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ZanzabumX [31]
3 years ago
11

How many magnesium atoms are contained in 3.75 moles of magnesium?

Chemistry
1 answer:
Katarina [22]3 years ago
7 0
In 1 mole of magnesium there are Avogadro's number of atoms are present.
Avogadro's number = 6.023 x 10²³ 
1 mole = 6.023 x 10²³ atoms
3.75 moles = 3.75 x 6.023 x 10²³ 
=2.26 x 10²⁴
So, in 3.75 moles of magnesium there are 2.26 x 10²⁴ atoms of magnesium are present.
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Answer:

2) Gas molecules do not have preferred direction of motion, their motion is completely random. 3) Gas molecules travels in straight line. 4) The time interval of collision between any two gas molecules is very small. 5) The collision between gas molecules and the walls of container is perfectly elastic.

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2 years ago
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Conduction, convection, and radiation are three means by which _____________ may be transferred.
Vaselesa [24]
The correct answer to this question would be heat energy

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Determine whether each statement describes an electromagnetic wave, a mechanical wave, or both types of wave.
KiRa [710]

The question is incomplete, the complete question is;

Which statement describes a difference between electromagnetic and mechanical waves?

A. Mechanical waves cannot be longitudinal, but electromagnetic waves can.

B. Electromagnetic waves cannot move particles, but mechanical waves can.

C. Electromagnetic waves do not require a medium, but mechanical waves do.

D. Mechanical waves do not transfer energy, but electromagnetic waves do.

Answer:

Electromagnetic waves do not require a medium, but mechanical waves do.

Explanation:

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8 0
3 years ago
A solution is prepared by dissolving 215 grams of methanol, ch3oh, in 1000. grams of water. what is the freezing point of this s
Leya [2.2K]

Answer : The freezing point of the solution is, 260.503 K

Solution : Given,

Mass of methanol (solute) = 215 g

Mass of water (solvent) = 1000 g = 1 kg       (1 kg = 1000 g)

Freezing depression constant = 1.86^oC/m=1.86Kkg/mole

Formula used :

\Delta T_f=K_f\times m\\T^o_f-T_f=K_f\times \frac{w_{solute}}{M_{solute}\times w_{solvent}}

where,

T^o_f = freezing point of water = 100^oC=273K

T_f = freezing point of solution

K_f = freezing point constant

w_{solute} = mass of solute

w_{solvent} = mass of solvent

M_{solute} = molar mass of solute

Now put all the given values in the above formula, we get

273K-T_f=(1.86Kkg/mole)\times \frac{215g}{(32g/mole)\times (1kg)}

By rearranging the terms, we get the freezing point of solution.

T_f=260.503K

Therefore, the freezing point of the solution is, 260.503 K

6 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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