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3 years ago

What would barium do to obtain a noble gas structure? gain 2 electrons lose 2 electrons gain 6 electrons lose 6 electrons

Chemistry

1 answer:

Tema [17]3 years ago

4 0

Answer:

Lose two electrons.

Explanation:

Barium is present in group 2.

It is alkaline earth metal.

Its atomic number is 56.

Its electronic configuration is Ba₅₆ = [Xe] 6s².

In order to attain the noble gas electronic configuration it must loses its two valance electrons.

When barium loses it two electron its electronic configuration will equal to the Xenon.

The atomic number of xenon is 54 so barium must loses two electrons to becomes equal to the xenon.

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How many moles of each reactant are needed to produce 3.60 x 10^2g ch3oh

zysi [14]

Methanol is prepared by reacting Carbon monoxide and Hydrogen gas,

CO + 2 H₂   → CH₃OH

Calculating Moles of CO:
According to equation,

32 g (1 mole) of CH₃OH is produced by = 1 Mole of CO
So,
3.60 × 10² g of CH₃OH is produced by = X Moles of CO

Solving for X,
X = (3.60 × 10² g × 1 Mole) ÷ 32 g

X = 11.25 Moles of CO

Calculating Moles of H₂:
According to equation,

32 g (1 mole) of CH₃OH is produced by = 2 Mole of H₂
So,
3.60 × 10² g of CH₃OH is produced by = X Moles of H₂

Solving for X,
X = (3.60 × 10² g × 2 Mole) ÷ 32 g

X =  22.5 Moles of H₂

Result:
  3.60 × 10² g of CH₃OH is produced by reacting 11.25 Moles of CO and 22.5 Moles of H₂.

3 0

4 years ago

How many formula units make up 24.2 g of magnesium chloride (MgCl2)?<br><br> 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 $0.254\; \rm mol$ :

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

6 0

3 years ago

The relative atomic mass of Chlorine is 35.45. Calculate the percentage abundance of the two isotopes of Chlorine, 35Cl and 37Cl

nata0808 [166]

Answer:

35Cl = 75.9 %

37Cl = 24.1 %

Explanation:

Step 1: Data given

The relative atomic mass of Chlorine = 35.45 amu

Mass of the isotopes:

35Cl = 34.96885269 amu

37Cl = 36.96590258 amu

Step 2: Calculate percentage abundance

35.45 = x*34.96885269 + y*36.96590258

x+y = 1 x = 1-y

35.45 = (1-y)*34.96885269 + y*36.96590258

35.45 = 34.96885269 - 34.96885269y +36.96590258y

0.48114731 = 1,99704989‬y

y = 0.241 = 24.1 %

35Cl = 34.96885269 amu = 75.9 %

37Cl = 36.96590258 amu = 24.1 %

3 0

3 years ago

A fluorine atom has 9 positive charge,9 negative charges and a mass of 19. Describe the structure of its atom.

Tanya [424]

The atomic structure of the atom contains 9 positively charged particles (protons) and 10 neutrally charged particles (neutrons) in the center of the atom in a clump called the nucleus. Those 9 negatively charged particles (electrons) are moving around outside of the nucleus.
There are 10 neutral charges, because the mass of 19 comes from the number of neutral charges plus the number of positive charges.
To calculate the number of neutral charges, subtract the positive charges from the mass (19 - 9), and you get the number of neutral charges (10).

6 0

3 years ago

Baking soda (NaHCO3) and vinegar (HC2H3O2) react to form sodium acetate, water, and carbon dioxide. If 42.00 g of baking soda re

Setler [38]

Answer:

0.5 mole of CO₂.

Explanation:

We'll begin by calculating the number of mole in 42 g of baking soda (NaHCO₃). This can be obtained as follow:

Mass of NaHCO₃ = 42 g

Molar mass of NaHCO₃ = 23 + 1 + 12 + (16×3)

= 23 + 1 + 12 + 48

= 84 g/mol

Mole of NaHCO₃ =?

Mole = mass / molar mass

Mole of NaHCO₃ = 42/84

Mole of NaHCO₃ = 0.5 mole

Next, balanced equation for the reaction. This is given below:

NaHCO₃ + HC₂H₃O₂ → NaC₂H₃O₂ + H₂O + CO₂

From the balanced equation above,

1 mole of NaHCO₃ reacted to produce 1 mole of CO₂

Finally, we shall determine the number of mole of CO₂ produced by the reaction of 42 g (i.e 0.5 mole) of NaHCO₃. This can be obtained as follow:

From the balanced equation above,

1 mole of NaHCO₃ reacted to produce 1 mole of CO₂.

Therefore, 0.5 mole of NaHCO₃ will also react to produce 0.5 mole of CO₂.

Thus, 0.5 mole of CO₂ was obtained from the reaction.

7 0

3 years ago