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4 years ago

What type of bond is most likely being represented in row 1?

Chemistry

2 answers:

Zina [86]4 years ago

8 0

The main types of bonds are as follows:

Ionic: These are bonds that are formed by the complete transfer of electrons from one atom to the other

Covalent : These are bonds that are formed by the sharing of electrons between two atoms

Metallic: These are bonds or interactions that occur between ionized metal atoms and a sea of electrons surrounding them

Ion-dipole: These are bonds or attractive forces that operate between a polar molecule and an ion.

The only 2 elements in row 1 of the periodic table: hydrogen (H) and helium (He)

The bond formed between H atoms in H₂ is covalent

He₂ is very weakly held by Van Der waals forces and is unstable under normal conditions.

Based on the given options, the type of bond most likely being represented by row 1 is that of a covalent bond.

Angelina_Jolie [31]4 years ago

5 0

ionic is the awnser have a blessed day

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3 years ago

In the laboratory a student combines 26.2 mL of a 0.234 M chromium(III) acetate solution with 10.7 mL of a 0.461 M chromium(III)

Natalka [10]

Answer: The molarity of $Cr^{3+}$ ions in the solution is 0.299 M

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For chromium (III) acetate:

Molarity of chromium (III) acetate solution = 0.234 M

Volume of solution = 26.2 mL

Putting values in equation 1, we get:

$0.234=\frac{\text{Moles of chromium (III) acetate}\times 1000}{26.2}\\\\\text{Moles of chromium (III) acetate}=\frac{0.234\times 26.2}{1000}=0.00613mol$

1 mole of chromium (III) acetate $(Cr(CH_3COO)_3)$ produces 1 mole of chromium $(Cr^{3+})$ ions and 3 moles of acetate $(CH_3COO^-)$ ions

Moles of $Cr^{3+}\text{ ions}=(1\times 0.00613)=0.00613moles$

For chromium (III) nitrate:

Molarity of chromium (III) nitrate solution = 0.461 M

Volume of solution = 10.7 mL

Putting values in equation 1, we get:

$0.461=\frac{\text{Moles of chromium (III) nitrate}\times 1000}{10.7}\\\\\text{Moles of chromium (III) nitrate}=\frac{0.461\times 10.7}{1000}=0.00493mol$

1 mole of chromium (III) nitrate $(Cr(NO_3)_3)$ produces 1 mole of chromium $(Cr^{3+})$ ions and 3 moles of nitrate $(NO_3^-)$ ions

Moles of $Cr^{3+}\text{ ions}=(1\times 0.00493)=0.00493moles$

For chromium cation:

Total moles of chromium cations = [0.00613 + 0.00493] = 0.01106 moles

Total volume of solution = [26.2 + 10.7] = 36.9 mL

Putting values in equation 1, we get:

$\text{Molarity of }Cr^{3+}\text{ cations}=\frac{0.01106\times 1000}{36.9}\\\\\text{Molarity of }Cr^{3+}\text{ cations}=0.299M/tex]Hence, the molarity of [tex]Cr^{3+}$ ions in the solution is 0.299 M

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