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skelet666 [1.2K]
3 years ago
11

Which of the following elements is a metalloid

Chemistry
2 answers:
klemol [59]3 years ago
6 0
A metalloid can be:
 - Boron (B)
 - Silicon (Si)
 - Germanium ( Ge)
 - Arsenic (As) 
 - Antimony ( Sb)
 - Tellurium (Te)
 - Polonium (Po)

Hope this helps :)
irina1246 [14]3 years ago
6 0

Answer : Silicon is the element of Metalliod.

Explanation :

Metals : These are the elements that can easily loose electrons and forms cations.

For example : All the alkali or alkaline earth metals that is group 1 and 2 are the metals.

Non-metals : These are the elements that can easily gain electrons and form an anion.

For example : All the p-block elements that is group 13 to 18 are the non-metals.

Metalloids : These are the elements that shows both the property of metals and non-metals.

For example : Boron, silicon, germanium, arsenic, antimony, tellurium, polonium and astatine are the metalloids.

As per question, silicon is the elements of metalloids.

While, rubidium and vanadium are metals and radon is a noble gas.

Hence, the silicon is the element of Metalliod.

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A solution is prepared by dissolving 10.0 g of NaBr and 10.0 g of Na2SO4 in water to make a 100.0 mL solution. This solution is
Colt1911 [192]

Answer:

M_{Na^+}=1.36M

M_{Br^-}=1.58M

Explanation:

Hello,

At first, it turns out convenient to compute the total moles of sodium that will be dissolved into the solution by considering the added amounts of sodium bromide and sodium sulfate:

n_{Na^+}=n_{Na^+,NaBr}+n_{Na^+,Na_2SO_4}\\n_{Na^+,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molNa^+}{1molNaBr}=0.0971molNa^+\\n_{Na^+,Na_2SO_4}=10.0gNa_2SO_4*\frac{1molNa_2SO_4}{142gNa_2SO_4}*\frac{2molNa^+}{1molNa_2SO_4} =0.141molNa^+\\n_{Na^+}=0.0971molNa^++0.141molNa^+\\n_{Na^+}=0.238molNa^+

Once we've got the moles we compute the final volume via:

V=100.0mL+75.0mL=175.0mL*\frac{1L}{1000mL}=0.1750L

Thus, the molarity of the sodium atoms turn out into:

M_{Na^+}=\frac{0.238mol}{0.1750L} =1.36M

Now, we perform the same procedure but now for the bromide ions:

n_{Br^-}=n_{Br^-,NaBr}+n_{Br^-,AlBr_3}\\n_{Br^-,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molBr^-}{1molNaBr}=0.0971molBr^-\\n_{Br^-,AlBr_3}=0.0750L*0.800\frac{molAlBr_3}{L} *\frac{3molBr^-}{1molAlBr_3}=0.180molBr^- \\n_{Br^-}=0.0971molBr^-+0.180molBr^-\\n_{Br^-}=0.277molBr^-

Finally, its molarity results:

M_{Br^-}=\frac{0.277molBr^-}{0.1750L}=1.58M

Best regards.

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Answer:

Diluted concentration is 0.5M

Explanation:

Let's solve this with rules of three, although there is a formula to see it easier

In 1000 mL (1L), we have 2 moles of NaOH

In 250 mL we must have (250 . 2) / 1000 = 0.5 moles of NaOH

These moles will be also in 1 L of the final volume of the diluted solution

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1 L of solution has 0.5 moles of NaOH

Then, molarity is 0.5 M

The formula is: Concentrated M . Conc. volume = Diluted M . Diluted volume

2 M . 0.250L = 1L . Diluted M

0.5M = Diluted M

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