Fats (triglycerides) that contain palmitic acid and stearic acid are therefore known as saturated fats. Fats made up of saturated fatty acids are solid at room temperature. ... Because it is polyunsaturated, it is liquid at room temperature.
2Fe2O3, reason is when we add 4Fe + 3O2, we get the same answer, but in a different form.
<h3><u>Answer</u>;</h3>
a. 3 molecules 3 carbon
b. 6 molecules 18 carbon
c. 6 molecules 18 carbon
d. 5 molecules 15 carbon
e. 3 molecules 15 carbon
f. 3 molecules 15 carbon
<h3><u>Explanation</u>;</h3>
- In the Calvin cycle, carbon atoms from CO2 are ncorporated into organic molecules and then used to build three-carbon sugars, a process that is fueled by, and dependent on, ATP and NADPH from the light reactions.
- Calvin cycle take place in the stroma. Reactions of Calvin cycle are divided into three main stages: carbon fixation, reduction, and regeneration of the starting molecule.
- During carbon fixation, a CO2 molecule combines with a five carbon acceptor molecule ribulose-1,5-bisphosphate. The result is a six carbon compound that splits to two three carbon compound, 3-PGA.
- During reduction; ATP and NADPH are used to convert the 3-PGA molecules into molecules of a three-carbon sugar, glyceraldehyde-3-phosphate.
- Finally during regeneration, some G3P molecules are used to make glucose while others are recycled to regenerate RuBP acceptor.
Answer:
D
Explanation:
You need both, independant and dependant variables for a hypothesis.
Answer:
See below.
Step-by-step explanation:
Ethers react with HI at high temperature to produce an alky halide and an alcohol.
R-OR' + HI ⟶ R-I + H-OR'
<em>Benzylic ethers</em> react by an Sₙ1 mechanism by forming the stable benzyl cation.
- PhCH₂-OR + HI ⟶ PhCH₂-O⁺(H)R + I⁻ Protonation of the ether
- PhCH₂-O⁺(H)R ⟶ PhCH₂⁺ + HOR Sₙ1 ionization of oxonium ion
- PhCH₂⁺ + I⁻ ⟶ PhCH₂-I Nucleophilic attack by I⁻
If there is excess HI, the alcohol formed in Step 2 is also converted to an alkyl iodide:
ROH +HI ⟶ R-I + H-OH
Thus, benzyl ethyl ether reacts to form benzyl iodide (a) and ethanol (b).
The ethanol reacts with excess HI in an Sₙ2 reaction to form ethyl iodide (c).
