1.Which element would most likely lose electrons to form positive ions when bonding with other elements?

1 answer:

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1) The element that will most likely lose electrons to form positive ions when bonding with other elements is rubidium (Rb).

2) The correct statement about sodium atoms is; "The sodium atom transfers electrons to the chlorine atoms to form ionic bonds."

3) Based on their location in the periodic table, nitrogen (N) and oxygen (O) are most likely to form covalent bonds with each other

4) Electronegativity is best described by the phrase; "the relative strength with which an element attracts electrons in a chemical bond"

Metals of group 1 and 2 are highly electropositive and are more likely to loose electrons in a bonding situation. Therefore, the element that will most likely lose electrons to form positive ions when bonding with other elements is rubidium (Rb).

Sodium chloride is an ionic compound. It is formed by transfer of electrons from sodium to chlorine. Sodium is highly electropositive while chlorine is highly electronegative. Therefore, sodium chloride is formed when sodium atom transfers electrons to the chlorine atoms to form ionic bonds.

Covalent bonds are formed between two nonmetals. Nitrogen and oxygen are non metals hence they form covalent bonds.

According to Linus Pauling, electronegativity refers to the ability of an element in a compound to draw electrons towards itself.

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A pump contains 1.5 L of air at 175 kPa. You draw back on the piston of the pump, expanding the volume until the pressure reads

Tresset [83]

Answer:

$\large \boxed{\text{86.8 L}}$

Explanation:

The temperature and amount of gas are constant, so we can use Boyle’s Law.

$p_{1}V_{1} = p_{2}V_{2}$

Data:

$\begin{array}{rclrcl}p_{1}& =& \text{0.579 atm}\qquad & V_{1} &= & \text{150 L} \\p_{2}& =& \text{1.00 atm}\qquad & V_{2} &= & ?\\\end{array}$

Calculations:

$\begin{array}{rcl}0.579 \times 150 & =& 1.00V_{2}\\86.85 & = & 1.00V_{2}\\V_{2} & = &\dfrac{86.85}{1.00}\\\\& = &\textbf{86.8}\\\end{array}\\\text{The new volume of the gas is } \large \boxed{\textbf{86.8 L}}$