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Helga [31]
3 years ago
7

I will give brainiest! PLEASE ANSWER ASAP.How does the law of conservation of mass apply to this reaction: C2H4 + 3O2 → 2H2O + 2

CO2 ?
Only the oxygen needs to be balanced. There are equal numbers of hydrogen and carbon.


The equation needs to be balanced. There are fewer oxygen atoms in the equation than hydrogen or carbon.


The law of conservation of mass has already been applied. There is an equal number of each element on both sides of the equation.


Each element needs to be balanced.
Chemistry
1 answer:
blondinia [14]3 years ago
7 0
<h3>Answer:</h3>

The law of conservation of mass has already been applied. There is an equal number of each element on both sides of the equation.

<h3>Explanation:</h3>

<u>We are given;</u>

  • The equation;

C₂H₄ + 3O₂ → 2H₂O + 2CO₂

We need to ask ourselves the following questions;

Is the equation balanced?

  • Yes, the equation because the number of atoms of each element is equal on both sides of the equation.
  • That is; Reactants: 2 carbon atoms, 4 hydrogen atoms and 6 oxygen atoms.
  • Products: 2 carbon atoms, 4 hydrogen atoms and 6 oxygen

What is the law of conservation of mass?

  • According to the law of conservation of mass, the mass of reactants should always be equal to the mass of the products in chemical reactions.

What makes the equation obey the law of conservation of mass?

  • For an equation to obey the law of conservation of mass, it must be balanced.
  • In this case, <u>the equation given is already balanced and therefore obeys the law of conservation of mass</u>.
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Answer:

The products are Calcium oxide and Carbon dioxide.

Explanation:

When calcium carbonate is heated, thermal decomposition occurs.

Calcium calcium → Calcium oxide + Carbon dioxide

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A scientist observed a 5-mL sample of an unknown substance. It is a blue liquid with a density of 1.2 g/cm. the substance reacts
LenaWriter [7]
Characteristic properties can be used to describe and identify the substances, while non-characteristic properties, although can be used to describe the substances, cannot be used to identify them.

Temperature, mass, color, shape and volume are examples of non-characteristic properties.

Density, boiling point, melting point, chemical reactivity are examples of characteristic properties.

List of the properties observed by the scientist:
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Property                                     Type of property
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Volume: 5 ml                              non-characteristic
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Color: blue                                 non-characteristic
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State: liquid                                characteristic
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density: 1.2 g/cm                        characteristic
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5 0
3 years ago
Ca(s) + O₂(g) + S(s) → CaSO4(s)
san4es73 [151]

The mass of Calcium required to complete this reaction is 4.008 g.

  • Law of conservation of mass states that In a closed system, mass cannot be produced or destroyed, but it can be changed from one form to another.
  • The mass of the chemical constituents before a chemical reaction is equal to the mass of the constituents after the reaction.
  • In several disciplines, including chemistry, mechanics, and fluid dynamics, the idea of mass conservation is widely applied.

In the given reaction mass of product after completion of reaction is 13.614 g that means total mass of constituents before reaction should also be 13.614.

So,

mass of Ca + mass of O₂ + mass of S = mass of CaSO4

Ca + 6.400 g + 3.206 g = 13.614 g

mass of Ca = 13.614 - 9.606 = 4.008 g

Therefore, by law of conservation of mass 4.008 g of Ca is required for the completion of the reaction.

Learn more about mass conservation here:

brainly.com/question/2030891

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5 0
2 years ago
CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7
padilas [110]

<u>Answer:</u> The \Delta G for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)

We are given:

\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol

To calculate \Delta G^o_{rxn} for the reaction, we use the equation:

\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]

For the given equation:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J

Conversion factor used = 1 kJ = 1000 J

The expression of K_p for the given reaction:

K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}

We are given:

p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm

Putting values in above equation, we get:

K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85

To calculate the gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_p

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = 8.314J/K mol

T = Temperature = 25^oC=[25+273]K=298K

K_p = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol

Hence, the \Delta G for the reaction is 54.425 kJ/mol

7 0
3 years ago
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