Friedrich Wohler was working with Ammonium Cyanate and found crystals of urea in his beakers
Answer:
43.47 g
Explanation:
The <em>boiling point elevation</em> is described as:
Where ΔT is the difference in boiling points: 120.6-118.4 = 2.2 °C
K is the boiling point elevation constant, K= 2.40 °C·kg·mol⁻¹
and m is the molality of the solution (molality = mol solute/kg solvent).
So first we <u>calculate the molality of the solution</u>:
- 2.2 °C = 2.40 °C·kg·mol⁻¹ * m
Now we calculate the moles of benzamide (C₇H₇NO, MW=315g/mol), using the given mass of the liquid X.
- 150 g ⇒ 150/1000 = 0.150 kg
- 0.917 m = molC₇H₇NO / 0.150kg
Finally we convert moles of C₇H₇NO into grams, using its molecular weight:
- 0.138 molC₇H₇NO * 315g/mol = 43.47 g
Answer:
Zero-Nine
Explanation:
this is becasue these numbers are rather small and if you plug these numbers into an equation you will most likely get 0.
<u>Answer:</u>
The primary product of the photosystem I is NADPH.
<u>Explanation:
</u>
NADPH is the acronym of Nicotinamide adenine Dinucleotide Phosphate. This NADPH plays a vital role in many chemical reactions that take place during photosynthesis process. NADPH is released as a product in Photosystem I.
In the photosystem I process, the molecule present will absorb sunlight energy and transfers it to electrons to produce NADPH. This NADPH molecule is used as fuel for the chemical process that occurs during the 2nd stage of the photosynthesis process. Its molecular formula is 
. It acts as a fundamental metabolite.
The balanced chemical reaction is written as:
<span>2Al2O3 = 4Al + 3O2
We are given the amount of aluminum oxide used in the reaction. This will be the starting point for the calculations. We do as follows:
26.5 mol Al2O3 ( 3 mol O2 / 2 mol Al2O3 ) = 39.75 mol O2 produced</span>
