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2 years ago

Of the following metals, which is the LEAST reactive? A. K B. Ca C. Na D. Mg

Chemistry

2 answers:

emmainna [20.7K]2 years ago

6 0

The Answer Should Be D.Mg

sladkih [1.3K]2 years ago

3 0

Answer: D. Mg

Explanation:

Chemical reactivity is defined as the tendency of an element to loose of gain electrons.

Metals are the elements which lose electrons and hence, their chemical reactivity will be the tendency to lose electrons.

Their chemical reactivity increases as we move top to bottom in a group because the valence shell come gets far away from the nucleus. Thus, the loss of electron from the valence shell becomes easier due to lesser attraction between nucleus and valence electron.

Their chemical reactivity decreases as we move from left to right in a period .As, the electrons get added up in the same shell, the electron in the outermost orbital gets near to the nucleus. And hence, the electron will be difficult to lose.Thus, the chemical reactivity of metals decreases.

For the given options:

Potassium(K) belongs to Group 1 and period 4 of the periodic table.

Calcium (Ca) belongs to Group 2 and period 4 of the periodic table.

Sodium (Na)belongs to Group 1 and Period 3 of the periodic table.

Magnesium (Mg) belongs to Group 2 and Period 3 of the periodic table.

Order of reactivity of metals follow:

$K>Ca>Na>Mg$

Thus the LEAST reactive metal is Magnesium.

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2 years ago

Which of these phrases accurately describe disproportionation? Check all that apply.

Olin [163]

Answer:

A single compound is simultaneously oxidized and reduced.

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2 years ago

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2 years ago

Unit: Chemical Quantities

Vaselesa [24]

Answer:

(See explanation for further details)

Explanation:

1) The quantity of moles of sulfur is:

$n = \frac{1.20\times 10^{24}\,atoms}{6.022\times 10^{23}\,\frac{atoms}{mol} }$

$n = 1.993\,moles$

2) The number of atoms of helium is:

$x = (1.5\,moles)\cdot \left(6.022\times 10^{23}\,\frac{atoms}{mole} \right)$

$x = 9.033\times 10^{23}\,atoms$

3) The quantity of moles of carbon monoxide is:

$n = \frac{4.15\times 10^{23}\,molecules}{6.022\times 10^{23}\,\frac{molecules}{mol} }$

$n = 0.689\,moles$

4) The number of molecules of sulfur dioxide is:

$x = (2.25\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)$

$x = 1.355\times 10^{24}\,molecules$

5) The quantity of moles of sodium chloride is:

$n = \frac{2.4\times 10^{23}\,molecules}{6.022\times 10^{23}\,\frac{molecules}{mol} }$

$n = 0.399\,moles$

6) The number of formula units of magnesium iodide is:

$x = (1.8\,moles)\cdot \left(6.022\times 10^{23}\,\frac{f.u.}{mole} \right)$

$x = 1.084\times 10^{24}\,f.u.$

7) The quantity of moles of potassium permanganate is:

$n = \frac{3.67\times 10^{23}\,f.u.}{6.022\times 10^{23}\,\frac{f.u.}{mol} }$

$n = 1.214\,moles$

8) The number of molecules of carbon tetrachloride is:

$x = (0.25\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)$

$x = 1.506\times 10^{23}\,molecules$

9) The quantity of moles of aluminium is:

$n = \frac{3.67\times 10^{23}\,atoms}{6.022\times 10^{23}\,\frac{atoms}{mol} }$

$n = 0.609\,moles$

10) The number of molecules of oxygen difluoride is:

$x = (3.52\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)$

$x = 2.120\times 10^{24}\,molecules$

3 0

3 years ago

How would you prepare 1.00 L of a 0.50 M solution of each of the following?

Lunna [17]

Answer:

Take approx 41.7 mL of 12-M HCl in a 1.00-L flask and fill the rest of the volume with distilled water.

Explanation:

Hello,

In this case, for the dilution process from concentrated 12-M hydrochloric acid to 1.00 L of the diluted 0.50M hydrochloric acid, the volume of concentrated HCl you must take is computed by considering that the moles remain constant for all dilution processes as shown below:

$n_1=n_2$

Which can also be written in terms of concentrations and volumes:

$M_1V_1=M_2V_2$

Thus, solving for the initial volume or aliquot that must be taken from the 12-M HCl, we obtain:

$V_1=\frac{M_2V_2}{M_1} \\\\V_1=\frac{1.00L*0.5M}{12M}\\ \\V_1=0.0417L=41.7mL$

It means that you must take approx 41.7 mL of 12-M HCl in a 1.00-L flask and fill the rest of the volume with distilled water for such preparation.

Best regards.

6 0

3 years ago