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3 years ago

Write the electronic configuration of the following element or ion and circle on their valance electrons

Chemistry

1 answer:

agasfer [191]3 years ago

6 0

1s2,2s2,2p6

Explanation:

electron arrangements

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7 Which physical characteristic of a solution may

Nuetrik [128]

Answer:

2= its color

Explanation:

Transition elements are present in the middle of periodic table. These are d-block elements.

These are 38 elements.

All transition elements have partially filled d orbitals.

They showed color in compound because of d-d transition.

During the d-d transition electron absorbed the energy and emit the reminder energy. The emission is usually in the form of color light.

The color of ion is complementary to the absorbed color.

The transition elements are used as a catalyst in industries such as polymer, petroleum industries.

They are ductile, conduct heat and electricity.

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3 years ago

Which is an example of potential energy?

Setler79 [48]

Probably C hope this helps

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3 years ago

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What is the osmotic pressure of a solution of FeCl3(aq) in which 69.5 g of FeCl3(s) (molar mass = 162.20 g mol-1) are dissolved

ivolga24 [154]

The osmotic pressure of the solution is 25.48atm

Data;

Mass = 69.5g
Molar mass = 162.20g/mol
Volume = 2.35L
Temperature = 35.0^oC

<h3>Osmotic Pressure</h3>

The osmotic pressure of the solution is calculated as

$\pi = iMRT$

i = Vant Hoff Factor
M = molarity
R = gas constant
T = Temperature

The Vant hoff factor of the factor = 4

Let's calculate the molarity of the solution;

$M = \frac{number of moles}{volume}$

Number of moles = mass / molar mass

$n = \frac{mass}{molar mass} \\$

substitute the values into the formula and solve for it

$n = \frac{69.5}{162.20} \\n = 0.428 moles$

The molarity of the solution is calculated as

$M = \frac{number of moles}{volume} \\M = \frac{0.428}{2.35} = 0.182 mol/L$

The osmotic pressure of the solution is calculated as

$\pi = iMRT\\\pi = 4 * 0.182 * 35\\\pi = 25.48 atm$

The osmotic pressure of the solution is 25.48atm

Learn more on osmotic pressure here;

brainly.com/question/8195553

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2 years ago

An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

Brrunno [24]

Answer:

Approximately $81.84\%$ .

Explanation:

Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

$\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

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3 years ago

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