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3 years ago

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How many moles of CaCl2 are required to prepare 2.00 liters of 0.700 M CaCl2?

1 answer:

suter [353]3 years ago

5 0

The dissolution<span> of </span>calcium chloride<span> in </span>water<span>. </span>CaCl2(s<span>)--> </span>Ca^2+ (aq) +2 Cl^-1<span> (</span>aq<span>). A) </span>Use<span> the </span>data<span> from</span>appendix C<span>. and </span>below<span> to </span>determine Delta H<span> of </span>formation<span> in </span>kj/mol<span> for this </span>process<span> (</span>deltaH(Ca^2+(aq)=-542.8 kj/mol<span>).</span>

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Suppose a new group 5A element is discovered. Photoelectron spectroscopy (PES) reveals that its last two peaks occur at 1370 kJ/

iren2701 [21]

Answer: The average valence electron energy (AVEE) of this element =

1014.2 KJ/ mol or 1.0142mJ/mol.

Explanation:

The average valence electron energy = (number of electrons in s subshell x Ionization energy of that subshell) + (number of electrons in p subshell x Ionization energy of that subshell) / total number of electrons in both subshells of the valence shells.

The 5A elements are non-metals like Nitrogen and Phosphorus with the metallic character increasing as you go down the group, So a new 5A element will have characteristics of its group with 5 valence electron in its outermost shell represented as ns2 np3

Therefore the average valence electron energy (AVEE) of this element will be calculated as

The average valence electron energy = (2 x 1370 kJ/mol + 3 x 777 kJ/mol.) / 5

2740+2331/ 5 =5071/5

=1014.2 KJ/ mol or 1.0142mJ/mol.

5 0

2 years ago

Calcium and bromine have formed a bond. Leading up to this, calcium gave up electrons. It was a(n) ____.

Savatey [412]

Answer:

C. Positive Ion

Explanation:

7 0

2 years ago

Protons, neutrons, and electrons all have similar masses. explain why this statement is false.

yan [13]

It is false because electrons have no mass.

6 0

2 years ago

(6) Compare a CSTR with a PFR below. a. A flow of 0.3 m3/s enters a CSTR (volume of 200 m3) with an initial concentration of spe

Dmitry [639]

Answer:

Explanation:

Given that:

The flow rate Q = 0.3 m³/s

Volume (V) = 200 m³

Initial concentration $C_o$ = 2.00 ms/l

reaction rate K = 5.09 hr⁻¹

Recall that:

$time (t) = \dfrac{V}{Q}$

$time (t) = \dfrac{200}{0.3}$

$time (t) = 666.66 \ sec$

$time (t) = \dfrac{666.66 }{3600} hrs$

$time (t) = 0.185 hrs$

$\text{Using First Order Reaction:}$

$\dfrac{dc}{dt}=kc$

where;

$t = \dfrac{1}{k} \Big( \dfrac{C_o}{C_e}-1 \Big)$

$0.185 = \dfrac{1}{5.09} \Big ( \dfrac{200}{C_e}- 1 \Big)$

$0.942 = \Big ( \dfrac{200}{C_e}- 1 \Big)$

$1+ 0.942 = \Big ( \dfrac{200}{C_e} \Big)$

$\dfrac{200}{C_e} = 1.942$

$C_e = \dfrac{200}{1.942}$

$\mathbf{C_e = 102.98 \ mg/l}$

Thus; the concentration of species in the reactant = 102.98 mg/l

b). If the plug flow reactor has the same efficiency as CSTR, Then:

$t _{PFR} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{5.09} \Big [ In ( \dfrac{200}{102.96}) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} =0.196 \Big [ In ( 1.942) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} =0.196(0.663)$

$\dfrac{V_{PFR}}{0.3 hrs} =0.196(0.663)$

$\dfrac{V_{PFR}}{0.3*3600 sec} =0.196(0.663)$

$V_{PFR} =0.196(0.663)*0.3*3600$

$V_{PFR} = 140.34 \ m^3$

The volume of the PFR is ≅ 140 m³

3 0

2 years ago

Heat is the movement from areas of ___ temperature to areas of ___ temperature

crimeas [40]

Answer:

heat is the movement from areas of <u>high </u>temperature to areas of <u>low </u>temperature

4 0

2 years ago