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leva [86]
3 years ago
15

It is defined as the number of particles per mole of a substance. *

Chemistry
1 answer:
luda_lava [24]3 years ago
7 0

Answer:

Avogadro's number

Explanation:

Avogadro's number

defined as the number of elementary particles (molecules, atoms, compounds, etc.) per mole of a substance. It is equal to 6.022×1023 mol-1 and is expressed as the symbol NA.

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Elena makes the table below to determine the number of atoms of each element in the chemical formula 3(NH4)2SO4
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Answer:

she should not have multiplied the sulfur atoms by the subscript 4

Explanation:

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Calculate the volume of the gas when the pressure of the gas is 1.60 atm at a temperature of 298 K. There are 160. mol of gas in
LekaFEV [45]

Answer:

2445 L

Explanation:

Given:  

Pressure = 1.60 atm

Temperature = 298 K

Volume = ?

n =  160 mol

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 08206 L.atm/K.mol

Applying the equation as:

1.60 atm × V = 160 mol × 0.08206 L.atm/K.mol × 298 K  

<u>⇒V = 2445.39 L</u>

Answer to four significant digits, Volume = 2445 L

6 0
3 years ago
What are the examples of physical chemistry?
OverLord2011 [107]

ANSWER:

Physical chemistry is the branch of chemistry that deals with the physical structure of chemical compounds, the way they react with other matter and the bonds that hold their atoms together. An example of physical chemistry is nitric acid eating through wood.

3 0
2 years ago
2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2
nadezda [96]

Answer:

Kc = 3.90

Explanation:

CO reacts with H_2 to form CH_4 and H_2O. balanced reaction is:

CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)

No. of moles of CO = 0.800 mol

No. of moles of H_2 = 2.40 mol

Volume = 8.00 L

Concentration = \frac{Moles}{Volume\ in\ L}

Concentration of CO = \frac{0.800}{8.00} = 0.100\ mol/L

Concentration of H_2 = \frac{2.40}{8.00} = 0.300\ mol/L

                 CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)

Initial            0.100      0.300             0   0

equi.            0.100 -x    0.300 - 3x     x    x

It is given that,

at equilibrium H_2O (x) = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium H_2 = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium CH_4 = 0.0386 M

Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}

Kc=\frac{0.0386 \times 0.0386}{(0.184)^3 \times 0.0614} =3.90

8 0
4 years ago
Tính [H] và [OH] của dung dịch HC1 0,1M? Tính pH của dung dịch?
Brrunno [24]

Answer:

po jakiemu to jest XD

Explanation:

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