Why is density a characteristic property of matter?

Yo sup??

Balancing a chemical equation is more like solving mathematical equation.

You just need do the trial and error thing.

Just plug in some random numbers so that the number of atoms on both the sides become the same....that's all.....rest will come by practise.

Hope this helps

Answer:

Explanation:

Balancing Chemical Equations

Let's take a look at this scale. We can see that it is unbalanced, with the right (red) side, weighing more than the left (blue) side. In order for the two sides to be balanced, we need to put a little more mass on the left side until they are the same mass.

Unbalanced Scale

Just like we want the scale to be balanced on both sides, a chemical equation should also be balanced on both sides. A chemical equation shows us the substances involved in a chemical reaction - the substances that react (reactants) and the substances that are produced (products). In general, a chemical equation looks like this:

Chemical Equation: General Form

According to the law of conservation of mass, when a chemical reaction occurs, the mass of the products should be equal to the mass of the reactants. Therefore, the amount of the atoms in each element does not change in the chemical reaction. As a result, the chemical equation that shows the chemical reaction needs to be balanced. A balanced chemical equation occurs when the number of the atoms involved in the reactants side is equal to the number of atoms in the products side.

Let's take a look at an equation representing a chemical reaction:

Balancing Chemical Equations: Example  

In this chemical reaction, nitrogen (N2) reacts with hydrogen (H) to produce ammonia (NH3).The reactants are nitrogen and hydrogen, and the product is ammonia. If we look at this equation, we can see that the equation is not balanced.

Balancing Chemical Equations Example

The equation is not balanced because in the reactants side, there are 2 nitrogen (N) atoms and 2 hydrogen (H) atoms. In the products side, there are 1 nitrogen (N) atoms and 3 hydrogen (H) atoms. The number of the atoms is not balanced on both sides.

To balance the chemical equation above, we need to make use of coefficients. A coefficient is a number that we place in front of a chemical formula. In the chemical equation, to make the number of nitrogen (N) atoms equal on both sides, first, we place a coefficient of 2 in front of NH3.

Balancing Chemical Equations Example

Once we do that, the number of nitrogen (N) atoms on both sides is balanced. However, the number of hydrogen (H) atoms is not balanced on both sides. We need to make use of another coefficient in front of H2. This time, we put a coefficient of 3 in front of H2 to balance the chemical equation.

Balancing Chemical Equations Example

The equation above is now balanced. There are 2 nitrogen (N) atoms and 6 hydrogen (H) atoms on both the reactants and products side. Since there is no coefficient in front of N2, that means the coefficient is equal to 1.

Steps to Balance an Equation

Practice always makes perfect. In general, to balance an equation, here are the things we need to do:

Count the atoms of each element in the reactants and the products.

Use coefficients; place them in front of the compounds as needed.

The steps are simple, but it is a process of trial and error. Let's take a look at a few more example equations and techniques that can be used to balance each one.

Example 1

This is a reaction between methane (CH4) and oxygen (O2), producing carbon dioxide (CO2) and water (H2O).

Balancing Chemical Equations Example 1

The reaction shown is a combustion reaction: a compound reacts with oxygen and produces carbon dioxide and water. The technique is to balance the carbon (C) atoms first, then the hydrogen (H) atoms, and then the oxygen (O) atoms.

In this case, the carbon (C) atoms are already balanced. So now we look at the hydrogen (H) atoms. There are 4 hydrogen (H) atoms on the reactants side and 2 hydrogen (H) atoms on the products side. To balance them, we put a coefficient of 2 in front of H2O

Answer: You can apply the fact that water changes from liquid to gas when heated to its boiling point.

Explanation:

Salt solution contains salt dissolved in water. The water can be separated from the salt by evaporation of the solution to dryness. This exploits the fact that water changes from liquid to gas at it's boiling point. This is an example of change of state. When heated strongly, all the water present will be converted to vapour and the solid salt reappears.

Answer:

Part 1

Theoretical yield of MgO for trial 1 = 0.84 g

Theoretical yield of MgO for trial 2 = 1.01 g

Part 2

Percent yield trial 1 = 28.6 %

Percent yield trial 2 = 49.9 %

Part 3

Average percent yield of MgO for two trial = 39.25 %

Explanation:

Part 1.

