Answer:
Akash
Explanation:
it could be a magnet with the same poles facing eachoher
Answer:
Abundance of 32S is 94.41%
Explanation:
The average atomic mass is defined as the sum of the atomic masses of each isotope times its abundance:
Average atomic mass = ∑ Atomic mass istope*Abundance
For the sulfur:
32.07amu = 31.97207X + 32.97146Y + 33.96786*0.0422 <em>(1)</em>
<em>Where X is abundance of 32S and Y abundance of 33S</em>
Also we can write:
1 = X + Y + 0.0422 <em>(2)</em>
0.9578 - X = Y
Because the sum of the abundances = 1
Replacing (2) in (1):
32.07amu = 31.97207X + 32.97146(0.9578 - X) + 33.96786*0.0422
32.07 = 31.97207X + 31.58006 - 32.97146X + 1.43344
-0.9435 = -0.99939X
0.9441 =X
In percentage, abundance of 32S is 94.41%
Explanation:
Since the wheel moves up and down, the position that represents the potential energy is that which has the maximum height from the ground.
Potential energy is the energy at rest of a body.
It is given as:
Potential energy = m x g x h
m is the mass of the body
g is the acceleration due to gravity
h is the height of the body
We can see that mass and height are directly related to the potential energy a body exerts.
The higher the wheel from ground, the higher its potential energy.
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<u>Answer:</u> The element represented by M is Strontium.
<u>Explanation:</u>
Let us consider the molar mass of metal be 'x'.
The molar mass of MO will be = Molar mass of oxygen + Molar mass of metal = (16 + x)g/mol
It is given in the question that 15.44% of oxygen is present in metal oxide. So, the equation becomes:

The metal atom having molar mass as 87.62/mol is Strontium.
Hence, the element represented by M is Strontium.
Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.
See attached figure for the structure.
Vanillin have 3 functional groups:
1) aldehyde group: R-HC=O, in which the carbon is double bonded to oxygen
2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring
3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons
Now for the hybridization we have:
The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a <u>sp²</u> hybridization because they are involved in double bonds.
The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a <u>sp³</u> hybridization because they are involved only in single bonds.