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vivado [14]
3 years ago
10

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.

<u><em>Explanation</em></u>

  • <u><em> </em></u>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⁺)
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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<br> 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.

B)

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.

C)

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

C)

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
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