Data Given

                                                              Trial 1                     Trial 2

mass of empty crucible and lid:          26.679 g               26.685 g

mass of Mg metal, crucible and lid:    26.931 g               26.988 g

mass of MgO, crucible and lid:            27.090 g              27.179 g

Theoretical yield of MgO for trial 1 and 2 = ?

Solution:

As Mg is limiting reagent so amount of MgO depends on the amount of Mg.

So, now we will look for the reaction to calculate theoretical yield

MgO form by the following reaction:

               Mg  +  O₂ --------->  2 MgO

              1 mol                        2 mol

Convert moles to mass

Molar mass of Mg = 24 g/mol

Molar mass of MgO = 24 + 16 = 40 g/mol

So,

                     Mg        +         O₂      --------->     2 MgO

            1 mol (24 g/mol)                                  2 mol(40 g/mol)

                   24 g                                                    80 g

So,

24 g of Mg gives 80 g of MgO

To Calculate theoretical yield of MgO for Trial 1

First we look for the mass of Mg in the Crucible

• mass of Mg for trial 1

Mass of Mg = mass of Mg metal, crucible and lid - mass of empty crucible and lid

       Mass of Mg = 26.931 g - 26.679 g

       Mass of Mg = 0.252 g

As we come to know that 24 g of Mg gives 80 g of MgO, then amount of Mg from trial 1 that is 0.252 g will produce how many grams of MgO

Apply unity formula

               24 g of Mg ≅ 80 g of MgO

               0.252 g of Mg ≅ X g of MgO

Do cross multiplication

               X g of MgO = 0.252 g x 80 g / 24 g

               X g of MgO = 0.84 g

So the theoretical yield of MgO is  0.84 g

--------------

To Calculate theoretical yield of MgO for Trial 2

First we look for the mass of Mg in the Crucible

• mass of Mg for trial 2

Mass of Mg = mass of Mg metal, crucible and lid - mass of empty crucible and lid

      Mass of Mg = 26.988 g - 26.685 g

      Mass of Mg = 0.303 g

As we come to know that 24 g of Mg gives 80 g of MgO, then amount of Mg from trial 2 that is 0.303 g will produce how many grams of MgO

Apply unity formula

               24 g of Mg ≅ 80 g of MgO

                0.303 g of Mg ≅ X g of MgO

Do cross multiplication

               X g of MgO = 0.303 g x 80 g / 24 g

               X g of MgO = 1.01 g

So the theoretical yield of MgO is  1.01 g

__________________________

Part 2

percent yield of MgO for trial 1 and 2 = ?

Solution:

For trial 1

To calculate percent yield we have to know about actual yield of MgO

• mass of MgO for trial 1

Mass of MgO = mass of MgO, crucible and lid - mass of empty crucible and lid

    Mass of MgO =  27.090 g -  26.685 g

    Mass of MgO =  0.24 g

And we also know that

Theoretical yield of MgO for trial 1 = 0.84 g

Formula used

       Percent yield = actual yield / theoretical yield x 100

put values in above formula

       Percent yield =  0.24 g / 0.84 g x 100

       Percent yield = 28.6 %

--------------

For trial 2

To calculate percent yield we have to know about actual yield of MgO

• mass of MgO for trial 2

Mass of MgO = mass of MgO, crucible and lid - mass of empty crucible and lid

    Mass of MgO =  27.179 g -  26.685 g

    Mass of MgO =  0.494 g

And we also know that

Theoretical yield of MgO for trial 2 = 1.01 g

Formula used

       Percent yield = actual yield / theoretical yield x 100

put values in above formula

       Percent yield =   0.494 g/ 1.01 g x 100

       Percent yield = 49.9 %

--------------

Part 3

average percent yield of MgO for the two trials =?

Solution:

As we know

Percent yield trial 2 = 28.6 %

Percent yield trial 2 = 49.9 %

Formula used

Average percent yield = percent yield trial 1 + percent yield trial 2 / 2

Put values in above formula

           Average percent yield = 28.6 + 49.9  / 2

           Average percent yield = 78.5 / 2

           Average percent yield = 39.25 %

Average percent yield of MgO for two trial = 39.25 %