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4 years ago

as a solid, Cr adopts a body-centered cubic unit cell. How many unit cells are present per cubic centimeter of Cr?

1 answer:

8 0

D(Cr) = 8,96 g/cm3

M(Cr) = 63,546 g/mol 

In a cell face centred cubic, there are 4 atoms of Cur . . . ! 

n(Cr in 1 cm3) = m(Cu) / M(Cr) 
n(Cr in 1 cm3) = 8,96 / 63,546 
n(Cr in 1 cm3) = 0,141 mol of Cr 

Avogadro's number is : NA = 6,02•10^23 mol^-1 

N(Cr) = n(Cu) x NA 
N(Cr) = 0,141 x 6,02•10^23 
N(Cr) = 8,488•10^22 atoms of Cr 

N(cell) = N(Cu) / 4 
N(cell) = 8,488•10^22 / 4 
N(cell) = 2,122•10^22 cells in 1 cm3

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The half-cell is a chamber in the voltaic cell where one half-cell is the site of the oxidation reaction and the other half-cell

marysya [2.9K]

Answer:

$\boxed{\rm \text{Fe(s) $\rightleftharpoons$ Fe$^{2+}$(aq) + 2e$^{-}$}}$

Explanation:

The half-cell reduction potentials are

Ag⁺(aq) + e⁻ ⇌ Ag(s) E° = 0.7996 V

Fe²⁺(aq) + 2e⁻ ⇌ Fe(s) E° = -0.447 V

To create a spontaneous voltaic cell, we reverse the half-reaction with the more negative half-cell potential.

The anode is the electrode at which oxidation occurs.

The equation for the oxidation half-reaction is

$\boxed{\rm \textbf{Fe(s) $\rightleftharpoons$ Fe$^{2+}$(aq) + 2e$^{-}$}}$

4 0

3 years ago

2) A common "rule of thumb" -- for many reactions around room temperature is that the

babunello [35]

The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

$T_1=300 \ K$

$T_2=310 \ K$

The rate constant = $K_1 \text{ and } K_2$ respectively.

The activation energy and the Arhenius factor is same.

So by the arhenius equation,

$K_1 = Ae^{-\frac{E_a}{RT_1}}$ and $K_2 = Ae^{-\frac{E_a}{RT_2}}$

$\Rightarrow \frac{K_1}{K_2}= \frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}}$

$\Rightarrow \frac{K_1}{K_2}= e^{-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

Given, $E_a = 0.269$ J/mol

R = 8.314 J/mol/K

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \left(\frac{1}{300} -\frac{1}{310} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \times \frac{10}{300 \times 310}$

$\Rightarrow \ln \frac{K_2}{K_1}= 3.479 \times 10^{-6}$

$\Rightarrow \frac{K_2}{K_1}= e^{3.479 \times 10^{-6}}$

$\Rightarrow \frac{K_2}{K_1}= 1$

∴ $K_2=K_1$

So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.

8 0

3 years ago

A solution prepared by mixing 10 ml of 1 m hcl and 10 ml of 1.2 m naoh has a ph of

blagie [28]

Answer: pH of resulting solution will be 13

Explanation:

pH is the measure of acidity or alkalinity of a solution.

Moles of $H^+$ ion = $Molarity\times {\text {Volume in L}}=1M\times 0.01L=0.01mol$

Moles of $OH^-$ ion = $Molarity\times {\text {Volume in L}}=1.2M\times 0.01L=0.012mol$

$HCl+NaOH\rightarrow NaCl+H_2O$

For neutralization:

1 mole of $H^+$ ion will react with 1 mole of $OH^-$ ion

0.01 mol of $H^+$ ion will react with = $\frac{1}{1}\times 0.01mole$ of $OH^-$ ion

Thus (0.012-0.01)= 0.002 moles of $OH^-$ are left in 20 ml or 0.02 L of solution.

$[OH^-]=\frac{0.002}{0.02L}=0.1M$

$pOH=-log[OH^-]$

$pOH=-log[0.1]=1$

$pH+pOH=14$

$pH=14-1=13$

Thus the pH of resulting solution will be 13

7 0

3 years ago

2. what is nappening in this diagram:

natka813 [3]

Answer:

I'm sure but send thru this picture for the question so I can help.

7 0

3 years ago

What is the specific heat of a substance that absorbs 2500 joules of heat when a sample of 100g of the substance increases in te

Ierofanga [76]

Answer:

0.417 J/gºC

Explanation:

From the question given above, the following data were obtained:

Heat (Q) absorbed = 2500 J

Mass (M) of substance = 100 g

Initial temperature (T1) = 10 °C

Final temperature (T2) = 70 °C

Specific heat capacity (C) =?

Next, we shall determine the change in temperature (ΔT). This can be obtained as follow:

Initial temperature (T1) = 10 °C

Final temperature (T2) = 70 °C

Change in temperature (ΔT) =?

Change in temperature (ΔT) = T2 – T1

Change in temperature (ΔT) = 70 – 10

Change in temperature (ΔT) = 60 °C

Finally, we shall determine the specific heat capacity of the substance as follow:

Heat (Q) absorbed = 2500 J

Mass (M) of substance = 100 g

Change in temperature (ΔT) = 60 °C

Specific heat capacity (C) =?

Q = MCΔT

2500 = 100 × C × 60

2500 = 6000 × C

Divide both side by 6000

C = 2500 / 6000

C = 0.417 J/gºC

Therefore, the specific heat capacity of substance is 0.417 J/gºC

5 0

3 years ago