Answer:
If you were to plug your nose, it would block out more than 80% of taste, causing all three items to have no taste- therefore making them taste the same.
Explanation:
Answer:
In a neutral molecule, the sum of the bonding valance electrons must be equal. So the products of the negative element and its charges and the positive element and its charge must be equal.
Explanation:
C1×N1 = C2×N2
If we have a 3 valance electrons , the 'A' charge will be either +3 or -5 for a full octet and valance electron in 'B' atoms will mostly result in acquisition of additional electrons (2) for an octet and relative charge of -2.
Balancing the two,
3 × A = -2 × B
To be equal, A = 2 and B = 3
Therefore, A²B³
A. The molecules of solids are close together and compact, liquids are spread out but not too far apart, and gas molecules are really far apart.
B. Increase in temperature causes pressure to go up. Decrease in temperature cause pressure to go down
In order to find molarity, you must first find the number of moles that was dissolved.
Now, Moles = Mass ÷ Molar Mass
⇒ Moles of NaCl = 2.922 g ÷ 58.44 g/mol
= 0.05 moles
∴ the Molarity of the NaCl is 0.05 M [Option 1]
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.
