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stiv31 [10]
1 year ago
6

consider the fourth period elements ca, mn, co, se, and kr. which of these atoms are paramagnetic? a) ca, mn, co b) mn, co, kr c

) ca, kr, se d) mn, co, se e) ca, kr
Chemistry
1 answer:
Andrei [34K]1 year ago
6 0

Option (a) Ca, Mn, Co are paramagnetic elements in the fourth period of the periodic table due to unpaired electrons.

A substance's electron configuration can be used to assess if it has magnetic properties: The material is paramagnetic if any of its electrons are unpaired, and it is diamagnetic if all of its electrons are paired.

Materials with hindered electrons that are drawn to magnetic fields are considered to be paramagnetic. In the absence of a magnetic field, they become less magnetic. The magnetic moment of the material and, consequently, the paramagnetism, increase with the number of unpaired electrons.

Mn^{2+} is the ion with the most paramagnetic behaviour because it has the most unpaired electrons when two electrons are lost.

In this instance, calcium is a paramagnetic element as well. This is an anomaly. Due to the ability of one of calcium's electrons in the S orbital to jump to the D orbital, the element is paramagnetic. Since calcium is in Period 4, its D orbital is unoccupied. D orbitals exist in elements with period 4. Thus, this provides that electron with a destination.

Three electrons are not paired in cobalt. It is hence paramagnetic. Three unpaired electrons are present in the outermost subshell.

Learn more about Paramagnetic elements here:

brainly.com/question/1594086

#SPJ4

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What is the volume of 0.640 grams of Oz gas at Standard Temperature and Pressure (STP)?
Ilia_Sergeevich [38]

Answer: The volume of 0.640 grams of O_{2} gas at Standard Temperature and Pressure (STP) is 0.449 L.

Explanation:

Given: Mass of O_{2} gas = 0.640 g

Pressure = 1.0 atm

Temperature = 273 K

As number of moles is the mass of substance divided by its molar mass.

So, moles of O_{2} (molar mass = 32.0 g/mol) is as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{0.640 g}{32.0 g/mol}\\= 0.02 mol

Now, ideal gas equation is used to calculate the volume as follows.

PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

PV = nRT\\1.0 atm \times V = 0.02 mol \times 0.0821 L atm/mol K \times 273 K\\V = 0.449 L

Thus, we can conclude that the volume of 0.640 grams of O_{2} gas at Standard Temperature and Pressure (STP) is 0.449 L.

6 0
3 years ago
The ksp of lead iodide is 7.1 × 10-9. a chemical engineer adds 0.0025 mol of ki to a solution of 0.00004 mol pb(no3)2 in 500 ml
inn [45]
Answer is: n<span>o, because the ion product is less than the Ksp of lead iodide. </span>

Chemical dissociation 1: KI(s) → K⁺(aq) + I⁻(aq).
Chemical dissociation 2: Pb(NO₃)₂(s) → Pb²⁺(aq) + 2NO₃⁻(aq).
Chemical reaction: Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s).
Ksp(PbI₂) = 7.1·10⁻⁹.
V = 500 mL ÷ 1000 mL/L = 0.5 L.
c(KI) = c(I⁻) = 0.0025 mol ÷ 0.5 L.
c(I⁻) = 0.005 M.
c(Pb(NO₃)₂) = c(Pb²⁺) = 0.00004 mol ÷ 0.5 L.
c(Pb²⁺) = 0.00008 M.
Q = c(Pb²⁺) · c(I⁻)².
Q = 8·10⁻⁵ M · (5·10⁻³ M)².
Q = 2·10⁻⁹; <span> the ion product.</span>

8 0
3 years ago
This is the question i need help on
mash [69]
I thank that your answer is C.
8 0
3 years ago
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