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3 years ago

Rank the nonmetals in each set from most reactive (1) to least reactive (3). Bromine: Chlorine: Iodine:

2 answers:

6 0

The most reactive element of this list is Chlorine, the next most reactive is bromine, and the least reactive is iodine.

All of these three elements are group 7 elements in the periodic table. It is known that the reactivity of group 7 elements decreases down the group. The most reactive element in this group is Flourine with reactivity decreasing down the group.

The reason for this decrease in reactivity is that as you go down the group, the distance between the positive nucleus that attracts valence electrons increases, decreasing the electrostatic attraction between the nucleus and the outer electrons. The other reason is that the electrons in lower energy levels closer to the nucleus repel and shield the electrons in the outermost shell or energy level of the atom.

Chlorine>Bromine>Iodine.

igor_vitrenko [27]3 years ago

6 0

Answer: The order of reactivity of non-metals from most reactive to least reactive is $\text{Chlorine}>\text{Bromine}>\text{Iodine}$

Explanation:

Reactivity of a non-metal is defined as the tendency of an element to gain electrons. The reactivity increases as we move across a period and it decreases as we move down the group.

When the size of an element increases, the valence electrons gets away from the nucleus and the tendency of an element to gain electrons decreases.

In a group, the size of an element increases because there is an addition of new shell and electron is added in that shell.

The given elements belong to the same group which is Group 17.

Chlorine has the smallest size, then bromine and then iodine.

Hence, the order of reactivity of non-metals from most reactive to least reactive is $\text{Chlorine}>\text{Bromine}>\text{Iodine}$

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A 9-μC positive point charge is located at the origin and a 6 μC positive point charge is located at x = 0.00 m, y = 1.0 m. Find

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Answer:

The coordinates of the point is (0,0.55).

Explanation:

Given that,

First charge $q_{1}=9\times10^{-6}\ C$ at origin

Second charge $q_{2}=6\times10^{-6}\ C$

Second charge at point P = (0,1)

We assume that,

The net electric field between the charges is zero at mid point.

Using formula of electric field

$E=\dfrac{kq}{r^2}$

$0=\dfrac{k\times9\times10^{-6}}{d^2}+\dfrac{k\times6\times10^{-6}}{(1-d)^2}$

$\dfrac{(1-d)}{d}=\sqrt{\dfrac{6}{9}}$

$\dfrac{1}{d}=\dfrac{\sqrt{6}}{3}+1$

$\dfrac{1}{d}=1.82$

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3 years ago

Igneous rocks weather more easily than sedimentary rocks.

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3 years ago

Solution A has a speciﬁc heat of 2.0 J/g◦C. Solution B has a speciﬁc heat of 3.8 J/g◦C. If equal masses of both solutions start

fgiga [73]

Answer: 2. Solution A attains a higher temperature.

Explanation: Specific heat simply means, that amount of heat which is when supplied to a unit mass of a substance will raise its temperature by 1°C.

In the given situation we have equal masses of two solutions A & B, out of which A has lower specific heat which means that a unit mass of solution A requires lesser energy to raise its temperature by 1°C than the solution B.

Since, the masses of both the solutions are same and equal heat is supplied to both, the proportional condition will follow.

We have a formula for such condition,

$Q=m.c.\Delta T$ .....................................(1)

where:

$\Delta T$ = temperature difference

Q= heat energy

m= mass of the body

c= specific heat of the body

Proving mathematically:

According to the given conditions

we have equal masses of two solutions A & B, i.e. $m_A=m_B$

equal heat is supplied to both the solutions, i.e. $Q_A=Q_B$

specific heat of solution A, $c_{A}=2.0 J.g^{-1} .\degree C^{-1}$

specific heat of solution B, $c_{B}=3.8 J.g^{-1} .\degree C^{-1}$

$\Delta T_A$ & $\Delta T_B$ are the change in temperatures of the respective solutions.

Now, putting the above values

$Q_A=Q_B$

$m_A.c_A. \Delta T_A=m_B.c_B . \Delta T_B\\\\2.0\times \Delta T_A=3.8 \times \Delta T_B\\\\ \Delta T_A=\frac{3.8}{2.0}\times \Delta T_B\\\\\\\frac{\Delta T_{A}}{\Delta T_{B}} = \frac{3.8}{2.0}>1$

Which proves that solution A attains a higher temperature than solution B.

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