Answer:
<u>Yes</u>
Explanation:
Remember, <u>Newton's third law of motion;</u> which says in part that <em>"Every action has an equal and opposite reaction."</em>
Hence, in this case, the fact that the doorbell rang out implies that there was another force that was exerted on it; which is, John's finger pressing the doorbell.
In other words, when John uses his fingers to press the doorbell button he applies a force (a mechanical force), and that force results in an opposite reaction; the ringing of the doorbell.
By Boyles Law (P1V1=P2V2), substituting values in and solving for V2, we find that the new volume is 3.6 L<span />
Answer:
0.00369 moles of HCl react with carbonate.
Explanation:
Number of moles of HCl present initially = moles = 0.00600 moles
Neutralization reaction (back titration):
According to above equation, 1 mol of NaOH reacts with 1 mol of 1 mol of HCl.
So, excess number of moles of HCl present = number of NaOH added for back titration = moles = 0.00231 moles
So, mole of HCl reacts with carbonate = (Number of moles of HCl present initially) - (excess number of moles of HCl present) = (0.00600 - 0.00231) moles = 0.00369 moles
Hence, 0.00369 moles of HCl react with carbonate.
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.