CHEMISTRY MIDDLE SCHOOL

What chemicals are produced during the use of e-cigarettes

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Answer 1
Answer:

Answer:

tar and nicotine which are produced when the tobacco leaves are burning.

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What’s the compound name of Ca2N

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Calcium and Nitrogen

Why aren't the atomic masses of most elements whole numbers

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Isotopes and most average aren't whole numbers

A white powder is added to a solution. The images show observations made before the powder is added, just after the powder has been added, and a little while later. (The liquid in the small beaker is phenol red solution.) What evidence shows that a chemical change has taken place?

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Change in colour as well as liberating gas are the evidences that chemical reaction or change taken place.

The change in colour of the solution provides us information or evidence is that chemical change has taken place. There are many other ways which provides evidence of taking place of chemical change is the release of gas from the solution after adding powder or any substance.

There are some reactions in which new substance is formed which is also an identification of chemical change so we can conclude that change in colour as well as liberating gas are the evidences that chemical reaction or change taken place.

Learn more: brainly.com/question/12017204

D.All of above. Hope it helps!!!

How to do balancing equations

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It's quite hard to explain without an example, but I'll give it a go:

Basically, you have to count the number of an element on one side of the equation, then count it on the other, and the aim of balancing the equation is to get the 2 values to be the same (as mass cannot be created or destroyed).

This involves multiplication of the molecule(s) on either side of the equation in order to balance the whole thing. Note this is not like maths, you dont have to do the same to both sides, but you do have to multiply out entire molecules - you cannot multiply just the H2 in H2O for example.

A reasonably basic example of this balancing is:

Fe + Cl2 = FeCl3

The first thing I notice is that there is an even number of Cl atoms on the left, and an odd number on the right, so the first thing I'll do is multiply the one on the right to get it to an even number, ie by 2.

Fe + Cl2 = 2(FeCl3)

Expanding that gives Fe2Cl6 (although you dont write it like that). So we need to get 6 Cl atoms and 2 Fe atoms by multiplying the left hand molecules by amounts if possible. If you look at that, we can achieve that, by multiplying Cl2 by 3, and Fe by 2.

2Fe + 3Cl2 = 2FeCl3
And there you have your balanced equation, you'll notice that the number of any given element has the same number of atoms on both sides of the equation.

1. Magnesium is the limiting reactant in this experiment. Calculate the theoretical yield of MgO for trial 1 and 2.2. Determine the percent yield of MgO for your experiment for trial 1 and 2.

3. Determine the average percent yield of MgO for the two trials.​

Answers

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 %

How many grams of water can be produced when 11.7 moles of ethane (C2H6) react with excess oxygen gas? Unbalanced equation: C2H6 + O2 → CO2 + H2O

Answers

 2 C2H6 + 7 O2 → 4 CO2 + 6 H2O
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