Answer:
The given atom is of Ca.
Explanation:
Given data:
Speed of atom = 1% of speed of light
De-broglie wavelength = 3.31×10⁻³ pm (3.31×10⁻³ / 10¹² = 3.31×10⁻¹⁵ m)
What is element = ?
Solution:
Formula:
m = h/λv
m = mass of particle
h = planks constant
v = speed of particle
λ = wavelength
Now we will put the values in formula.
m = h/λv
m = 6.63×10⁻³⁴kg. m².s⁻¹/3.31×10⁻¹⁵ m ×( 1/100)×3×10⁸ m/s
m = 6.63×10⁻³⁴kg. m².s⁻¹/ 0.099×10⁻⁷m²/s
m = 66.97×10⁻²⁷ Kg/atom
or
6.69×10⁻²⁶ Kg/atom
Now here we will use the Avogadro number.
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
Now in given problem,
6.69×10⁻²⁶ Kg/atom × 6.022 × 10²³ atoms/ mol × 1000 g/ 1kg
40.3×10⁻³×10³g/mol
40.3 g/mol
So the given atom is of Ca.
A change in the Earth's magnetic<span> field resulting in the </span>magnetic<span> north being aligned with the geographic south, and the </span>magnetic<span> south being aligned with the geographic north. Also called </span>geomagnetic reversal<span>.</span>
The calculation for such a question can be achieved via Avogadro hypothesis
We know molar mass of CO2 is 44g/mole which is the sum of atomic masses i.e; C and 2 oxygen atoms
Molar mass of CO2 =12(C)+2*16(O) = 44 g/mole will contain 6.023 ※10^23 CO2 molecules ..
44g/mole = 6.023 ※10^23 CO2 molecules
=> 1g = (6.023/44) ※10^23 CO2 molecules
==> 8.80g = 8.80(6.023÷44)10^23 = 1.2046 ※10^23 molecules of CO2….
Thus there r 1.2046 ※10^23 molecules of CO2 in 8.80g
if u need to calculate no. of carbon atoms then multiply result by 1 and if u need no of oxygen atoms in 8.80g of co2 then multiply the result by 2 ….
Answer:
Carbonyl
Explanation:
While the diagram is slightly unclear, the molecule most likely being shown is a carbonyl. A molecule is a carbonyl when there is a carbon double-bonded to an oxygen.
