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allsm [11]
3 years ago
5

The most active metals on the periodic table are found on the: lower left upper right middle upper left lower right

Chemistry
2 answers:
loris [4]3 years ago
7 0

Answer: Option (a) is the correct answer.

Explanation:

Elements which are most electropositive or electronegative are the most reactive in nature. Therefore, they are active.

Elements of group 1 are also known as alkali metals. Elements of group 1 are lithium, sodium, potassium, rubidium, cesium, and francium.

All these are highly electropositive in nature and thus, are active. These are placed at the top left column of periodic table.

Thus, we can conclude that out of the given options, the most active metals on the periodic table are found on the lower left.

andreev551 [17]3 years ago
4 0

<u>Answer:</u> The correct answer is lower left.

<u>Explanation:</u>

Metals are defined as the elements which loose electrons to attain stable electronic configuration.

Reactivity of metals is defined as the tendency of metals to loose electrons. If a metal looses electrons easily, it means that it is more reactive.

Reactivity increases down the group and decreases across a period for metals.

In the periodic table attached,

  • The elements present in lower left are metals which are most reactive.
  • The elements present in upper right are non-metals which are most reactive.
  • The elements present in middle are metals which are less reactive.
  • The elements present in upper left are metals which are least reactive reactive.
  • The elements present in lower right are non-metals which are least reactive.

Hence, the correct answer is lower left.

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Trava [24]
The correct answer is oceanic crust, 80 km, Hope this helps let me know.
8 0
3 years ago
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
3 years ago
Explain why 2-bromopropane reacts with sodium iodide in acetone over 104 times faster than bromocyclopropane. hint: examine the
Mama L [17]

Answer:-

The reaction of 2-bromopropane reacts with sodium iodide in acetone is an example of Sn2 reaction.

The I - attacks from backside to give the transition state for both.

If we compare the transition state for cyclobromopropane 2-bromopropane then we see in case of cyclobromopropane transition state, one of the H is very close to the incoming I -.

This results in steric strain and less stability of the transition state. Hence 2-bromopropane reacts with sodium iodide in acetone over 104 times faster than bromocyclopropane.

7 0
3 years ago
The synthesis of nitrogen trihydride from nitrogen gas and hydrogen gas is shown by which balanced chemical equation?
Bumek [7]
 The correct answer is:  [B]:  
___________________________________________________________
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___________________________________________________________
Note of interest:
___________________________________________________________
This particular reaction is known as the "</span>Haber process" .
___________________________________________________________
7 0
3 years ago
Read 2 more answers
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I am Lyosha [343]

<span><span>m1</span>Δ<span>T1</span>+<span>m2</span>Δ<span>T2</span>=0</span>

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<span>50.0g×<span>(<span>Tf</span>l–l25.0 °C)</span>+23.0g×<span>(<span>Tf</span>l–l57.0 °C)</span>=0</span>

<span>50.0<span>Tf</span>−1250 °C+23.0<span>Tf</span> – 1311 °C=0</span>

<span>73.0<span>Tf</span>=2561 °C</span>

<span><span>Tf</span>=<span>2561 °C73.0</span>=<span>35.1 °C</span></span>

8 0
3 years ago
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