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3 years ago

Which electron configurations represent the first two elements in Group 17 (7A) of the Periodic Table?

Chemistry

2 answers:

melisa1 [442]3 years ago

7 0

Answer:a

Explanation:

SpyIntel [72]3 years ago

7 0

Answer:

Option C is correct

1s22s22p5 and [Ne]3s23p5

Explanation:

Step 1: Data given

Group 17 = the halogens

Halogens have 7 valence electrons

The first element of group 17 = Fluorine (F)

Fluorine has 9 electrons

The electron configuration is

1s2 2s2 2p5

The second element of group 17 = Chlorine (Cl)

Chlorine has 17 electrons

The electron configuration is

1s2 2s2 2p6 3s2 3p5

Option C is correct

1s22s22p5 and [Ne]3s23p5

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3 years ago

Arrange the steps in order to describe how photons and carbon dioxide contribute to the warming of earth

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Answer:

The steps are arranged in the following order, describing how carbon dioxide and photons impact the earth's warming.

Explanation:

The sequence is in the following order;

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4 years ago

Calculate by (a)% weight and (b) %mole each of the elements present in sugar

Musya8 [376]

Explanation:

Molecular mass of sugar = $C_{12}H_{22}O_{11}$ : = 432 g/mol

Atomic mass of carbon atom = 12 g/mol

Atomic mass of hydrogen atom = 1 g/mol

Atomic mass of oxygen atom = 16 g/mol

a) Percentage of an element in a compound:

$\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{molecular mass of compound}}\times 100$

Percentage of carbon by weight in $C_{12}H_{22}O_{11}$ :

$\frac{12\times 12 g/mol}{342 g/mol}\times 100=42.10\%$

Percentage of hydrogen by weight in $C_{12}H_{22}O_{11}$ :

$\frac{22\times 1g/mol}{342 g/mol}\times 100=6.43\%$

Percentage of oxygen by weight in $C_{12}H_{22}O_{11}$ :

$\frac{11\times 16g/mol}{342 g/mol}\times 100=51.46\%$

b) Percentage of mole each of the elements present in sugar:

= $\frac{\text{Moles of atoms of an element}}{\text{total moles of all types of atoms}}\times 100$

In mole of sugar we have 12 moles of carbon atom , 22 moles of hydrogen atoms and 11 moles of oxygen atoms.

Percentage of carbon by mole in $C_{12}H_{22}O_{11}$ :

$\frac{12 mol}{45 mol}\times 100=26.66\%$

Percentage of hydrogen by mole in $C_{12}H_{22}O_{11}$ :

$\frac{22 mol}{45 mol}\times 100=48.88\%$

Percentage of oxygen by mole in $C_{12}H_{22}O_{11}$ :

$\frac{11 mol}{45 mol}\times 100=24.44\%$

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4 years ago

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3 years ago

A 13.0% solution of K2CO3 by mass has a density of 1.09 g/cm3. Calculate the molality of the solution.

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