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stiv31 [10]
2 years 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]2 years 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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The limiting reactant determines what the actual yield is. (T/F)
g100num [7]

Answer:

True

Explanation:

Limiting reactant - the reactant which get completely consumed in a chemical reaction , is known as the limiting reactant .

As, the concentration of limiting reactant after the completion of the reaction will be zero , hence, it is used to determine the concentration of other reactants .

For example,

for a general reaction -

A + B ---> 3C

Assuming B to be the limiting reactant ,

hence, the concentration of C and A can be determined as -

1 mol of B can give 3 mol of C and 1 mol of A is used for the reaction.

6 0
3 years ago
Use the Ref An aqueous solution of chromium(II) acetate has a concentration of 0.260 molal. The percent by mass of chromium(II)
poizon [28]

Answer:

The percent by mass of chromium(II) acetate in the solution is 4.42%.

Explanation:

Molality of the chromium(II) acetate solution = 0.260 m = 0.260 mol/kg

This means that in 1 kg of solution 0.260 moles of chromium(II) acetate are present.

1000 g = 1 kg

So, in 1000 grams of solution 0.260 moles of chromium(II) acetate are present.

Then in 100 grams of solution = \frac{0.260 mol}{1000}\times 100=0.0260 mol

Mass of 0.0260 moles chromium(II) acetate:

= 0.0260 mol × 170 g/mol = 4.42 g

(w/w)\%=\frac{\text{mass of solute in 100 gram solution}}{100}\times 100

=\frac{4.42 g}{100 g}\times 100=4.42\%

The percent by mass of chromium(II) acetate in the solution is 4.42%.

7 0
3 years ago
The specific heat capacity of titanium is 0.523j/g c what is the heat capacity of 2.3g of titanium
kodGreya [7K]

Answer:

1.2029 J/g.°C

Explanation:

Given data:

Specific heat capacity of titanium = 0.523 J/g.°C

Specific heat capacity of 2.3 gram of titanium = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

1 g of titanium have 0.523 J/g.°C specific heat capacity

2.3  × 0.523 J/g.°C

1.2029 J/g.°C

8 0
3 years ago
What are at least five major systems that make up the human body
bezimeni [28]

Muscular, respiratory, skeletal, digestive,
7 0
3 years ago
when 4.008 g of calcium metal is heated in the air, the sample gains 1.600 g of oxygen in forming the oxide. Calculate the empir
horsena [70]

Answer:

CaO

Explanation:

Firstly, let’s get the percentage compositions.

The total mass is now = 4.008 + 1.600 = 5.608g

The percentage composition of carbon is thus 4.008/5.608 * 100 = 71.47%

The percentage composition of oxygen is 100 - 71.4 = 28.53%

Now, we divide each percentage by the atomic masses. The atomic mass of calcium is 40 while that of oxygen is 16

Ca = 71.47/40 = 1.78675

O = 28.53/16 = 1.783125

Both values are quite similar and dividing by the smallest will yield same values of 1.

Hence , the empirical formula is caO

3 0
3 years ago
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