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kramer
3 years ago
14

What do the Roman numerals in a cation's name indicate?

Chemistry
2 answers:
Agata [3.3K]3 years ago
6 0
The Roman numerals in a cation's name indicate: THE POSITIVE CHARGE ON THE CATION

Cations are metallic atoms that loosely hold it electrons, making it easy to lose electrons.

The Roman numerals in a cation's name not only indicates the charge on the cation but it makes it easier to distinguish cations that share the same metal name. 
erik [133]3 years ago
6 0

Answer: Option (c) is the correct answer.

Explanation:

When an element gain electrons then it acquires a negative charge. For example, Cl + 1e^{-} \rightarrow Cl^{-}

Whereas when an element loses electrons then it acquires a positive charge.

For example, Fe \rightarrow Fe^{2+} + 2e^{-1}

The atom Fe^{2+} can also be written using roman numerals as Fe(II).

Therefore, we can conclude that roman numerals in a cation's name indicate the positive charge on the cation.

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Explanation:

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3 years ago
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Given that the molar mass of NaCl is 58.44 g/mol, Solve for the molarity of a solution that contains 87.75 g of NaCl in 500 mL o
aliina [53]

Molarity is defined as the number of moles of solute per 1 L of solvent.

the mass of NaCl in the solution is 87.75 g

number of moles of NaCl is calculated by dividing mass present by molar mass

number of NaCl moles = 87.75 g / 58.44 g/mol = 1.502 mol

the number of NaCl moles in 500 mL is - 1.502 mol

therefore number of NaCl moles in 1000 mL is - 1.502 mol/ 500 mL x 1000 mL/L = 3.004 mol

molarity of NaCl is - 3.004 M

answer is D. 3.00 M

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3 years ago
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Which of the following is the preferred degree for a seismologist?
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5 0
3 years ago
If I have 4.5 liters of gas at a temperature of 33 0C and a pressure of 6.54 atm, what will be the pressure of the gas if I rais
Nikitich [7]

This is an exercise in the general or combined gas law.

To start solving this exercise, we must obtain the following data:

<h3>Data:</h3>
  • V₁ = 4.5 l
  • T₁ = 33 °C + 273 = 306 k
  • P₁ = 6.54 atm
  • T₂ = 94 °C + 273 = 367 k
  • V₂ = 2.3 l
  • P₂ = ¿?

We use the following formula:

  • P₁V₁T₂ = P₂V₂T₁ ⇒ General Formula

Where

  • P₁ = Initial pressure
  • V₁ = Initial volume
  • T₂ = Initial temperature
  • P₂ = Final pressure
  • V₂ = Final volume
  • T₁ = Initial temperature

We clear the general formula for the final pressure.

\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{P_{1}V_{1}T_{2} }{V_{2}T_{1}}  \ \to \ Clear \ formula \end{gathered}$}

We solve by substituting our data in the formula:

\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{(6.54 \ atm)(4.5 \not{l})(367 \not{K}) }{(2.3 \not{l})(306 \not{k})}  \end{gathered}$}

\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{10800.81}{ 703.8 } \ atm  \end{gathered}$}

\boxed{\large\displaystyle\text{$\begin{gathered}\sf P_{2}=15.346 \ atm \end{gathered}$} }

If I raise the temperature to 94°C and decrease the volume to 2.3 liters, the pressure of the gas will be 15,346 atm.

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2 years ago
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