What is the concentration of a solution with a volume of 1 L that contain 200 grams of Fe(OH)3?

- 5. Describe the general trend in atomic radil across a row of the periodic table.

sattari [20]

Answer:

Across a row, atomic radii decreases.

Explanation:

It decreases because nuclear charge increases across a row, electrons are added to the same shell and shielding remains constant. Therefore the outer electrons are more attracted to the nucleus making the atom smaller.

8 0

2 years ago

Air is compressed from an inlet condition of 100 kPa, 300 K to an exit pressure of 1000 kPa by an internally reversible compress

ElenaW [278]

Answer:

(a) $W_{isoentropic}=8.125\frac{kJ}{mol}$

(b) $W_{polytropic}=7.579\frac{kJ}{mol}$

(c) $W_{isothermal}=5.743\frac{kJ}{mol}$

Explanation:

Hello,

(a) In this case, since entropy remains unchanged, the constant $k$ should be computed for air as an ideal gas by:

$\frac{R}{Cp_{air}}=1-\frac{1}{k} \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\$

$0.2856=1-\frac{1}{k}\\\\k=1.4$

Next, we compute the final temperature:

$T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K$

Thus, the work is computed by:

$W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}$

(b) In this case, since $n$ is given, we compute the final temperature as well:

$T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K$

And the isentropic work:

$W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}$

(c) Finally, for isothermal, final temperature is not required as it could be computed as:

$W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}$

Regards.

8 0

3 years ago

Help on my homework

Illusion [34]

Bronchi

Bronchioles

Alveoli

3 0

3 years ago

97

alex41 [277]

B

Please help

Sorry I need 20 characters to submit this Answer that’s why I’m adding more words

5 0

3 years ago

What is the daughter nucleus (nuclide) produced when 64 Cu Cu64 undergoes beta decay by emitting an electron? Replace each quest

kenny6666 [7]

<u>Answer:</u> The daughter nuclide formed by the beta decay of given isotope is $_{30}^{64}\textrm{Zn}$

<u>Explanation:</u>

Beta decay is defined as the process in which beta particle is emitted. In this process, a neutron gets converted to a proton and an electron.

The released beta particle is also known as electron.

$_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta$

We are given:

Parent isotope = $_{29}^{64}\textrm{Cu}$

The chemical equation for the beta decay process of $_{29}^{64}\textrm{Cu}$ follows:

$_{29}^{64}\textrm{Cu}\rightarrow _{30}^{64}\textrm{Zn}+_{-1}^0\beta$

Hence, the daughter nuclide formed by the beta decay of given isotope is $_{30}^{64}\textrm{Zn}$

4 0

3 years ago