Answer:
1. 0.0500 moles K, 0.0250 moles Cr and 0.1000 moles O.
2. 9.00 moles ammonium ions are present.
Explanation:
1. In 1mol of K2CrO4 there are 2 moles of K, 1 mole of Cr and 4 moles of O.
The moles in 0.0250 moles K2CrO4 are:
0.0250mol * 2 = 0.0500 moles K
0.0250mol * 1 = 0.0250 moles Cr
0.0250mol * 4 = 0.1000 moles O
2. In 1 mole of (NH4)2CO3 there are 2 moles of ammonium ions, NH₄⁺.
In 4,50 moles are:
4.50 moles * 2 = 9.00 moles ammonium ions are present
The value of delta H° for aniline = 6415 Kj/mol
<u><em>calculation</em></u>
Step 1: find heat
Q (heat) = C (specific heat capacity) x ΔT (change in temperature)
C= 4.25 kj/c°
ΔT = 69.8-29.5 = 40.3 c°
Q= 4.25 kj/c x 40.3 c = 171.28 kj
Step 2: find the moles of aniline
moles = mass/molar mass
= 2.49 g/ 93.13 g/mol =0.0267 moles
Step 3: find delta H
171.28 kj/0.0267 mol = 6415 kj/mol
since the reaction is exothermic delta H = 6415 Kj/mol
Answer: Sulfur dioxide and nitrogen oxides dissolve very easily in water and can be carried very far by the wind. These pollutants cause acid rain.
If an Atom has 12 Electrons then if it is a Neutral atom then the Atomic Number of the Atom will be 12.
The Electronic Configuration of Element with Atomic Number 12 will be :
⇒ 1s² 2s² 2p⁶ 3s²
We can notice from the Electronic Configuration that :
Given atom has 2 electrons in the 1st shell.
⇒ In the Shell : n = 1
Number of Electrons = 2
In the similar way : Number of Electrons in 2nd shell = 2 + 6 = 8
⇒ In the Shell : n = 2
Number of Electrons = 8
In the similar way : Number of Electrons in 3rd shell = 2
⇒ In the Shell : n = 3
Number of Electrons = 2
PV = nRT
1.026atm * 42l = n * 0.0821L atm/mol K * 305K
n=1.72.
1.72 molecules.
