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OLga [1]
1 year ago
11

The magnet below is cut in half. What will be the result

Chemistry
2 answers:
guapka [62]1 year ago
3 0

the cut pieces will have their own north and south pole

Stells [14]1 year ago
3 0
It’s like cutting a snake with a head at the ends of the body. You cut it, it will still have a head(north) and body(south).
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BRAINLIESTTT ASAP!!! PLEASE HELP ME :)
SCORPION-xisa [38]

Every science experiment should follow the basic principles of proper investigation so that the results presented at the end are seen as credible.

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4 0
3 years ago
Which of the following are properties of metalloids?
harkovskaia [24]

Answer:

all of these are properties of metalloids

3 0
2 years ago
Read 2 more answers
Which of the following statements explain why the van der Waals equation must be used to describe real gases? X. interactions be
8090 [49]

Answer:

Statements Y and Z.

Explanation:

The Van der Waals equation is the next one:

nRT = (P + \frac{an^{2}}{V^{2}})(V -nb) (1)

The ideal gas law is the following:

nRT = PV (2)

<em>where n: is the moles of the gas, R: is the gas constant, T: is the temperature, P: is the measured pressure, V: is the volume of the container, and a and b: are measured constants for a specific gas.  </em>

As we can see from equation (1), the Van der Waals equation introduces two terms that correct the P and the V of the ideal gas equation (2),<u> by the incorporation of the intermolecular interaction between the gases and the gases volume</u>. The term an²/V² corrects the P of the ideal gas equation since the measured pressure is decreased by the attraction forces between the gases. The term nb corrects the V of the ideal gas equation, <u>taking into account the volume occuppied by the gas in the total volume, which implies</u> a reduction of the total space available for the gas molecules.          

So, the correct statements are the Y and Z: the non-zero volumes of the gas particles effectively decrease the amount of "empty space" between them and the molecular attractions between gas particles decrease the pressure exerted by the gas.            

Have a nice day!

5 0
3 years ago
How many grams of sodium chloride decompose to yield 15 grams of chlorine gas?
defon

Answer:

24.6g of NaCl

Explanation:

Expression of the reaction:

            2NaCl →  2Na  +  Cl₂

Given parameters:

Mass of Cl₂  = 15g

Unknown:

Mass of NaCl  = ?

Solution:

To solve this problem, we have to use mole relationships.

 Find the number of moles of the mass of the given specie;

    Number of moles  = \frac{mass}{molar mass}  

     Molar mass of Cl₂  = 2(35.5) = 71g/mol

   Number of moles  = \frac{15}{71}   = 0.21mole

Now;

 From the balanced reaction equation;

        1 mole of Cl₂ is produced from 2 moles of NaCl;

      0.21 mole of Cl₂ will be produced from 0.21 x 2 = 0.42mole of NaCl

So,

  Mass of NaCl = number of moles x molar mass

    Molar mass of NaCl  = 23 + 35.5 = 58.5g/mol

 Mass of NaCl  = 0.42 x 58.5 = 24.6g of NaCl

8 0
3 years ago
Calculate the solubility product constant, Ksp, of lead(II) chloride, PbCl2, which has a
V125BC [204]

Answer:

0.0159m

Explanation:

9 M

Explanation:

Lead(II) chloride,  

PbCl

2

, is an insoluble ionic compound, which means that it does not dissociate completely in lead(II) cations and chloride anions when placed in aqueous solution.

Instead of dissociating completely, an equilibrium rection governed by the solubility product constant,  

K

sp

, will be established between the solid lead(II) chloride and the dissolved ions.

PbCl

2(s]

⇌

Pb

2

+

(aq]

+

2

Cl

−

(aq]

Now, the molar solubility of the compound,  

s

, represents the number of moles of lead(II) chloride that will dissolve in aqueous solution at a particular temperature.

Notice that every mole of lead(II) chloride will produce  

1

mole of lead(II) cations and  

2

moles of chloride anions. Use an ICE table to find the molar solubility of the solid

 

PbCl

2(s]

 

⇌

 

Pb

2

+

(aq]

 

+

 

2

Cl

−

(aq]

I

 

 

 

−

 

 

 

 

 

0

 

 

 

 

 

0

C

 

 

x

−

 

 

 

 

(+s)

 

 

 

 

(

+

2

s

)

E

 

 

x

−

 

 

 

 

 

s

 

 

 

 

 

2

s

By definition, the solubility product constant will be equal to

K

sp

=

[

Pb

2

+

]

⋅

[

Cl

−

]

2

K

sp

=

s

⋅

(

2

s

)

2

=

s

3

This means that the molar solubility of lead(II) chloride will be

4

s

3

=

1.6

⋅

10

−

5

⇒

s

=  √ 1.6 4 ⋅ 10 − 5  = 0.0159 M

8 0
3 years ago
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