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ziro4ka [17]
3 years ago
6

Determine the number of grams in each of the quantities 1.39.0 x 1024 molecules Cl2

Chemistry
1 answer:
STatiana [176]3 years ago
8 0

Mass of Cl₂ : 164.01 g

<h3>Further explanation</h3>

A mole is a number of particles(atoms, molecules, ions)  in a substance

This refers to the atomic total of the 12 gr C-12  which is equal to 6.02.10²³, so 1 mole = 6.02.10²³ particles  

Can be formulated :

N = n x No

N = number of particles

n = mol

No = 6.02.10²³ = Avogadro's number

mol Cl₂ :

\tt n=\dfrac{N}{No}\\\\n=\dfrac{1.39.10^{24}}{6.02.10^{23}}\\\\n=2.31

mass Cl₂(MW=71 g/mol) :

\tt mass=mol\times MW\\\\mass=2.31\times 71=164.01

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Perform the following
Dmitry [639]

Answer:

0.5474

Explanation:

3.22 x 0.17

     \/

322 x 17 = 5,474

[] There are four places "after" the decimal in 3.22 and 0.17 combined, meaning we move the decimal to the right four times in our current answer

-> 5,474 becomes 0.5474

Have a nice day!

     I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

4 0
2 years ago
C3H8+3O2 = 3CO2+4H2O what is the enthalpy combustion
dezoksy [38]

Answer:

\Delta H_{comb}=2043.85kJ/mol

Explanation:

Hello there!

In this case, according to the given chemical reaction, it possible for us to set up the expression for the calculation of the enthalpy change as shown below:

\Delta H_r=-\Delta H_{comb}=3\Delta _fH_{CO_2}+4\Delta _fH_{H_2O}-\Delta _fH_{C_3H_8}-3\Delta _fH_{O_2}

Thus, given the values of the enthalpies of formation on the attached file, we obtain:-\Delta H_{comb}=3(-393.5kJ/mol)+4(-241.8kJ/mol)-(-103.85kJ/mol)-3(0kJ/mol)\\\\-\Delta H_{comb}=-2043.85kJ/mol\\\\\Delta H_{comb}=2043.85kJ/mol

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8 0
2 years ago
Suppose it took 108 joules of energy to raise a bar of gold from 25 °C to 29.7°C. Given that the specific heat capacity of gold
Lady bird [3.3K]

Answer:

m = 180 g

Explanation:

Given data:

Energy absorbed = 108 J

Mas of gold = ?

Initial temperature = 25°C

Final temperature = 29.7 °C

Specific heat capacity of gold = 0.128 J/g.°C

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT =29.7 °C - 25°C

ΔT = 4.7 °C

108 J = m ×0.128 J/g.°C ×4.7 °C

108 J = m ×0.60 J/g

m = 108 J/0.60 J/g

m = 180 g

3 0
3 years ago
Physical properties can be observed
Alika [10]

Answer:

Physical properties include: appearance, texture, color, odor, melting point, boiling point, density, solubility, polarity, and many others. That is your answer! Thanks! :)

Explanation:

4 0
3 years ago
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Mercury’s natural state is where the atoms are close to each other but are still free to pass by each other. In which state(s) c
kvv77 [185]

Mercury naturally exists in Liquid state.

On Condensing it can exist in Solid state as well.

Hope it helps...

Regards;

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3 0
3 years ago
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