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WITCHER [35]
3 years ago
14

Which of the following statements about metalloids is true?

Chemistry
2 answers:
Klio2033 [76]3 years ago
7 0
Metalloids had properties that fall between those of metals and nonmetals (I believe that to be correct-.-)
aleksklad [387]3 years ago
3 0

Answer: Option (c) is the correct answer.

Explanation:

Metalloids are the substances or elements that show properties of both metals and non-metals.

Metals are good conductors of heat and electricity. They have shiny surface and can be drawn into thin wires and sheets.

Whereas non-metals are bad conductors of heat and electricity. They are brittle in nature and are not shiny.

Similarly, metalloids are shiny but they are brittle at the same time with a relatively good electrical conductivity.

Thus, we can conclude that the statement, metalloids have properties that fall between those of metals and nonmetals, is true about metalloids.

You might be interested in
What is the molar out of a solution that contains 33.5g of CaCl2 in 600.0mL of water
omeli [17]

Answer:

Here's what I got.

Explanation:

Interestingly enough, I'm not getting

0.0341% w/v

either. Here's why.

Start by calculating the percent composition of chlorine,

Cl

, in calcium chloride, This will help you calculate the mass of chloride anions,

Cl

−

, present in your sample.

To do that, use the molar mass of calcium chloride, the molar mass of elemental chlorine, and the fact that

1

mole of calcium chloride contains

2

moles of chlorine atoms.

2

×

35.453

g mol

−

1

110.98

g mol

−

1

⋅

100

%

=

63.89% Cl

This means that for every

100 g

of calcium chloride, you get

63.89 g

of chlorine.

As you know, the mass of an ion is approximately equal to the mass of the neutral atom, so you can say that for every

100 g

of calcium chloride, you get

63.89 g

of chloride anions,

Cl

−

.

This implies that your sample contains

0.543

g CaCl

2

⋅

63.89 g Cl

−

100

g CaCl

2

=

0.3469 g Cl

−

Now, in order to find the mass by volume percent concentration of chloride anions in the resulting solution, you must determine the mass of chloride anions present in

100 mL

of this solution.

Since you know that

500 mL

of solution contain

0.3469 g

of chloride anions, you can say that

100 mL

of solution will contain

100

mL solution

⋅

0.3469 g Cl

−

500

mL solution

=

0.06938 g Cl

−

Therefore, you can say that the mass by volume percent concentration of chloride anions will be

% m/v = 0.069% Cl

−

−−−−−−−−−−−−−−−−−−−

I'll leave the answer rounded to two sig figs, but keep in mind that you have one significant figure for the volume of the solution.

.

ALTERNATIVE APPROACH

Alternatively, you can start by calculating the number of moles of calcium chloride present in your sample

0.543

g

⋅

1 mole CaCl

2

110.98

g

=

0.004893 moles CaCl

2

To find the molarity of this solution, calculate the number of moles of calcium chloride present in

1 L

=

10

3

mL

of solution by using the fact that you have

0.004893

moles present in

500 mL

of solution.

10

3

mL solution

⋅

0.004893 moles CaCl

2

500

mL solution

=

0.009786 moles CaCl

2

You can thus say your solution has

[

CaCl

2

]

=

0.009786 mol L

−

1

Since every mole of calcium chloride delivers

2

moles of chloride anions to the solution, you can say that you have

[

Cl

−

]

=

2

⋅

0.009786 mol L

−

1

[

Cl

−

]

=

0.01957 mol L

−

This implies that

100 mL

of this solution will contain

100

mL solution

⋅

0.01957 moles Cl

−

10

3

mL solution

=

0.001957 moles Cl

−

Finally, to convert this to grams, use the molar mass of elemental chlorine

0.001957

moles Cl

−

⋅

35.453 g

1

mole Cl

−

=

0.06938 g Cl

−

Once again, you have

% m/v = 0.069% Cl

−

−−−−−−−−−−−−−−−−−−−

In reference to the explanation you provided, you have

0.341 g L

−

1

=

0.0341 g/100 mL

=

0.0341% m/v

because you have

1 L

=

10

3

mL

.

However, this solution does not contain

0.341 g

of chloride anions in

1 L

. Using

[

Cl

−

]

=

0.01957 mol L

−

1

you have

n

=

c

⋅

V

so

n

=

0.01957 mol

⋅

10

−

3

mL

−

1

⋅

500

mL

n

=

0.009785 moles

This is how many moles of chloride anions you have in

500 mL

of solution. Consequently,

100 mL

of solution will contain

100

mL solution

⋅

0.009785 moles Cl

−

500

mL solution

=

0.001957 moles Cl

−

So once again, you have

0.06938 g

of chloride anions in

100 mL

of solution, the equivalent of

0.069% m/v

.

Explanation:

i think this is it

8 0
2 years ago
Plz help (it’s a picture)
Feliz [49]
Umm I’ll figure it out rn! Will come back in 1 min
7 0
3 years ago
What two properties affect the<br> density of ocean water?
mihalych1998 [28]

Answer: temperature and salinity.

Explanation:

5 0
2 years ago
What is the advantage of having valves in the
fenix001 [56]
The valves stop the blood from flowing backwards
6 0
3 years ago
Rank the solutions below in order of increasing acidity. (Drag and drop into the appropriate area)
Vladimir79 [104]

Answer:

0.1 M NaOH, 3 M NH3, 0.01 M CH3COOH, 0.01 M H2SO4, 0.1 M HCl

Explanation:

Strong acids are more acids than weak acids. In the same way, strong bases are more basic than weak bases that are in the same concentration.

Then, the more concentrated acid or base will be more acidic or basic.

CH3COOH. Weak acid

NaOH. Strong base

H2SO4. Strong acid

NH3. Weak base.

HCl. Strong acid

The less acid (More basic):

<h3>0.1 M NaOH, 3 M NH3, 0.01 M CH3COOH, 0.01 M H2SO4, 0.1 M HCl</h3>

Strong base, weak base, weak acid, diluted strong acid, undiluted strong acid

7 0
3 years ago
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