The AP Biology teacher is measuring out 638.0 g of dextrose (C6H12O6) for a lab the moles of dextrose is this equivalent to is 3.6888 moles.
<h3>What are moles?</h3>
A mole is described as 6.02214076 × 1023 of a few chemical unit, be it atoms, molecules, ions, or others. The mole is a handy unit to apply due to the tremendous variety of atoms, molecules, or others in any substance.
To calculate molar equivalents for every reagent, divide the moles of that reagent through the moles of the restricting reagent. The calculation is follows:
- 655/12 x 6 + 12+ 16 x 6
- = 655/ 180 = 3.6888 moles.
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Answer:
2.53 L is the volume of H₂ needed
Explanation:
The reaction is: C₁₈H₃₀O₂ + 3H₂ → C₁₈H₃₆O₂
By the way we can say, that 1 mol of linolenic acid reacts with 3 moles of oxygen in order to produce, 1 mol of stearic acid.
By stoichiometry, ratio is 1:3
Let's convert the mass of the linolenic acid to moles:
10.5 g . 1 mol / 278.42 g = 0.0377 moles
We apply a rule of three:
1 mol of linolenic acid needs 3 moles of H₂ to react
Then, 0.0377 moles will react with (0.0377 . 3 )/1 = 0.113 moles of hydrogen
We apply the Ideal Gases Law to find out the volume (condition of measure are STP) → P . V = n . R . T → V = ( n . R .T ) / P
V = (0.113 mol . 0.082 L.atm/mol.K . 273.15K) 1 atm = 2.53 L
The patch area in square kilometers is 1.59*10⁻⁹ km₂
Why?
This is an unit conversion problem. We have to convert from cm² to km². We can do that by knowing that there are 100 cm in 1 m, and 1000 m in 1 km, so 100000 cm=1km. Knowing that we can apply the following conversion factor:
Now to convert this value to scientific notation, we have to move the decimal point to the right until we get a whole number, and the exponent of the number 10 is going to be the number of spaces we moved to the right (negative), so the final answer is:
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D.) a transformation of light energy into chemical energy
Thomson suggested the model of atom which was a sphere of positive matter within which electronic forces determined the positioning of the corpuscles. The corpuscles were distributed in a uniform sea of positive charge. This was so-called "plum pudding" model.
Answer: C ) Thomson
