Answer:
Noble gas Electronic configuration of arsenic:
As₃₃ = [Ar] 3d¹⁰ 4s² 4p³
Explanation:
Arsenic is metalloid.
Its atomic number is 33.
Its atomic mass is 75 amu.
Its symbol is As.
It is usually present in combine with sulfur and metals.
it is used in bronzing.
It is also used for hardening.
Electronic configuration:
As₃₃ = Is² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p³
Noble gas Electronic configuration:
As₃₃ = [Ar] 3d¹⁰ 4s² 4p³
Noble gas electronic configuration is shortest electronic configuration by using the noble gas elements full octet electronic configuration.
Answer:
Complete ionic:
.
Net ionic:
.
Explanation:
Start by identifying species that exist as ions. In general, such species include:
- Soluble salts.
- Strong acids and strong bases.
All four species in this particular question are salts. However, only three of them are generally soluble in water:
,
, and
. These three salts will exist as ions:
- Each
formula unit will exist as one
ion and one
ion. - Each
formula unit will exist as one
ion and two
ions (note the subscript in the formula
.) - Each
formula unit will exist as one
and two
ions.
On the other hand,
is generally insoluble in water. This salt will not form ions.
Rewrite the original chemical equation to get the corresponding ionic equation. In this question, rewrite
,
, and
(three soluble salts) as the corresponding ions.
Pay attention to the coefficient of each species. For example, indeed each
formula unit will exist as only one
ion and one
ion. However, because the coefficient of
in the original equation is two,
alone should correspond to two
ions and two
ions.
Do not rewrite the salt
because it is insoluble.
.
Eliminate ions that are present on both sides of this ionic equation. In this question, such ions include one unit of
and two units of
. Doing so will give:
.
Simplify the coefficients:
.
Answer: 27.09 ppm and 0.003 %.
First, <u>for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. </u>Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.
Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.
So, according to the <em>law of ideal gases,</em>
PV = nRT
where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)
The moles of CO will be,
n = 35 mg x
x
→ n = 0.00125 mol
We clear V from the equation and substitute P = 0.92 atm and
T = -30 ° C + 273.15 K = 243.15 K
V = 
→ V = 0.0271 L
As 1000 cm³ = 1 L then,
V = 0.0271 L x
= 27.09 cm³
<u>Then the acceptable concentration </u><u>c</u><u> of CO in ppm is,</u>
c = 27 cm³ / m³ = 27 ppm
<u>To express this concentration in percent by volume </u>we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:
c = 27.09
x
x 100%
c = 0.003 %
So, <u>the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is </u>27.09 ppm and 0.003 %.
Answer:

Explanation:
Hello,
In this case, for the given information, we can compute the rate of disappearance of NO₂ by using the following rate relationship:

Whereas it is multiplied by the the inverse of the stoichiometric coefficient of NO₂ in the reaction that is 2. Moreover, the subscript <em>f</em> is referred to the final condition and the subscript <em>0</em> to the initial condition, thus, we obtain:

Clearly, it turns out negative since the concentration is diminishing due to its consumption.
Regards.
Answer:
C. 1.35
Explanation:
2NH3 (g) <--> N2 (g) + 3H2 (g)
Initial concentration 2.2 mol/0.95L 1.1 mol/0.95L 0
change in concentration 2x x 3x
-0.84 M +0.42M +1.26M
Equilibrium 1.4 mol/0.95L=1.47M 1.58 M 1.26 M
concentration
Change in concentration(NH3) = (2.2-1.4)mol/0.95 L = 0.84M
Equilibrium concentration (N2) = 1.1/0.95 +0.42=1.58 M
Equilibrium concentration(NH3) = 1.4/0.95 = 1.47M
K = [N2]*{H2]/[NH3] = 1.58M*1.26M/1.47M = 1.35 M