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3 years ago

How does the law of conservation of mass apply to this reaction: C2H4+3O2----> 2H2O+2CO2

Chemistry

2 answers:

Alexeev081 [22]3 years ago

5 0

The answer is; C

There are 2 Carbons, 4 Hydrogens, and 6 oxygens on both sides

All the moles of the elements on the right of the reaction equation are balanced with those on the left of the equation. These moles are indicated by the number in front of the symbols of the elements. This ensures that the first law of thermodynamics is observed. This law states that no matter or energy can be destroyed or created; It can only be transformed.

aliina [53]3 years ago

3 0

Answer:

the answer is C

Explanation:

I took the quiz on k12 and got a hundred percent.

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Determine molar mass! help please

bezimeni [28]

Answer:

the molar mass is the mass of a given chemical element or chemical compound divided by the amount of substance

8 0

2 years ago

Explain your observations in term of how tempurature affects paritcle motion and how liquid changes to a solid

Rus_ich [418]

Answer:

See explanation

Explanation:

Temperature is defined as a measure of the average kinetic energy of the molecules of a body.

When a substance is heated, the kinetic energy of its molecules increases as the temperature increases; hence the particles of the substance moves faster with increasing temperature.

When heat is withdrawn from a liquid, the temperature decreases and the average kinetic energy of the molecules decreases. The molecules become less energetic hence the liquid changes into solid

4 0

3 years ago

Which probably best indicates that a compound contains an ionic bond

alex41 [277]

I’m thinking

It conducts electricity when Molten.

5 0

3 years ago

Vsevolod [243]

Answer:

2 half-lives=0.8

6 half-lives= 0.05

Explanation

Half-lives are constant and always decrease by half, implying that the concentration decreases by half at a consistent rate.

3.2/2= 1.6/2= 0.8 is two half-lives

3.2/2= 1.6/2= 0.8/2= 0.4/2= 0.2/2= 0.1/2=0.05 is six half-lives

3 0

2 years ago

A 30.0-g piece of metal at 203oC is dropped into 400.0 g of water at 25.0 oC. The water temperature rises to 29.0 oC. Calculate

oee [108]

The specific heat of the metal is 2.4733 J/g°C.

Given the following data:

Initial temperature of water = 25.0°C

Final temperature of water = 29.0°C

Mass of water = 400.0 g

Mass of metal = 30.0 g

Temperature of metal = 203.0°C

We know that the specific heat capacity of water is 4.184 J/g°C.

To find the specific heat of the metal (J/g°C):

Heat lost by metal = Heat gained by water.

$Q_{metal} = Q_{water}$

Mathematically, heat capacity or quantity of heat is given by the formula;

$Q = mc\theta$

<u>Where:</u>

Q is the heat capacity or quantity of heat.

m is the mass of an object.

c represents the specific heat capacity.

∅ represents the change in temperature.

Substituting the values into the formula, we have:

$30(203)c = 400(4.184)(29 \; -\; 20)\\\\6090c = 1673.6(9)\\\\6090c = 15062.4\\\\c = \frac{15062.4}{6090}$

Specific heat capacity of metal, c = 2.4733 J/g°C

Therefore, the specific heat of the metal is 2.4733 J/g°C.

8 0

1 year ago