Please show information to support your answerWhich of the following is a sign of a chemical reaction?


A. A liquid changes to a solid.


B. A solid changes to a gas.


C. A gas is released from a liquid.


D. A liquid changes to a gas.

Answer:

Gay-Lussac’s law

Explanation:

Gay's law states that the pressure of a given mass of gas varies directly with the absolute temperature of the gas, when the volume is kept constant.

The correct option is C.

When writing chemical equations, the chemical symbols of the element are usually used together with numbers to indicate the ratio of each element in the chemical compounds. The additional sign is usually used in between the reactants and the products. The arrow sign in used to show the direction in which the reaction is proceeding. In the correct option given above, the arrow sign is used to indicate that the reaction is moving in the forward direction.

Answer:

C) C+ O -> CO2

Explanation:

I took the test k12

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

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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 %

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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 %

Answer:

Explanation:

Q1. For assuring accuracy and precision we have to take into account statistical values such as absolut error, relative error and standard deviation. Errors(absolute and relative) are related to accuracy, and they tell you how much you are getting away from the correct answer. Standard deviation tells you how precisa your measure was, that means how many times you hit on the objective.

standard deviation can be added to the result as

Q2.

Q1: sort your numbers into numerical order so you can determine the highest and lowest measured values. and then subtract the lowest measured value from the highest measured value. Now determine that the answer is the precision.

Q2: In one meter there are 100 centemeter. Now you got 5.8 miles per hour which will become 580 centemeter per hour. In addition, there are 60 minutes in an hour. Based on what we know, 580 centemeters per hour will and should become 580/60 cm/min

Answer: 9.27 g of Na

Explanation:Please see attachment for explanation