Which of these equations is balanced, and which is not balanced? Explain how you can tell.

Pls, Help!!!!!!!

yKpoI14uk [10]

Answer:

Element A = Oxygen

Element H =

Element B = Aluminum

Element J = Magnesium

Element C = Selenium

Element L = Carbon

Element D = Sodium

Element Q = Francium

Element F = Antimony

Element R = Calcium

Element G = Chlorine

Element S = Tellurium

Explanation:

Element A is Oxygen because: oxygen 6 valence electrons ; is a gas at room temperature ; and is transported in blood to cells.

Element H is Neon because: Neon is a noble gas ; qppears as red light when charged with electricity (Neon light signs) and it has the second highest Ionization energy of the elements

Element B is Aluminum because: Aluminum is a metal and its ion has charge of +3. It is also located on the borders of the Metalloid staircase .

Element J is Magnesium because its ion has charge of 2+ and is isoelectronic with Neon because it loses two electrons to now have 10 electrons.

Element C is Selenium because its ion that has a charge of -2 is formed by gaining two electrons in order to have 36 electrons which is isoelectronic with Kr ypton

Element L is Carbon because carbon has the smallest atomic radius of any member in the Carbon family because it is the first member of the family and atomic radius increases on going down the group.

Element D is Sodium because its ion has charge of +1 and it has 2 inner core levels , the 1 and 2 energy levels.

Element Q is Francium because it has the largest radius and lowest ionization energy of any element

Element F is Antimony. It is a member of Nitrogen family and has the second highest ionization energy level in family .

Element R is calcium because its on has charge of +2 which is isoelectronic with Argon . Calcium also has atomic radius is larger than Ar gon.

Element G is Chlorine. It has the second to the smallest radius of elements in the 3rd period as the second to the last element in the period because atomic radius decreases across a period from left to right.

Element S is Tellurium. It has atomic mass larger than Iodine just to the right of it and is found in the 5th period

4 0

3 years ago

An isotope of potassium with a half-life of roughly one billion years is often used in radioactive dating. This isotope decays t

melisa1 [442]

Answer : The chemical symbol for the element that results from this process is, (Ar) for argon.

Explanation :

Electron capture : In this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. Simply, a proton and an electron combines together to form a neutron. Mass number does not change in this process.

$_Z^A\textrm{X}+_{-1}^0\textrm{e}\rightarrow _{Z-1}^A\textrm{Y}$

The equation for the given reaction is,

$_{19}^{40}\textrm{K}+_{-1}^0\textrm{e}\rightarrow _{18}^{40}\textrm{Ar}$

Thus, the chemical symbol for the element that results from this process is, argon (Ar).

7 0

3 years ago

When you jump-starts a car, what happens to the stored energy in the working battery? PLEASE ANSWER

Yuri [45]

The stored energy in the battery does not work anymore.

4 0

2 years ago

There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.

Similarly, for the second compound

$m(\text{O}) = m(\text{loss}) = 0.016 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012 \; g$

$n = m/M$ ; $0.4172 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol$
$n(\text{Hg atoms}) = 0.4012 \; g / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol$

$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

$\text{Hg}_{2} \text{O}$ .

8 0

2 years ago

rusting of steel can be prevented by coating the steel with a layer of zinc. Explain, in terms of electron transfer, why steal d

Luba_88 [7]

Answer:

Wouldn't rust because zinc will lose electrons more readily than iron and will therefore oxidize first.

Explanation:

This process whereby rusting of steel is prevented by coating the steel with a layer of zinc is known as galvanization.

Now, in this process, the steel object will be coated in a thin layer of zinc. This coating will prevent oxygen and water from reaching the underneath metal since the zinc will also act as a sacrificial metal.

Now, Zinc is used because it has a lower reduction potential than iron and thus it will get easily more oxidized than iron. Which means the zinc will lose electrons more readily than iron.

Also, since zinc has a lower reduction potential, it is therefore the more active metal. Thus, even if the zinc coating is scratched and the steel is exposed to moist air, the zinc will still get to oxidize before the iron.

7 0

2 years ago