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2 years ago

Which isolated gaseous atom has a total of five electrons occupying spherically shaped orbitals?

1 answer:

3 0

Let's find the electron configuration of each atom

Atom atomic number electron configuration

Bo 5 1s2 2s2 2p
F 9 1s2 2s2 2p5
Na 11 1s2 s2 2p6 3s
K 19 1s2 2s2 2p6 3s2 3p6 4s

The spherically shaped orbitals are the s orbitals.

You can se in the table that I wrote above that sodium has 2 electrons at 1s, 2 electrons at 2s and 1 electron at 3s orbitals, that is a total of five electrons at s (spherically shaped orbtials).

So the answer is the option A) sodium.

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To determine the concentration of ethanol in cognac a 5.00 mL sample of the cognac is diluted to 0.500 L. Analysis of the dilute

julia-pushkina [17]

Answer : The molar concentration of ethanol in the undiluted cognac is 8.44 M

Explanation :

Using neutralization law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molar concentration of undiluted cognac = ?

$M_2$ = molar concentration of diluted cognac = 0.0844 M

$V_1$ = volume of undiluted cognac = 5.00 mL = 0.005 L

$V_2$ = volume of diluted cognac = 0.500 L

Now put all the given values in the above law, we get molar concentration of ethanol in the undiluted cognac.

$M_1\times 0.005L=0.0844M\times 0.500L$

$M_1=8.44M$

Therefore, the molar concentration of ethanol in the undiluted cognac is 8.44 M

4 0

2 years ago

5. How many mm are equal to 4.75 x 10-2 m?

liq [111]

Answer:

47.5 mm are equal to 4.75 × 10⁻² m.

Explanation:

Meter and millimeter both are units of length. Millimeter is smaller unit while meter is larger unit. The one meter consist of thousand millimeter or we can say that one thousand millimeters are equal to one meter. In order to convert the meter value into millimeter we have to multiply the values with thousand.

we know that one meter is equal to 1000 millimeter.

1 m = 1000 mm

4.75 × 10⁻² × 1000 = 47.5 mm

47.5 mm are equal to 4.75 × 10⁻² m.

6 0

2 years ago

Use the drop‑down menus to label the statements as either true of false. Dichloromethane is flammable. true false Methanol is fl

Flauer [41]

Explanation:

Dichloromethane is flammable - FALSE

Methanol is flammable. - TRUE

Concentrated sulfuric acid is corrosive. - TRUE

10% sodium carbonate solution must be used in the fume hood. - FALSE

Benzoyl chloride is a lachrymator. - TRUE

5 0

2 years ago

A student reacts 20.0 mL of 0.248 M H2SO4 with 15.0 mL of 0.195 M NaOH. Write a balanced chemical equation to show this reaction

Molodets [167]

Answer: The concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For NaOH:

Initial molarity of NaOH solution = 0.195 M

Volume of solution = 15.0 mL = 0.015 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.195M=\frac{\text{Moles of NaOH}}{0.015L}\\\\\text{Moles of NaOH}=(0.195mol/L\times 0.015L)=2.925\times 10^{-3}mol$

For sulfuric acid:

Initial molarity of sulfuric acid solution = 0.248 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.248M=\frac{\text{Moles of }H_2SO_4}{0.020L}\\\\\text{Moles of }H_2SO_4=(0.248mol/L\times 0.020L)=4.96\times 10^{-3}mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, $2.925\times 10^{-3}$ moles of KOH will react with = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sulfuric acid

As, given amount of sulfuric acid is more than the required amount. So, it is considered as an excess reagent.

Thus, NaOH is considered as a limiting reagent because it limits the formation of product.

Excess moles of sulfuric acid = $(4.96-1.462)\times 10^{-3}=3.498\times 10^{-3}mol$

By Stoichiometry of the reaction:

2 moles of KOH produces 1 mole of sodium sulfate

So, $2.925\times 10^{-3}$ moles of KOH will produce = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sodium sulfate

For sodium sulfate:

Moles of sodium sulfate = $1.462\times 10^{-3}moles$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sodium sulfate}=\frac{1.462\times 10^{-3}}{0.035}=0.0418M$

For sulfuric acid:

Moles of excess sulfuric acid = $3.498\times 10^{-3}mol$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sulfuric acid}=\frac{3.498\times 10^{-3}}{0.035}=0.0999M$

For NaOH:

Moles of NaOH remained = 0 moles

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of NaOH}=\frac{0}{0.050}=0M$

Hence, the concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

8 0

2 years ago

How many moles of H2O are produced when 64.0 g C2H2 burns in oxygen

Gnoma [55]

The balanced chemical reaction will be:

C2H2 + 5/2O2 = 2CO2 + H2O

We are given the amount of C2H2 being burned. This will be our starting point.

64.0 g C2H2 (1 mol C2H2 / 26.02 g C2H2) (1 mol H2O/1mol C2H2) ( 18.02 g H2O/1mol H2O) = 44.32 g H2O.

Thus, the answer is 44.32 g H2O.

6 0

2 years ago