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belka [17]
3 years ago
12

What do the atoms in Group 1 of the periodic table have in common

Chemistry
1 answer:
kari74 [83]3 years ago
8 0

Explanation:

The group 1 elements have similar properties because of the electronic structure of their atoms- they all have one electron in their outer shell.

I hope it's helpful!

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Write electron configurations for the following ion: Cd2 Cd2 . Express your answer in order of increasing orbital energy. For ex
Mashutka [201]

Answer:

Cd2+ : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10 or [Kr] 4d¹⁰

Explanation:

Before proceeding to write out the electron configuration of Cd2+, we have to obtain the electron configuration of Cadmium (Cd),

Cadmium has an atomic number of 48, this means that a neutral cadmium atom will have a total of  48  electrons surrounding its nucleus.

The electronic configuration of Cadmium is;

Cd: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10

The shorthand notation is given as;

Cd: [Kr] 4d¹⁰5s²

Cd2+ means that it has two less electrons, hence it's electron configuration is given as;

Cd2+ : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10 or [Kr] 4d¹⁰

5 0
3 years ago
78.6 grams of O2 and 67.3 grams of F2 are placed in a container with a volume of 40.6 L. Find the total pressure if the gasses a
saul85 [17]

1) List the known and unknown quantities.

<em>Sample: O2.</em>

Mass: 78.6 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

<em>Sample: F2.</em>

Mass: 67.3 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

2) Find the pressure of O2.

<em>2.1- List the known and unknown quantities.</em>

<em>Sample: O2.</em>

Mass: 78.6 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

<em>2.2- Convert grams of O2 to moles of O2.</em>

The molar mass of O2 is 31.9988 g/mol.

mol\text{ }O_2=78.6\text{ }g*\frac{1\text{ }mol\text{ }O_2}{31.9988\text{ }g\text{ }O_2}=2.46\text{ }mol\text{ }O_2

<em>2.3- Set the equation.</em>

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)

PV=nRT

<em>2.4- Plug in the known quantities and solve for P.</em>

(P)(40.6\text{ }L)=(2.46\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)

<em>.</em>

P_{O_2}=\frac{(2.46\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)}{40.6\text{ }L}P_{O_2}=1.57\text{ }atm

<em>The pressure of O2 is 1.57 atm.</em>

3) Find the pressure of F2.

<em>3.1- List the known and unknown quantities.</em>

<em>Sample: F2.</em>

Mass: 67.3 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

3.2- <em>Convert grams of F2 to moles of F2.</em>

The mmolar mass of F2 is 37.9968 g/mol.

mol\text{ }F_2=67.3\text{ }g\text{ }F_2*\frac{1\text{ }mol\text{ }F_2}{37.9968\text{ }g\text{ }F_2}=1.77\text{ }mol\text{ }F_2

<em>3.3- Set the equation.</em>

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)

PV=nRT

<em>3.4- Plug in the known quantities and solve for P.</em>

(P)(40.6\text{ }L)=(1.77\text{ }mol\text{ }F_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)

<em>.</em>

P_{F_2}=\frac{(1.77molF_2)(0.082057L*atm*K^{-1}*mol^{-1})(316.28K)}{40.6\text{ }L}P_{F_2}=1.13\text{ }atm

<em>The pressure of F2 is 1.13 atm.</em>

4) The total pressure.

Dalton's law - Partial pressure. This law states that the total pressure of a gas is equal to the sum of the individual partial pressures.

<em>4.1- Set the equation.</em>

P_T=P_A+P_B

4.2- Plug in the known quantities.

P_T=1.57\text{ }atm+1.13\text{ }atmP_T=2.7\text{ }atm

<em>The total pressure in the container is </em>2.7 atm<em>.</em>

5 0
1 year ago
Drag the labels onto the table to indicate when each statement Is true. Labels can be used once, more than once, or not at all.
liq [111]

The labels the table to indicate when each statement Is true. Labels can be used once, more than once, or not at all, The orange dye moves independently of the purple dye. 2. Concentration gradients exist that drive diffusion of both dyes. : <u>free water, solute, free water, solute.</u>

In chemistry, attention is the abundance of a constituent divided by way of the total quantity of an aggregate. several types of mathematical descriptions may be prominent: mass concentration, molar concentration, range concentration, and quantity concentration.

it's miles the amount of solute dissolves in one hundred g solvent. If the attention of the answer is 20 %, we understand that there are 20 g solutes in one hundred g solution. instance: 10 g salt and 70 g water are mixed and the solution is ready. find awareness of the answer by means of percentage mass.

Learn more about concentration here:

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3 0
1 year ago
What is the first anmial
denpristay [2]
The first animal was called the comb jelly
8 0
4 years ago
1 point
blsea [12.9K]

Answer:

It freezes at 32 f

Explanation:

6 0
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