All categories

3 years ago

Of the elements below, __________ is the least metallic.

1 answer:

dsp733 years ago

8 0

Answer: last option: Ar

Justification:

The metallic character of the elements in the periodic table increase from right to left and from top to bottom.

That means that the elements of groups (rows) 1 and 2 and periods (columns) 6 and 7 are the most metallics, and the elements of groups 17 and 18 and periods 1 and 2 are the least metallics.

Na, Mg, Al, P and Ar are in the same period (row) so you can differentiate their metallic character by the rule that the more to the right on the row the less metallic the element is.

Na (left end the row) is the most metallic and Ar (right end of the row) is the least metallic.

P and Ar are less metallic.

Then, the list arranged in order of decreasing metallic character is:

Na > Mg > Al > P > Ar.

Therefore, the answser is Ar.

You might be interested in

A 3.301 mass % aqueous solution of potassium hydroxide has a density of 1.05 g/mL. Calculate the molality of the solution. Give

8_murik_8 [283]

Answer: The molality of potassium hydroxide solution is 0.608 m

Explanation:

We are given:

3.301 mass % of potassium hydroxide solution.

This means that 3.301 grams of potassium hydroxide is present in 100 grams of solution

Mass of solvent = Mass of solution - Mass of solute (KOH)

Mass of solvent = (100 - 3.301) g = 96.699 g

To calculate the molality of solution, we use the equation:

$\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams})}$

Where,

$m_{solute}$ = Given mass of solute (KOH) = 3.301 g

$M_{solute}$ = Molar mass of solute (KOH) = 56.1 g/mol

$W_{solvent}$ = Mass of solvent = 96.699 g

Putting values in above equation, we get:

$\text{Molality of KOH}=\frac{3.301\times 1000}{56.1\times 96.699}\\\\\text{Molality of KOH}=0.608m$

Hence, the molality of potassium hydroxide solution is 0.608 m

4 0

3 years ago

Hydrogen is prepared commercially by the reaction of methane and water vapor at elevated temperatures. CH₄(????) + H₂O(????) ⇌ 3

sammy [17]

Answer:

6.2846

Explanation:

Given that:-

Concentrations at equilibrium :-

$[CH_4]=0.126\ M$

$[H_2O]= 0.242\ M$

$[CO]= 0.126\ M$

$[H_2]= 1.15\ M$

The equilibrium reaction is:-

$CH_4+H_2O\rightleftharpoons 3H_2+CO$

The expression for equilibrium constant is:

$K_{c}=\frac {\left [ H_2 \right ]^3\left [ CO \right ]}{\left [ CH_4 \right ]\left [ H_2O \right ]}$

Applying the values as:-

$K_c=\frac{1.15^3\times 0.126}{0.126\times 0.242}=6.2846$

The equilibrium constant for the reaction is:- 6.2846

5 0

3 years ago

There are two groups of waves with the same amplitude. One contains waves of short wavelength and the other has waves of long wa

jolli1 [7]

Answer:

i think the long wavelength has more energy

Explanation:

plz correct me if i'm wrong

3 0

3 years ago

Read 2 more answers

Sometimes, during the DNA replication process, mistakes are made. These are called

wolverine [178]

Answer: IT'S MUTATION I TOOK THE TEST ITS RIGHT HOPE THIS HELPED :)

8 0

3 years ago

Read 2 more answers

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)× $\frac{140,116g}{1mol}$ = 0,0625 g of Ce(IV)

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

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

I hope it helps!

5 0

3 years ago