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4 years ago

For the following reaction, 15.2 grams of chlorine gas are allowed to react with 8.46 grams of water .

1 answer:

5 0

To solve this problem, you must first balance the chemical equation and then identify the limiting reactant. The balanced reaction is 3Cl2 + 3H2O = 5HCl + HClO3. Next is finding the limiting reactant. To do this, you have to compare each reactant to the same product. 15.2 g of chlorine gas (Cl2), is equal to 0.214 mols of chlorine gas, and the mol to mol ratio of chlorine gas to hydrochloric acid is 3:5. So, for every 3 mols of Cl2 gas, there are 5 mols of HCl. 0.214mols Cl2 times (5/3) gives you 0.357 mols of HCl. Next you have to go through the same process except with water. 8.46g water divided by 18.016g water per mol = 0.470 mols H2O. The mol to mol ration is still 3:5, so 0.470 times (5/3), which gives you 0.783mols HCl. A chemical reaction is limited based on the amount of reactants present, and so the reaction is limited by whichever reactant runs out first. In this case, the limiting reactant is chlorine gas because it produced a smaller amount of HCl, and therefore the “limit” of the reaction. This means that the maximum amount of HCl that can form is 0.357 mols, or 13.016 grams of HCl if the answer is supposed to be given in grams.

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Which of the following elements has the lowest ionization energy?

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Ionization energy is defined required energy to eliminate or remove an electron from an ion or atom.

From the given elements Rb or Rubidium has the lowest ionization energy as it has lowest shielding effect, so it easy to remove electron from it's shell.

The ionization energy generally decreases from top to bottom in groups due to lower shielding effect and outer electrons are loosely packed so easy to remove and increases from left to right across a period because of valence shell stability.

Rubidium has atomic number 37 and lies below than other given elements and also placed in the left side in the periodic table.

Hence, the correct option is Rb

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3 years ago

Select correct examples of linear molecules for five electron groups. Select correct examples of linear molecules for five elect

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Explanation:

Formula to calculate hybridization is as follows.

Hybridization = $\frac{1}{2}[V+N-C+A]$

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

So, hybridization of $BeCl_{2}$ is as follows.

Hybridization = $\frac{1}{2}[V + N - C + A]$

= $\frac{1}{2}[2 + 2]$

= 2

Hybridization of $BeCl_{2}$ is sp. Therefore, $BeCl_{2}$ is a linear molecule. There will be only two electron groups through which Be is attached.

Similarly, hybridization of $XeF_{2}$ is calculated as follows.

Hybridization = $\frac{1}{2}[V + N - C + A]$

= $\frac{1}{2}[8 + 2]$

= 5

Therefore, hybridization of $XeF_{2}$ is $sp^{3}d. Therefore, [tex]XeF_{2}$ is also a linear molecule. Though there are three lone pair of electrons present on a xenon atom and it is further attached with fluorine atoms through two electron pairs. Hence, there are in total five electron groups.

Thus, we can conclude that out of the given options $XeF_{2}$ is the correct examples of linear molecules for five electron groups.

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3 years ago

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3 years ago

How many formula units make up 24.2 g of magnesium chloride (MgCl2)? Help!!

NNADVOKAT [17]

Answer:

Approximately $1.53 \times 10^{23}$ formula units ( $0.254\; \rm mol$ ).

Explanation:

Refer to a modern periodic table for the relative atomic mass of magnesium ( $\rm Mg$ ) and chlorine ( $\rm Cl$ ):

$\rm Mg$ : $24.305$ .
$\rm Cl$ : $35.45$ .

In other words, the mass of $1\; \rm mol$ of $\rm Mg$ atoms would be (approximately) $24.305\; \rm g$ .

Likewise, the mass of $1\; \rm mol$ of $\rm Cl$ atoms would be approximately $35.45\; \rm g$ .

One formula unit of the ionic compound $\rm MgCl_{2}$ includes exactly as many atoms as there are in the given formula. The formula mass of a compound is the mass of $1\; \rm mol$ of the formula units of this compound.

The formula $\rm MgCl_{2}$ includes one $\rm Mg$ atom and two $\rm Cl$ atoms.

Hence, every formula unit of $\rm MgCl_{2} \!$ would include the same number of atoms: one $\rm Mg\!$ atom and two $\rm Cl\!$ atoms. There would be $1\; \rm mol$ of $\rm Mg$ atoms and $2\; \rm mol$ of $\rm Cl$ atoms in $1\; \rm mol\!$ of $\rm MgCl_{2}$ formula units.

Thus, the mass of $1\; \rm mol\!$ of $\rm MgCl_{2}$ formula units would be equal to the mass of $1\; \rm mol$ of $\rm Mg$ atoms plus the mass of $2\; \rm mol$ of $\rm Cl$ atoms. (The mass of $1\; \rm mol\!\!$ of each atom could be found from the relative atomic mass of each element.)

$\begin{aligned}& M({\rm MgCl_{2}}) \\ =\; & 24.305\; {\rm g \cdot mol^{-1}} + 2\times {\rm 35.45 \; \rm g \cdot mol^{-1}} \\ =\; & 95.205\; \rm g \cdot mol^{-1}\end{aligned}$ .

In other words, the formula mass of $\rm MgCl_{2}$ is $95.205\; \rm g \cdot mol^{-1}$ .

Therefore, the number of formula units in $m = 24.2\; \rm g$ of $\rm MgCl_{2}$ would be:

$\begin{aligned}n &= \frac{m({\rm MgCl_{2}})}{M({\rm MgCl_{2}})} \\ &= \frac{24.2\; \rm g}{95.205\; \rm g\cdot mol^{-1}} \\ & \approx 0.254\; \rm mol\end{aligned}$ .

Multiple $n$ by Avogadro's Number $N_{A} \approx 6.022 \times 10^{23}\; \rm mol^{-1}$ to estimate the number of formula units in $0.254\; \rm mol$ :

$\begin{aligned}N &= n \cdot N_{A} \\ &\approx 0.254\; \rm mol \times 6.022 \times 10^{23}\; \rm mol^{-1} \\ &\approx 1.53\times 10^{23}\end{aligned}$ .

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3 years ago