All categories

3 years ago

The electronic configuration of an element is given below.

Chemistry

1 answer:

Amiraneli [1.4K]3 years ago

4 0

The statement which is true about the reactivity of element with 1S²2S²2P⁶3S¹ is

it is reactive because it has to lose one electron to have a full outermost energy level.

Explanation

Element with 1S²2S²2P⁶3S¹ electron configuration is a sodium metal.

sodium has one electron in the outermost energy level.

for sodium to have a full outermost energy level ( 8 electrons) it loses the 1 electron in 3S¹ to form a positively charged ion. (Na⁺)

You might be interested in

15.0 mL of 0.050 M Ba(NO3)2 M and 100.0 mL of 0.10 M KIO3 are added together in a 250 mL erlenmeyer flask. In this problem, igno

timama [110]

Answer:

Yes, precipitation of barium iodate will occur.

Explanation:

Molarity of barium nitrate solution = 0.050 M

Volume of barium nitrate solution =15.0 mL = 0.0150 L

1 mL = 0.001 L

Moles of barium nitrate = n

$n=0.050 M\times 0.0150L=0.00075 mol$

$Ba(NO_3)_2(aq)\rightarrow Ba^{2+}(aq)+2NO_3^{-}(aq)$

Moles of barium ions: $1\times 0.00075 mol=0.00075 mol$

Molarity of potassium iodate solution = 0.10 M

Volume of potassium iodate solution =100.0 mL = 0.1000 L

1 mL = 0.001 L

Moles of potassium iodate = n'

$n'=0.10 M\times 0.1000 L=0.01 mol$

$KIO_3(aq)\rightarrow K^{+}(aq)+IO_3^{-}(aq)$

Moles of iodate ions = $1\times 0.01 mol=0.01 mol$

After mixing of both solution in 250 mL in erlenmeyer flask

Volume of the final solution = 250 mL = 0.250 L

Concentration of barium ions in 250 mL solution :

$[Ba^{2+}]=\frac{0.00075 mol}{0.250 L}=0.003 M$

Concentration of iodate ions:

$[IO_3^{-}]=\frac{0.01 mol}{0.250 L}=0.04 M$

Solubility product of barium iodate, $K_{sp}=4.01\times 10^{-9}$

Ionic product of the barium iodate in solution : $K_i$

$Ba(IO_3)_2\rightleftahrpoons Ba^{2+}+2IO_3^{-}$

$K_i=[Ba^{2+}][IO_2^{-}]^2$

$K_i=0.003 M\times (0.04 M)^2=4.8\times 10^{-6}$

$K_{sp}$ ( precipitation)

As we can see, the ionic product of the barium iodate is greater than the solubility product of the barium iodate precipitation of barium iodiate will occur in 250 mL of final solution.

5 0

3 years ago

Which of these is NOT an example of how your community uses natural resources?

Reil [10]

Can you give use an clue of what it is all about

3 0

3 years ago

Read 2 more answers

Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic

ale4655 [162]

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

$\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9$

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = $164.5\times 10^{-9} g$

$1 ng=10^{-9} g$

Volume of the sample = V = $15.3 cm^3$

Density of the lake water sample ,d= $1.00 g/cm^3$

Mass of sample = M = $d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g$

$ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75$

10.75 is the concentration of arsenic in the sample in parts per billion.

Mass of arsenic in $1 cm^3$ of lake water = $\frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g$

Mass of arsenic in $0.710 km^3$ lake water be m.

$1 km^3=10^{15} cm^3$

Mass of arsenic in $0.710\times 10^{15} cm^3$ lake water :

$m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g$

1 g = 0.001 kg

7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg

7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

$\frac{2.74}{41.90} days$

Then days required to remove 7,633.66 kg of arsenic from the lake water :

$=7,633.66\times \frac{2.74}{41.90} days=499.19 days$

1 year = 365 days

499.19 days = $\frac{499.19}{365} years = 1.367 years\approx 1.37 years$

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

$\frac{2.74}{41.90} days$

Then days required to remove 7,633.66 kg of arsenic from the lake water :

$=7,633.66\times \frac{2.74}{41.90} days=499.19 days$

1 year = 365 days

499.19 days = $\frac{499.19}{365} years = 1.367 years\approx 1.37 years$

It will take 1.37 years to remove all of the arsenic from the lake.

6 0

3 years ago

8. Why would it be a bad idea to make coins made of sodium metal?

Elina [12.6K]

Answer:

Sodium reacts with the oxygen in air. It reacts vigorously with oxygen and the moisture that is already present in the air and thus catches fire.

5 0

3 years ago

According to their locations on the Periodic Table, argon (Ar) and neon (Ne) are

a_sh-v [17]

Your answer would be (D), Noble Gases, or Inert Gases - The far right on the periodic table is also known as Group(0)., or Group 18 on the periodic table. Elements in this group includes, Neon(NE), Argon(AR), Krypton(KR), Xenon(XE), and Radon(RN).

Hope that helps!!!

3 0

3 years ago

Read 2 more answers