It depends on the context iorn is a transition metal so it can hold a charge from 1-8
Answer:
6.17 g/cm³
Explanation:
Data given:
one side of cube = 0.53 cm
mass of the cube is 0.92 g
density of the cube = ?
Solution:
First we will calculate for volume the cube
As we know all the sides or edges of a cube are equal so volume equation will be
So,
V = length x width x height
V = e³
as on side = 0.53 cm
then
V = (0.53 cm)³
V = 0.149 cm³
Now we will calculate density of cube
To calculate density, formula will be used
d = m/v . . . . . (1)
where
d = density
m = mass
v = volume
put values in above formula 1
d = 0.92 g / 0.149 cm³
d = 6.17 g/cm³
so. the density of cube = 6.17 g/cm³
Answer:
Product: ethyl L-valinate
Explanation:
If we want to understand what it is the molecule produced we have to an<u>alyze the reagents</u>. We have valine an <u>amino acid</u>, in this kind of compounds we have an <em>amine group</em> (
) and a <em>carboxylic acid</em> group (
). Additionally, we have an <u>alcohol </u>(
) in the presence of HCl (a <u>strong acid</u>) in the first step, and a base (
).
When we have an acid and an alcohol in a vessel we will have an <u>esterification reaction</u>. In other words, an ester is produced. As the <em>first step,</em> the oxygen in the C=O (in the carboxylic acid group) would be protonated. In the <em>second step</em>, the ethanol attacks the carbon in the C=O of the carboxylic acid group producing a new bond between the oxygen in the ethanol and the carbon in the carboxylic acid. In <em>step 3</em>, a proton is transferred to produce a better leaving group (
). In <em>step 4</em>, a water molecule leaves the main structure to produce again the double bond C=O. <em>Finally</em>, a base () removes the hydrogen from the C=O bond to produce ethyl L-valinate
See figure 1
I hope it helps!
The answer would be MgO, as you spilt the two.
