Answer:
82.18% is the mass percentage of bromine in the original compound.
Explanation:
Mass of AgBr = 1.1166 g
Moles of AgBr =
According to reaction, 1 mole of AgBr is obtained from 1 mole of bromide ions , then 0.0059466 moles of AgBr will be formed from :
of bromide ions
Mass of 0.005939 moles of bromide ions :
0.0059466 mol × 79.90 g/mol = 0.4751 g
Mass of the sample = 0.5781 g
Mass percentage of bromine in sample :
82.18% is the mass percentage of bromine in the original compound.
Quantity of K2S m = 0.105 m
Number of ions i = 2(K) + 1(S) = 3
Freezing point depression constant of water Kf = 1.86
delta T = i x m x Kf = 3 x 0.105 x 1.86 = 0.586
Freezing point = 0 - 0.586 = 0.586 C
Boiling point constant of water Kb = 0.512
delta T = i x m x Kb = 3 x 0.105 x 0.512 = 0.161
Boiling point = 100 + 0.161 = 100.161 C
1) Na3PO4 (aq) + 3 KOH (aq) → 3 NaOH (aq) + K3PO4 (aq)
Displacement Reaction
2)MgF2 (s) + Li2CO3 (s) → MgCO3 (s) + 2 LiF (s)
Displacement Reaction
3) I don't know
4)2 RbNO3 + BeF2 → Be(NO3)2 + 2 RbF
5)2 AgNO3 (aq) + Cu (s) → Cu(NO3)2 (aq) + 2 Ag (s)
Answer:
The ¹³C-NMR Spectrum of <em>tert</em>-butyl alcohol will show only two signals.
(i) Signal at around 31 ppm:
This signal towards upfield is for the carbon atoms which are more shielded and are having rich electron surroundings. The height of peak at y-axis shows the number of carbon atoms as compared to other peaks. In this case it is three times the height of second signal hence, it shows that this peak corresponds to three carbon atoms.
(ii) Signal at around 70 ppm:
This signal towards downfield is for the carbon atom which is more deshielded and is having electron deficient surrounding. As compared to the second signal the height of this peaks corresponds to only one carbon. And the deshielded environment shows that this carbon is directly attached to an electronegative element.