Given :
Mass of oxygen containing carbon monoxide (CO) is 2.666 gram .
To Find :
How many grams of carbon (C) would be present in carbon monoxide (CO) that contains 2.666 grams of oxygen (O) .
Solution :
By law of constant composition , a given chemical compound always contains its component elements in fixed ratio (by mass) and does not depend on its source and method of preparation.
So , volume of solution does not matter .
Moles of oxygen , 
.
Now , molecule of CO contains 1 mole of C .
So , moles of C is also 0.167 mole .
Mass of carbon , 
.
Therefore , mass of carbon is 2 grams .
Hence , this is the required solution .
<span>Answer is: pH of solution of sodium cyanide is 11.3.
Chemical reaction 1: NaCN(aq) → CN</span>⁻(aq)
+ Na⁺<span>(aq).
Chemical reaction 2: CN</span>⁻ +
H₂O(l) ⇄ HCN(aq) + OH⁻<span>(aq).
c(NaCN) = c(CN</span>⁻<span>)
= 0.021 M.
Ka(HCN) = 4.9·10</span>⁻¹⁰<span>.
Kb(CN</span>⁻) = 10⁻¹⁴ ÷
4.9·10⁻¹⁰ = 2.04·10⁻⁵<span>.
Kb = [HCN] · [OH</span>⁻]
/ [CN⁻<span>].
[HCN] · [OH</span>⁻<span>] =
x.
[CN</span>⁻<span>] = 0.021 M - x..
2.04·10</span>⁻⁵<span> = x² / (0.021 M
- x).
Solve quadratic equation: x = [OH</span>⁻<span>] = 0.00198 M.
pOH = -log(0.00198 M) = 2.70.
pH = 14 - 2.70 = 11.3.</span>
Answer:
63.55
Explanation:
relative atomic mass=(mass of isotope1×relative abundance)+(mass of isotope 2×relative abundance)/100
r.a.m=(62.93×69.09)+(64.93×30.91)/100
=(4347.8337)+(2006.9863)/100
=6354.82/100
=63.55
Answer:
At equilibrium, reactants predominate.
Explanation:
For every reaction, the equilibrium constant is defined as the ratio between the concentration of products and reactants. Thus, for the reaction N2 (g) + O2 (g) ⇌ 2NO the expression of its equilibrium constant is:
![Keq = \frac{[NO]^{2}}{[O_{2} ][N_{2}]}](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5BNO%5D%5E%7B2%7D%7D%7B%5BO_%7B2%7D%20%5D%5BN_%7B2%7D%5D%7D)
Since the equilibrium constant is Keq = 4.20x10-31 the concentration of reactants O2 and N2 must be much higher than products to obtain such a small number as 4.20x10-31 at the equilibrium. Hence, at equilibrium reactants predominate.
1. option B
2. ground state Fe : [Ar] 4s² 3d⁶
<h3>
Further explanation</h3>
Aufbau rule
<em>Electrons occupy orbitals of the lowest energy level</em>
Charging electrons in the sub-shell uses the following sequence:
1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s², etc.
So the correct choice of answer is B.
Ground state is the state of electrons filling skins with the lowest energy levels.
Excited state is the state of electrons which occupy a higher energy level
The atomic number of Fe = 26, so electrons=26
Ground state configuration of Fe : [Ar] 4s² 3d⁶ or
1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d⁶
