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xenn [34]
2 years ago
13

Ionization energy generally decreases down a group because as one moves down a group, the outermost electron moves _____ the nuc

leus and it takes _____ energy to remove it.
Chemistry
1 answer:
GarryVolchara [31]2 years ago
7 0

Ionization energy generally decreases down a group because as one moves down a group, the outermost electron moves <u>further away </u>from the nucleus and it takes <u>less</u> energy to remove it.

Ionization energy, also known as ionization energy, would be the minimal amount of energy needed to free an isolated gaseous atom's or molecule's least loosely bonded electron.

First ionization energy often drops as you advance down a group on the periodic table. This occurs even though the outermost electron would be typically held less securely and can be removed with less energy since it travels farther away from the nucleus.

Therefore, Ionization energy generally decreases down a group because as one moves down a group, the outermost electron moves <u>further away </u>from the nucleus and it takes <u>less</u> energy to remove it.

To know more about Ionization energy

brainly.com/question/16243729

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How many kilograms of potassium iodide (ki) are needed to make 1.25 l of a 4.41 m ki solution?
Lapatulllka [165]
C = n/V
n = C×V
n = 4,41M × 1,25L
n = 5,5125 mol

mKI: 39+127 = 166 g/mol

1 mol --------- 166g
5,5125 mol --- X
X = 166×5,5125 = 915,075g KI

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7 0
3 years ago
A calorimeter contains 22.0 mL of water at 14.0 ∘C . When 2.50 g of X (a substance with a molar mass of 82.0 g/mol ) is added, i
Alik [6]

Answer:

The enthalpy change in the the reaction is -47.014 kJ/mol.

Explanation:

X(s)+H_2O(l)\rightarrow X(aq)

Volume of water in calorimeter = 22.0 mL

Density of water = 1.00 g/mL

Mass of the water in calorimeter = m

m=1.00 g/mL\times 22.0 mL=22 g

Mass of substance X = 2.50 g

Mass of the solution = M = 2.50 g + 22 g = 24.50 g

Heat released during the reaction be Q

Change in temperature =ΔT = 28.0°C - 14.0°C = 14.0°C

Specific heat of the solution is equal to that of water :

c = 4.18J/(g°C)

Q=Mc\times \Delta T

Q=24.50 g\times 4.18 J/g ^oC\times 14.0^oC=1,433.74 J=1.433 kJ

Heat released during the reaction is equal to the heat absorbed by the water or solution.

Heat released during the reaction =-1.433 kJ

Moles of substance X= \frac{2.50 g}{82.0 g/mol}=0.03048 mol

The enthalpy change, ΔH, for this reaction per mole of X:

\Delta H=\frac{-1.433 kJ}{0.03048 mol}=-47.014 kJ/mol

5 0
3 years ago
Estimate the ligand fi eld splitting for (a) [CrCl6]3 (max 740 nm), (b) [Cr(NH3)6]3 (max 460 nm), and (c) [Cr(OH2)6]3 (max 575 n
igor_vitrenko [27]

Answer:

NH3>H2O>Cl-

Explanation:

The given wavelengths of maximum absorption for each complex can be used to estimate the magnitude of field splitting of the respective ligands as shown in the image attached. The field splitting is reported in the unit kilojoule per mole (KJmol-1).

It can be seen from the calculation in the image attached that ammonia shows the highest crystal field splitting followed by water and lastly the chloride anion. This corresponds to the respective positions of these species in the spectrochemical series. Water and the chloride ion are weak field ligands.

4 0
3 years ago
CAN SOMEONE HELP ME PLZ AND THANKS WILL MARK U AS BRAINLIEST
jekas [21]

Explanation:

2. 2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O

First, we need to find the number of moles of CO_2 at 300K and 1.5 atm using the ideal gas law:

n= \dfrac{PV}{RT}= \dfrac{(1.5\:\text {atm})(33\:L)}{(0.082\:\text{L-atm/mol-K})(300K)}

=2.0\:\text{mol}\:CO_2

Now use the molar ratios to find the number of moles of ethane to produce this much CO_2.

2.0\:\text{mol}\:CO_2 \times \left(\dfrac{2\:\text{mol}\:C_2H_6}{4\:\text{mol}\:CO_2}\right)

=1.0\:\text{mol}\:C_2H_6

Finally, convert this amount to grams using its molar mass:

1.0\:\text {mol}\:C_2H_6 \times \left(\dfrac{30.07\:\text g\:C_2H_6}{1\:\text{mol}\:C_2H_6} \right)

=30.1\:g\:C_2H_6

3. 3Zn + 2H_3PO_4 \rightarrow 3H_2 + Zn_3(PO_4)_2

Convert 75 g Zn into moles:

75\:\text g\:Zn \times \left(\dfrac{65.38\:\text g\:Zn}{1\:\text{mol}\:Zn}\right)=1.1\:\text{mol}\:Zn

Then use the molar ratios to find the amount of H2 produced.

1.1\:\text{mol}\:Zn \times \left(\dfrac{3\:\text{mol}\:H_2}{3\:\text{mol}\:Zn}\right)=1.1\:\text{mol}\:H_2

Now use the ideal gas law PV=nRT to find the volume of H2 produced at 23°C and 4 atm:

V= \dfrac{nRT}{P}= \dfrac{(1.1\:\text{mol}\:H_2)(0.082\:\text{L-atm/mol-K})(296K)}{4\:\text{atm}}

=8.9\:\text L\:H_2

8 0
3 years ago
What is the molarity of a solution containing 56 grams of LiF in 959 mL of solution?
andrezito [222]

Answer:

2.25 M     <========= I do not see this in your selection of answers !

Explanation:

Mole weight of  Li F

      ( from periodic table ) = 6.94 + 18.998  = 25.938 gm/mole

56 gm / ( 25.938 gm /mole) = 2.15899 moles

2.15899 mole / (.959) L = 2.25 M

7 0
2 years ago
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