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3 years ago

How many total atoms are in 0.450 moles of carbon dioxide?

1 answer:

4 0

Avogadro's number tells you how many atoms are in 1 mol of every element.

So,
1 mol : 6.022x10^23 atoms = 0.450 moles : x atoms

There are 2.7099x10^23 atoms in 0.450 moles

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Magma can partially crystallize at depth and then rise to shallow depths where the remaining magma solidifies. The early-formed

Doss [256]

<h2>Phenocrysts and Porphyritic texture </h2>

Explanation:

The early formed crystals are of phenocrysts and the texture of these crystals is porphyritic texture.
This crystallization occurs when early-forming plagioclase crystals which are rich in calcium start coating with plagioclase crystals which are rich in sodium.
On cooling, the magma is then processed in a volcanic eruption, after the eruption the liquid which is left behind will start cooling and forms a porphyritic texture.

5 0

3 years ago

What are to long term impacts that the concept of matter cannot be destroyed or created

Tcecarenko [31]

As long as matter cannot be destroyed or created , nothing can be gained or lost.

there is zero impact and hence one cannot numerate the impact

8 0

3 years ago

In a given substance, as the temperature of a substance increases, the volume of that substance

steposvetlana [31]

C. Increases. Increasing temperature=Increasing Volume

3 0

3 years ago

What is the entropy change in the environment when 5.0 MJ of energy is transferred thermally from a reservoir at 1000 K to one a

Leni [432]

Answer:

The entropy change in the environment is 3.62x10²⁶.

Explanation:

The entropy change can be calculated using the following equation:

$\Delta S = \frac{Q}{k_{B}}(\frac{1}{T_{f}} - \frac{1}{T_{i}})$

Where:

Q: is the energy transferred = 5.0 MJ

$k_{B}$ : is the Boltzmann constant = 1.38x10⁻²³ J/K

$T_{i}$ : is the initial temperature = 1000 K

$T_{f}$ : is the final temperature = 500 K

Hence, the entropy change is:

$\Delta S = \frac{5.0 \cdot 10^{6} J}{1.38 \cdot 10^{-23} J/K}(\frac{1}{500 K} - \frac{1}{1000 K}) = 3.62 \cdot 10^{26}$

Therefore, the entropy change in the environment is 3.62x10²⁶.

I hope it helps you!

7 0

3 years ago

The heat of reaction for the combustion of propane is –2,045 kJ. This reaction is: C3H8(g) +502 (g) 3CO2 (g) + 4H2O (g). Determi

Ede4ka [16]

Answer:

$\Delta _fH_{C_3H_8}=-102.7kJ/mol$

Explanation:

Hello,

In this case, we can consider that the given heat of combustion is indeed the heat of reaction since it corresponds to the combustion of propane, which is computed by using the heat formation of all the involved species as shown below:

$\Delta _cH=3*\Delta _fH_{CO_2}+4*\Delta _fH_{H_2O}-\Delta _fH_{C_3H_8}-5*\Delta _fH_{O_2}$

Thus, since the heat of formation of gaseous carbon dioxide is -393.5 kJ/mol, water -241.8 kJ/mol and oxygen 0 kJ/mol, the heat of formation of propane is:

$\Delta _fH_{C_3H_8}=3*\Delta _fH_{CO_2}+4*\Delta _fH_{H_2O}-5*\Delta _fH_{O_2}-\Delta _cH\\\\\Delta _fH_{C_3H_8}=3*(-393.5)+4*(-241.8)-5*0-(-2045)\\\\\Delta _fH_{C_3H_8}=-102.7kJ/mol$

Best regards.

8 0

3 years ago