As the elements in Period 2 of the Periodic Table are considered in succession from left to right, there is a decrease in atomic

HNO₃ → H⁺ + NO₃⁻

v₁=35.0 mL
c₁=0.255 mmol/mL
n₁(NO₃⁻)=v₁c₁

Mg(NO₃)₂ → Mg²⁺ + 2NO₃⁻

v₂=45.0 mL
c₂=0.328 mmol/mL
n₂(NO₃⁻)=2c₂v₂

c₃={n₁+n₂}/(v₁+v₂)={c₁v₁ + 2c₂v₂}/(v₁+v₂)

c₃={35.0*0.255+2*0.328*45.0}/(35.0+45.0)≈0.481 mmol/mL

a. 0.481 m

Although multiple values are given, our focus is on HCl.

<span>We have 215 mL (0.215 L) of 0.300 M HCl fully consumed in the reaction. It's important to recall that the number of moles is found by multiplying volume by molarity:</span>

 

moles = 0.215 L × 0.300 M

<span>moles = 0.0645 moles of HCl</span>

The oxidation state numbercan aid in identifying the unknown element present in both compounds. They denote the number of electrons that are either donated, received, or shared to yield compounds.

Remember the fundamental principles governing oxidation numbers.

1. In a neutral compound, the total of all oxidation numbers is zero.
2. Chlorine, bromine, and iodine typically exhibit an oxidation number of -1(unless paired with fluorine and oxygen)

Assume the oxidation state for element Mis designated as x.
Referring to rule 2, chlorine possesses an oxidation state of -1.

Now, for the compound MCl₂ (which is neutral), the equation can be formulated as
x + (2 * -1)= 0 ⇒ x₁= +2

For MCl₃, the corresponding equation is
x + (3 * -1)=  0 ⇒ x₂= +3

This indicates that the elementhas two distinct oxidation statesin its compounds, which are +2and +3.

The identified element is iron (Fe), as it shows +2 and +3 oxidation states across these compounds.

Memorizing this is essential. Regrettably, there isn't a simpler method to tackle these oxidation states.

The final answer is iron (Fe).

Utilize the ideal gas law:

n = PV / RT

P = 100kPa = 100 x 1000 x (9.8 x 10^{-6}) = 0.98 atm
Convert kPa to atm, where 1 Pa = 9.8 x 10^{-6} atm.
T = 293 K
V = 6.8 L
R = 1/12
Substituting all values leads to:
n = 0.272

Answer:

Explanation:

The relationship between the new temperature scale and the absolute temperature scale is defined as follows

Aw = 2 K

for K = 273.15 (the freezing point of water on the absolute scale)

Aw = 2 x 273.15 = 546.3 K

Each division of the new scale is equivalent to half that of each division on the absolute scale

each division of the new scale is minimal.

The value of R = 8.314 J per mole per K

Here, per K corresponds to 2Aw

Hence, the value of R in the new scale = 8.314/2 J per mole per Aw

= 4.157 J per mole per Aw

k = R / N

= 4.157 / 6.02 x 10²³

= .69 x 10⁻²³

= 6.9 x 10⁻²⁴ J per molecule per Aw .