Answer:
Why ? Because 1 molecule of H2SO4 gives 2 H+ ions per molecule while only one H+ ion is required to neutralize 1 molecule of KOH. So, 1 molecule of H2SO4 can neutralize 2 molecules of KOH. Hence, we would require 525 ml of 0.03 M H2SO4 to neutralize 525 ml of 0.06 M KOH. How will we prepare 525 ml of 0.03 M H2SO4 ?
Explanation:
Now, we have 0.025 M H2SO4 and we do not know how much volume we have.
We will use the standard N1 X V1 = N2 X V2 for this calculation.
N1=0.025 M; V1=unknown; N2=0.03 M and V2=525 ml.
So V1= (0.03 X 525)/(0.025) = 630 ml.
Answer:
240 seconds
Explanation:
Data Given:
Distance Radio travel to earth = 7.19 ×10⁷ km
Speed of this radio signal = 3.00 ×10⁸ m/s
Solution:
First convert Km to m
1 km = 1000 m
7.19 ×10⁷ km = 7.19 ×10⁷ x 1000 = 7.19 × 10¹⁰m
Formula used to calculate time
Speed (S) = distance (d) / time (t)
rearrange above equation for time
time (t) = distance (d) / Speed (S) . . . . . . . . .(1)
Put values in equation 1
time (t) = 7.19 × 10¹⁰/ 3.00 ×10⁸ m/s
time (t) = 240 s
So 240 seconds it take for the signal to reach the earth
Answer:
By using renewable energy sources.
Explanation:
You need to satisfy a quota of energy for (whatever country you live in). People do this by using the cheapest way of producing the most energy, the most efficient. Sounds great right? Wrong! It stuffs our atmosphere with harmful gasses like carbon dioxide. You can reduce the use of these fossil fuels by using renewable energy sources such as windmills, watermills, and most notable, solar panels!
The first dissociation for H2X:
H2X +H2O ↔ HX + H3O
initial 0.15 0 0
change -X +X +X
at equlibrium 0.15-X X X
because Ka1 is small we can assume neglect x in H2X concentration
Ka1 = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15)
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
HX + H2O↔ X^2 + H3O
8.2x10^-4 Y 8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2 = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m
1-PRIMARY ALKANOL 2-SECONDARY ALKANOL 3-TERTIARY ALKANOL
