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vfiekz [6]
3 years ago
5

If you are given the number of molecules in an unidentified chemical compound can you calculate the number of moles in this samp

le
Chemistry
2 answers:
Sever21 [200]3 years ago
8 0
Yes.
Mols x Avogadro's Constant (6.02x10^23) = number of molecules.
You can rearrange that formula to find the number of Mols. Hope this helped
alexira [117]3 years ago
7 0

<u>Answer:</u> By using mole concept

<u>Explanation:</u>

We are given:

An unknown compound having some number of molecules.

To calculate the number of moles that are contained in a sample, we use mole concept:

According to mole concept:

1 mole of any compound contains 6.022\times 10^{23} number of molecules.

Using this relation and applying unitary method, we can easily calculate the number of moles of a compound.

<u>For Example:</u> A sample of water contains 10.8396\times 10^{23} number of molecules.

So, by using mole concept:

6.022\times 10^{23} number of molecules are contained in 1 mole of a compound.

So, 10.8396\times 10^{23} number of molecules will be contained in \frac{1}{6.022\times 10^{23}}\times 10.8396\times 10^{23}=1.8 moles of water

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maksim [4K]
The answer is B because of the calculator
3 0
2 years ago
What is the molar mass of an unknown gas with a density of 2.00 g/L at 1.00 atm and 25.0 °C?
soldier1979 [14.2K]

Answer:

Explanation:Explanation:

Your starting point here will be the ideal gas law equation

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

P

V

=

n

R

T

a

a

∣

∣

−−−−−−−−−−−−−−−

, where

P

- the pressure of the gas

V

- the volume it occupies

n

- the number of moles of gas

R

- the universal gas constant, usually given as

0.0821

atm

⋅

L

mol

⋅

K

T

- the absolute temperature of the gas

Now, you will have to manipulate this equation in order to find a relationship between the density of the gas,

ρ

, under those conditions for pressure and temperature, and its molar mass,

M

M

.

You know that the molar mass of a substance tells you the mass of exactly one mole of that substance. This means that for a given mass

m

of this gas, you can express its molar mass as the ratio between

m

and

n

, the number of moles it contains

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

M

M

=

m

n

a

a

∣

∣

−−−−−−−−−−−−−

(

1

)

Similarly, the density of the substance tells you the mass of exactly one unit of volume of that substance.

This means that for the mass

m

of this gas, you can express its density as the ratio between

m

and the volume it occupies

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

ρ

=

m

V

a

a

∣

∣

−−−−−−−−−−−

(

2

)

Plug equation

(

1

)

into the ideal gas law equation to get

P

V

=

m

M

M

⋅

R

T

Rearrange to get

P

V

⋅

M

M

=

m

⋅

R

T

P

⋅

M

M

=

m

V

⋅

R

T

M

M

=

m

V

⋅

R

T

P

Finally, use equation

(

2

)

to write

M

M

=

ρ

⋅

R

T

P

Convert the temperature of the gas from degrees Celsius to Kelvin then plug in your values to find

M

M

=

1.02

g

L

⋅

0.0821

atm

⋅

L

mol

⋅

K

⋅

(

273.15

+

37

)

K

0.990

atm

M

M

=

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

26.3 g mol

−

1

a

a

∣

∣

−−−−−−−−−−−−−−−−

I'll leave the answer rounded to three

7 0
3 years ago
HELP! 10 pts!! || Which does NOT describe a water molecule?
AVprozaik [17]

Answer:

C.)

Explanation:

Hope this helps:)

5 0
3 years ago
A 40.2 g sample of a metal heated to 99.3°C is placed into a calorimeter containing 120 g of water at 21.8°C. The final temper
aliina [53]

Answer:

B) Iron (c=0.45 J/g°C)

Explanation:

Given that:-

Heat gain by water = Heat lost by metal

Thus,  

m_{water}\times C_{water}\times (T_f-T_i)=-m_{metal}\times C_{metal}\times (T_f-T_i)

Where, negative sign signifies heat loss

Or,  

m_{water}\times C_{water}\times (T_f-T_i)=m_{metal}\times C_{metal}\times (T_i-T_f)

For water:

Mass = 120 g

Initial temperature = 21.8 °C

Final temperature = 24.5 °C

Specific heat of water = 4.184 J/g°C

For metal:

Mass = 40.2 g

Initial temperature = 99.3 °C

Final temperature = 24.5 °C

Specific heat of metal = ?

So,  

120\times 4.184\times (24.5-21.8)=40.2\times C_{metal}\times (99.3-24.5)

40.2C_{metal}\left(99.3-24.5\right)=120\times \:2.7\times \:4.184

40.2C_{metal}\left(99.3-24.5\right)=1355.616

C_{metal}=0.45\ J/g^0C

<u>This value corresponds to iron. Thus answer is B.</u>

3 0
3 years ago
What makes noble gases stable?
inn [45]
The correct answer to your question is noble gases are stable <span>due to having the maximum number of valence electrons their outer shell can hold. Meaning their outer shells are stable. 

Hope this helps let me know!</span>
8 0
3 years ago
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