A convergent boundary is when two plates push against each other. a transform boundary is when they slide past each other. both can couse mid ocean ridges.
<span>The old car left on the side of the road showed signs of corrosion. We call this common chemical reaction rusting. The equation for this reaction can be written as 4Fe + 3O2 → 2Fe2O3. What are the reactants in the rusting process?
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C. Iron and Oxygen
To know the number of atoms, it will be necessary to use the formula of avogadro. We calculate the number of moles then we can calculate the number of avogadro.
The molar mass of:
Phosphorus = 31g / mol
Mercury = 200 g / mol
Bismuth = 209 g / mol
Strontium = 87 g / mol.
The number of avogadro (N) is 6.023 10 ^ 23
n = m (mass) / M (molar mass)
number of atoms = n. N = m / M . N
number of phosphorus atoms = 5.14/31 . 6.023 10^23 = 9.98 10 ^ 22 atoms
Number of atoms of mercury = 2.16/200 . 6.023 10^23 = 6.5 10 ^ 21 atoms
Number of bismuth atoms = 1.8/209 . 6.023 10^23 = 5.18 10 ^ 21 atoms
Number of strontium atoms = 8.8 x 10-2 /87 . 6.023 10^23 = 6.09 10 ^ 20 atoms
Answer:The structures of 3-brominated products are available in attachment. Kindly find in attachment.
Explanation:
1-ethyl-4-methylbenzene undergoes a radical substitution reaction when it is treated with NBS(N-bromosuccinimide).
There are 3 products which are produced in the reaction.
There are 2 positions available where bromination can occur one at 1 position and other at 4 position.
There is a ethyl group present at 1-position and a methyl group is present at 4-position.
At the 1-position where ethyl group is present 1-phenylethyl radical is generated on irradiation with UV-light. Now since this 1-phenylethyl radical is planar and a chiral centre as all groups attached to the carbon center is different hence two products of bromination can occur from this position. One from below the plane and other from above the plane. These 2 products can form with equal probability.
In one product the Bromine radical can combine with 1-phenylethyl radical form above the plane and in other product bromine radical can combine with 1-phenyl radical form below the plane.
Hence 1-phenylethyl radical would lead to products which would be 2 stereiosomers and would be known as enantiomers (mirror images).
Radical generated at 4 position that is at methyl position would be a benzyl radical and this would also be planar but since it is not a chiral center hence both the sides would be equivalent so only one product would be generated.
Kindly find in attachment for the structures of the products and reactants.