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2 years ago

Which properties are characteristic of metalloids?

Chemistry

2 answers:

ziro4ka [17]2 years ago

8 0

Answer:

Yes the answer is C.) intermediate conductivity and a high melting point

Explanation:

I just took the test and got 100%

Setler [38]2 years ago

7 0

<h2>Answer:</h2>

<u>The right option is</u><u> (C) intermediate conductivity and a high melting point</u>

<h2>Explanation:</h2>

Metalloids usually look like metals but behave largely like nonmetals. Metalloids are shiny, brittle solids with intermediate good electrical conductivity. Their properties lie between metals and non metals. All metalloids exist as solids at room temperature and they have very high melting points. The physical properties of metalloids are more likely to be metallic, but their chemical properties tend to be non-metallic

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What is the concentration of no3- ions in a solution prepared by dissolving 15.0 g of ca(no3)2 in enough water to produce 300. m

r-ruslan [8.4K]

Ca(NO3)2 -------> Ca²⁺ +2NO3⁻

M(Ca(NO3)2)= M(Ca) + M(N) + 6M(O)= 40.0 +14.0 +6*16.0 = 150 g/mol

15.0 g Ca(NO3)2 * 1mol/150 g = 0. 100 mol Ca(NO3)2

Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
1 mol 2 mol
0.100 mol 0.200 mol

We have 0.2 mol NO3⁻ in 300. mL=0.300 L of solution,
so
0.200 mol NO3⁻ / 0.300 L solution ≈ 0.667 mol NO3⁻ /L solution = 0.667 M

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3 years ago

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WARRIOR [948]

Answer:

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2 years ago

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hram777 [196]

Answer:

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Explanation:

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2 years ago

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AnnZ [28]

Answer:

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3 years ago

A solution contains an unknown mass of dissolved barium ions. When sodium sulfate is added to the solution, a white precipitate

AlladinOne [14]

The given question is incomplete. The complete question is as follows.

A solution contains an unknown mass of dissolved barium ions. When sodium sulfate is added to the solution, a white precipitate forms. The precipitate is filtered and dried and then found to have a mass of 212 mg. What mass of barium was in the original solution? (Assume that all of the barium was precipitated out of solution by the reaction.)

Explanation:

When $Ba^{2+}$ and $Na_{2}SO_{4}$ are added then white precipitate forms. And, reaction equation for this is as follows.

$Ba^{2+} + SO^{2-}_{4} \rightarrow BaSO_{4}$

It is given that mass (m) is 212 mg or 0.212 g (as 1 g = 1000 mg). Molecular weight of $BaSO_{4}$ is 233.43.

Now, we will calculate the number of moles as follows.

No. of moles = mass × M.W

= $\frac{0.212}{233.43}$

= 0.00091 mol of $BaSO_{4}$

Hence, it means that 0.00091 mol of $Ba^{2+}$ . Now, we will calculate the mass as follows.

Mass = moles × MW

= $0.00091 \times 137.327$

= 0.124 grams or 124 mg of barium

Thus, we can conclude that mass of barium into the original solution is 124 mg.

8 0

3 years ago