0

3 years ago

The decomposition of acetaldehyde is a second order reaction with a rate constant of 4.71 10−8 L/mol/s. What is the instantaneou

pochemuha

Answer: Rate of decomposition of acetaldehyde in a solution is $1.45\times 10^{-14}mol/Ls$

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

For a reaction : $A\rightarrow products$

$Rate=k[A]^x$

k= rate constant

x = order of the reaction = 2

$Rate=4.71\times 10^{-8}L/mol/s\times {5.55\times 10^{-4}M}^2$

$Rate=1.45\times 10^{-14}$

Thus rate of decomposition of acetaldehyde in a solution is $1.45\times 10^{-14}mol/Ls$

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3 years ago