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3 years ago

For metals, reactivity _ as you go from _ in a period.

1 answer:

8 0

Reactivity - Reactivity refers to how likely or vigorously an atom is to react with other substances. This is usually determined by how easily electrons can be removed (ionization energy) and how badly they want to take other atom's electrons (electronegativity) because it is the transfer/interaction of electrons that is the basis of chemical reactions.

Metals

Period - reactivity decreases as you go from left to right across a period.

Group - reactivity increases as you go down a group

Why? The farther to the left and down the periodic chart you go, the easier it is for electrons to be given or taken away, resulting in higher reactivity.

Non-metals

Period - reactivity increases as you go from the left to the right across a period.

Group - reactivity decreases as you go down the group.

Why? The farther right and up you go on the periodic table, the higher the electronegativity, resulting in a more vigorous exchange of electron

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Calcium hydroxide is a strong base but is not very soluble ( Ksp = 5.02 X 10-6 ). What is the pH of a saturated solution of Ca(O

Misha Larkins [42]

Answer : The pH of a saturated solution is, 12.33

Explanation : Given,

$K_{sp}$ = $5.02\times 10^{-6}$

First we have to calculate the solubility of $OH^-$ ion.

The balanced equilibrium reaction will be:

$Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-$

Let the solubility will be, 's'.

The concentration of $Ca^{2+}$ ion = s

The concentration of $OH^-$ ion = 2s

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ca^{2+}][OH^-]^2$

Let the solubility will be, 's'

$K_{sp}=(s)\times (2s)^2$

$K_{sp}=(4s)^3$

Now put the value of $K_[sp}$ in this expression, we get the solubility.

$5.02\times 10^{-6}=(4s)^3$

$s=1.079\times 10^{-2}M$

The concentration of $Ca^{2+}$ ion = s = $1.079\times 10^{-2}M$

The concentration of $OH^-$ ion = 2s = $2\times (1.079\times 10^{-2}M)=2.158\times 10^{-2}M$

First we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (2.158\times 10^{-2})$

$pOH=1.67$

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-1.67=12.33$

Therefore, the pH of a saturated solution is, 12.33

7 0

3 years ago

How can you make a solution more saturated?

NNADVOKAT [17]

Heat the solution till all the water molecules are no longer in the solution.
Or add more solute till it stops dissolving.(maximum point of dissolvtion.)

7 0

3 years ago

Suppose an adult is encased in a thermally insulating barrier so that all the heat evolved by metabolism of foodstuffs is retain

mihalych1998 [28]

Given:
Q = 9.4 kJ/(kg-h), the heat production rate
c = 4.18 J/(g-K), the heat capacity
t = 2.5 h, amount of time

Note that
c = 4.18 J/(g-K) = 4180 J/(kg-K) = 4.18 kJ/kg-K)

Consider 1 kg of mass.
Then
Qt = cΔT
where ΔT is the increase in temperature (°K)
(1 kg)*(9.4 kJ/(kg-h))*(2.5 h) = (1 kg)*(4.18 kJ/(kg-K))*(ΔT K)
23.5 = 4.18 ΔT
ΔT = 23.5/4.18 = 5.622 K = 5.622 °C

Answer: 5.62 K (or 5.62 °C)

7 0

3 years ago

The elements in any blank in periodic table have similar physical and chemical properties

Yakvenalex [24]

Elements in the same group have similar properties

8 0

3 years ago

Calculate the percent composition of the following:<br><br> A. VO3<br><br> B. V2O5

tino4ka555 [31]

A) Answer is: 51.48% V and 48.52% O.

Ar(V) = 50.94; atomic weight of vanadium.

Ar(O) = 16; atomic weight of oxygen.

Ar(VO₃) = 50.94 + 3 · 16.

Ar(VO₃) = 98.94; molecular weight of vanadium (VI) oxide.

ω(V) = Ar(V) ÷ Ar(VO₃) · 100%.

ω(V) = 51.48%; the percent composition of vanadium.

ω(O) = 100% - 51.48%.

ω(O) = 48.52%; the percent composition of oxygen.

B) Answer is: 67.80% V and 32.20% O.

Ar(V) = 50.94; atomic weight of vanadium.

Ar(O) = 16; atomic weight of oxygen.

Ar(V₂O₅) = 2 · 50.94 + 5 · 16.

Ar(V₂O₅) = 149.88; molecular weight of vanadium (V) oxide.

ω(V) = 2 · Ar(V) ÷ Ar(V₂O₅) · 100%.

ω(V) = 67.80%; the percent composition of vanadium.

ω(O) = 100% - 67.80%.

ω(O) = 32.20%; the percent composition of oxygen.

4 0

3 years ago

For metals, reactivity _____ as you go from _____ in a period.

For metals, reactivity _ as you go from _ in a period.