PLEASE HELP!!

Answer the following questions using the data below:

Data :Trial 1 :Trial 2

Mass of empty crucible with lid: 26.679 grams 26.698 grams

Mass of Mg metal, crucible, and lid: 26.934 grams 27.051 grams

Mass of MgO, crucible, and lid: 27.097 grams 27.274 grams

Balanced Chemical Equation for reaction: 2 MG(s) + O2(g) = 2 MGO(s)

Mass of magnesium for each trial:

Trial 1: 0.255g

Trial 2: 0.353g

Actual yield of magnesium oxide for each trial:

Trial 1: 0.418g

Trial 2: 0.576g

Question 1: Calculate the theoretical yield of MgO for each trial:

Question 2: Determine the percent yield of MgO for your experiment for each trial:

Question 3: Determine the average percent yield of MgO for the two trials:

Question 1:

Question 2:

Question 3:

Explanation:

1. Organize all the data in one table:

Data:

                                                           (masses in grams)

                                                             Trial 1      Trial 2

Mass of empty crucible with lid:          26.679       26.698  

Mass of Mg metal, crucible, and lid:    26.934       27.051

Mass of MgO, crucible, and lid:            27.097        27.274

Mass of Mg metal                                 0.255            0.353

Actual yield of MgO                             0.418g           0.576

2. Question 1: Calculate the theoretical yield of MgO for each trial

a) Balanced chemical equation

b) Mole ratio

c) Convert the mass of Mg into number of moles

i) Formula: number of moles = mass in grams / molar mass

ii) Molar mass of Mg (same atomic mass): 24.305 g/mol

iii) Moles of Mg available:

Trtial 1:

Trial 2:

d) Use the mol ratio to calculate the moles of MgO

As per the stoichiometric ratio, one mol of Mg yield one mol of MgO. Thus:

c) Convert the number of moles to mass:

i) Formula:

Trial 1: 0.01049 mol × 40.3044 g/mol = 0.4229g = 0.423g

Trial 2: 0.01452 mol × 40.3044 g/mol = 0.5854g = 0.585 g

Those are the theoretical yields fore ach trial: 0.423g and 0.585 g

3. Question 2. Determine the percent yield of MgO for each trial

a) Formula:

b) Trial 1

c) Trial 2:

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

You can add the two percent yields and divide by 2:

in 1783 cavendish was working with his hydrogen gas and found that when he burned it and condensed the vapor it was water. this was very significant since it disproved the greek theory of elements completely since water was not a simple substance but the product of two gases. it was lavoisier who hearing of the experiment gave cavendish's gas the name hydrogen ("water-producer") and added that hydrogen burned by combining with oxygen and that water was a hydrogen-oxygen combination.

explanation:

question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation:   if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

question answered by

(jacemorris04)