Speed (m/s) = distance (metres) ÷ time (seconds)
For Track 1: 0.2 ÷ 2 = 0.1m/s
For Track 2: 0.2 ÷ 2 = 0.1m/s
For Track 3: 0.6 ÷ 6 = 0.1m/s
For Track 4: 0.4 ÷ 6 = 0.07m/s
The train on Track 4 had the slowest speed because it's got the shortest speed and it's covering less distance per second therefore it is slower.
Answer:
The oxygen element in H2O2 is the specie that is reduced in H2O and oxidized into O2.
Explanation:
5 H2O2(aq) + 2 MnO4-(aq) + 6 H+(aq) → 2 Mn2+(aq) + 8 H2O(l) + 5 O2(g)
Oxidation is defined as the outright loss of electrons. Oxidation leads to an increase in an element's oxidation number.
If this reaction is broken down into reduction and oxidation halves
It is observed that, of the reactants above,
H202 becomes H2O and O2
MnO4- + H+ becomes Mn2+ and H2O
Oxidation number of Mn changes from +7 In MnO4- to +2 In Mn2+ (evidently reduction)
The Oxygen in MnO4- doesn't change oxidation numbers as its oxidation number stays at -2
Oxidation number of Oxygen changes from -1 in H2O2 to -2 In H2O and 0 in O2
The Hyrogen in H2O2 doesn't change oxidation numbers, itsoxidation number stays at +1 in H2O2 and H2O.
This indicates that H2O2 undergoes oxidation and reduction; more specifically, the oxygen element in H2O2 is the specie that is reduced in H2O and oxidized into O2.
Hope this Helps
The chemical reaction between the reactants:
3 AgNO₃ (aq) + FeCl₃ (aq) → 3 AgCl (s) + Fe(NO₃)₃ (aq)
Explanation:
We have the following chemical reaction:
3 AgNO₃ (aq) + FeCl₃ (aq) → 3 AgCl (s) + Fe(NO₃)₃ (aq)
Complete ionic equation:
3 Ag⁺ (aq) + 3 NO₃⁻ (aq) + Fe³⁺ (aq) + 3 Cl⁻ (aq) → 3 AgCl (s) + Fe³⁺ (aq) + 3 NO₃⁻ (aq)
We remove the spectator ions and we get the net ionic equation:
Ag⁺ (aq) + Cl⁻ (aq) → AgCl (s)
where:
(aq) - aqueous
(s) - solid
Learn more about:
net ionic equation
brainly.com/question/7018960
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The most commonly used transistor configuration is the NPN Transistor. We also learnt that the junctions of the bipolar transistor can be biased in one of three different ways – Common Base, Common Emitter and Common Collector.
In this tutorial about bipolar transistors we will look more closely at the “Common Emitter” configuration using the Bipolar NPN Transistor with an example of the construction of a NPN transistor along with the transistors current flow characteristics is given below.
To know this you pretty much do have to kind of memorize a few electronegativities. I don't recall ever getting a table of electronegativities on an exam.
From the structure, you have:
I remember the following electronegativities most because they are fairly patterned:
EN
H
=
2.1
EN
C
=
2.5
EN
N
=
3.0
EN
O
=
3.5
EN
F
=
4.0
EN
Cl
=
3.5
Notice how carbon through fluorine go in increments of
~
0.5
. I believe Pauling made it that way when he determined electronegativities in the '30s.
Δ
EN
C
−
Cl
=
1.0
Δ
EN
C
−
H
=
0.4
Δ
EN
C
−
C
=
0.0
Δ
EN
C
−
O
=
1.0
Δ
EN
O
−
H
=
1.4
So naturally, with the greatest electronegativity difference of
4.0
−
2.5
=
1.5
, the
C
−
F
bond is most polar, i.e. that bond's electron distribution is the most drawn towards the more electronegative compound as compared to the rest.
When the electron distribution is polarized and drawn towards a more electronegative atom, the less electronegative atom has to move inwards because its nucleus was previously favorably attracted to the electrons from the other atom.
That means generally, the greater the electronegativity difference between two atoms is, the shorter you can expect the bond to be, insofar as the electronegative atom is the same size as another comparable electronegative atom.
However, examining actual data, we would see that on average, in conditions without other bond polarizations occuring:
r
C
−
Cl
≈
177 pm
r
C
−
C
≈
154 pm
r
C
−
O
≈
143 pm
r
C
−
F
≈
135 pm
r
C
−
H
≈
109 pm
r
O
−
H
≈
96 pm
So it is not necessarily the least electronegativity difference that gives the longest bond.
Therefore, you cannot simply consider electronegativity. Examining the radii of the atoms, you should notice that chlorine is the biggest atom in the compound.
r
Cl
≈
79 pm
r
C
≈
70 pm
r
H
≈
53 pm
r
O
≈
60 pm
So assuming the answer is truly
C
−
C
, what would have to hold true is that:
The
C
−
F
bond polarization makes the carbon more electropositive (which is true).
The now more electropositive carbon wishes to attract bonding pairs from chlorine closer, thereby shortening the
C
−
Cl
bond, and potentially the
C
−
H
bond (which is probably true).
The shortening of the
C
−
Cl
bond is somehow enough to be shorter than the
C
−
C
bond (this is debatable).