Answer:
Rate of the reaction= 9.92× 10^-5 M² min-1
Explanation:
Using the equation of reaction
2N2O5 ⟶ 4NO2+O2
Rate = k[N2O5]²
From the question k= 6.2×10-4
[N2O5]= 0.4
Rate = 6.2×10-4[0.4]²= 9.92×10-5M² min-1
Osmosis is the diffusion of water <span>across a semipermeable membrane (usually cell membrane) from a region of low solute concentration to a more concentrated solution so it can reach equilibrium (balance).
D</span>iffusion is <span>a spontaneous movement of particles from an area of high concentration to an area of low concentration.
Both results in particles moving and help balance out the concentrations.
Also, in osmosis, the water molecules are moving. In diffusion, it is the solutes moving.
I hope this helps and explains well.</span>
Answer:
(a) adding 0.050 mol of HCl
Explanation:
A buffer is defined as the mixture of a weak acid and its conjugate base -or vice versa-.
In the buffer:
1.0L × (0.10 mol / L) = 0.10 moles of HF -<em>Weak acid-</em>
1.0L × (0.050 mol / L) = 0.050 moles of NaF -<em>Conjugate base-</em>
-The weak acid reacts with bases as NaOH and the conjugate base reacts with acids as HCl-
Thus:
<em>(a) adding 0.050 mol of HCl:</em> The addition of 0.050moles of HCl produce the reaction of 0.050 moles of NaF producing HF. That means after the reaction, all NaF is consumed and you will have in solution just the weak acid <em>destroying the buffer</em>.
(b) adding 0.050 mol of NaOH: The NaOH reacts with HF producing more NaF. Would be consumed just 0.050 moles of HF -remaining 0.050 moles of HF-. Thus, the buffer <em>wouldn't be destroyed</em>.
(c) adding 0.050 mol of NaF: The addition of conjugate base <em>doesn't destroy the buffer</em>
1st one= CaO +H2O=Ca(OH)2
product side-
Ca=1
O=2
H=2
Reactant side-
Ca=1
O=2
H=2
The first one is balanced for you
There is 1 calcium on each side 2 oxygens on each side and 2 hydrogens on each side
Here we have to get the spin of the other electron present in a orbital which already have an electron which has clockwise spin.
The electron will have anti-clockwise notation.
We know from the Pauli exclusion principle, no two electrons in an atom can have all the four quantum numbers i.e. principal quantum number (n), azimuthal quantum number (l), magnetic quantum number (m) and spin quantum number (s) same. The importance of the principle also restrict the possible number of electrons may be present in a particular orbital.
Let assume for an 1s orbital the possible values of four quantum numbers are n = 1, l = 0, m = 0 and s = 
.
The exclusion principle at once tells us that there may be only two unique sets of these quantum numbers:
1, 0, 0, + and 1, 0, 0, -.
Thus if one electron in an orbital has clockwise spin the other electron will must be have anti-clockwise spin.
