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Daniel [21]
3 years ago
10

Potassium is kept in the s-blockof the periodic table. Why?​

Chemistry
2 answers:
makkiz [27]3 years ago
6 0

Answer:

Potassium is an s block element as the electrons of potassium are in the outermost s orbital in an electron shell. EXPLANATION: Alkali metals are known to be group 1 metals placed in s block.

UkoKoshka [18]3 years ago
5 0

Answer:

Because Pottasium's last electron enters in the s orbital of its last shell.

Explanation:

In electron configuration, the elements in which the last electron enters in the last s orbital are included in s block. The group 1 and 2 are in this block.

You might be interested in
balance by oxidation number method<br />KMnO4 + Na2O₂ + H2 so4 ko2so4 +<br />Mosou4 + Na2so<br />4 + O2 + H₂O​
Katarina [22]

Answer:

8H2SO4 + 2KMnO4 + 5Na2O2 => 8H2O + 2MnSO4 + 5Na2SO4 + 5O2 + K2SO4

have:

Explanation:

6 0
3 years ago
A sample of metal has a mass of 19.67 g, and a volume of 5.90 ml. what is the density of this metal?
pishuonlain [190]
Density
=mass÷volume
=19.67÷5.90
=3.33 g/ml
8 0
3 years ago
A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

5 0
3 years ago
Magnesium (average atomic mass = 24.3050 amu) consists of three isotopes with masses 23.9850 amu, 24.9858 amu, and 25.9826 amu.
iris [78.8K]

Answer: The first isotope has a relative abundance of 79% and last isotope has a relative abundance of 11%

Explanation: Given that the average atomic mass(M) of magnesium

= 24.3050amu

Mass of first isotope (M1) = 23.9850amu

Mass of middle isotope (M2)=24.9858amu

Mass of last isotope(M3)= 25.9826amu

Total abundance = 1

Abundance of middle isotope = 0.10

Let abundance of first and last isotope be x and y respectively.

x+0.10+y =1

x = 0.90-y

M = M1 × % abundance of first isotope + M2 × % of middle isotope +M3 ×% of last isotope

24.03050= 23.985× x + 24.9858 ×0.10 + 25.9826×y

Substitute x= 0.90-y

Then

y = 0.11

Since y=0.11, then

x= 0.90-0.11

x=0.79

Therefore the relative abundance of the first isotope = 11% and the relative abundance of the last isotope = 79%

5 0
3 years ago
(3 x 104) (4 x 1023)
Ksivusya [100]

Answer:

1 x 10^28

Explanation:

  • this what the scientific notati0N
3 0
3 years ago
Read 2 more answers
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