Answer:
34.3 g
Explanation:
Step 1: Write the balanced equation
2 CH₃CH₂OH ⇒ CH₃CH₂OCH₂CH₃ + H₂O
Step 2: Calculate the moles corresponding to 50.0 g of CH₃CH₂OH
The molar mass of CH₃CH₂OH is 46.07 g/mol.
50.0 g × 1 mol/46.07 g = 1.09 mol
Step 3: Calculate the theoretical moles of CH₃CH₂OCH₂CH₃ produced
The molar ratio of CH₃CH₂OH to CH₃CH₂OCH₂CH₃ is 2:1. The moles of CH₃CH₂OCH₂CH₃ theoretically produced are 1/2 × 1.09 mol = 0.545 mol.
Step 4: Calculate the real moles of CH₃CH₂OCH₂CH₃ produced
The percent yield of the reaction is 85%.
0.545 mol × 85% = 0.463 mol
Step 5: Calculate the mass corresponding to 0.463 moles of CH₃CH₂OCH₂CH₃
The molar mass of CH₃CH₂OCH₂CH₃ is 74.12 g/mol.
0.463 mol × 74.12 g/mol = 34.3 g
The biggest source of vitamin D is the sun.
the formation of cations by using electron dot structures are :
a) Al
.
Al . losing the three valence electrons makes the Al³⁺
.
b) Sr :
Sr : losing the two valence electrons makes Sr²⁺
c) Ba
: Ba , losing the two valence electrons makes it Ba²⁺
A Lewis electron dot diagram is a representation of the valence electrons of an atom that employments specks around the image of the element. The number of dots equals the number of valence electrons within the molecule. These dots are arranged to the right and left and over and underneath the symbol, with no more than two dots on a side. Cations are the positive ions shaped by the loss of one or more electrons. The foremost commonly shaped cations of the representative elements are those that include the loss of all of the valence electrons.
To know more about the lewis electron dot diagram refer to the link brainly.com/question/14191114?referrer=searchResults.
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3.8 Meters converts to 380 Centimeters.
Answer:
MnO- Manganese Oxide
Explanation:
Empirical formula: This is the formula that shows the ratio of elements
present in a
compound.
How to determine Empirical formula
1. First arrange the symbols of the elements present in the compound
alphabetically to determine the real empirical formula. Although, there
are exceptions to this rule, E.g H2So4
2. Divide the percentage composition by the mass number.
3. Then divide through by the smallest number.
4. The resulting answer is the ratio attached to the elements present in
a compound.
Mn O
% composition 72.1 27.9
Divide by mass number 54.94 16
1.31 1.74
Divide by the smallest number 1.31 1.31
1 1.3
The resulting ratio is 1:1
Hence the Empirical formula is MnO, Manganese oxide
