All categories

3 years ago

After the transfer of the electron, sodium will form an ion with __________.

Chemistry

2 answers:

ladessa [460]3 years ago

4 0

The most common pairing would be sodium forming an ion after transfering an electron to chlorine

Lerok [7]3 years ago

3 0

It will have a +1 charge. Sodium likes to give up electrons.

You might be interested in

write equations to show the chemical processes which occur when the first ionization and the second ionization energies of lithi

diamong [38]

Answer:

First ionization of lithium:

$\text{Li}\;(g)\to \text{Li}^{+} \; (g) + \text{e}^{-}$ .

Second ionization of lithium:

$\text{Li}^{+}\;(g) \to\text{Li}^{2+} \;(g) + \text{e}^{-}$ .

Explanation:

The ionization energy of an element is the energy required to remove the outermost electron from an atom or ion of the element in gaseous state. (Refer to your textbook for a more precise definition.) Some features of the equation:

Start with a gaseous atom (for the first ionization energy only) or a gaseous ion. Write the gaseous state symbol $(g)$ next to any atom or ion in the equation.
The product shall contain one gaseous ion and one electron. The charge on the ion shall be the same as the order of the ionization energy. For the second ionization energy, the ion shall carry a charge of +2.
Charge shall balance on the two sides of the equation.

First Ionization Energy of Li:

The products shall contain a gaseous ion with charge +1 $\text{Li}^{+}\;(g)$ as well as an electron $\text{e}^{-}$ .
Charge shall balance on the two sides. There's no net charge on the product side. Neither shall there be a charge on the reactant side. The only reactant shall be a lithium atom which is both gaseous and neutral: $\text{Li}\;(g)$ .

Hence the equation: $\text{Li}\;(g) \to \text{Li}^{+}\;(g) + \text{e}^{-}$ .

Second Ionization Energy of Li:

The product shall contain a gaseous ion with charge +2: $\text{Li}^{2+}\;(g)$ as well as an electron $\text{e}^{-}$ .
Charge shall balance on the two sides. What's the net charge on the product side? That shall also be the charge on the reactant side. What will be the reactant?

The equation for this process is $\text{Li}^{+} \; (g) \to \text{Li}^{2+}\;(g) + \text{e}^{-}$ .

5 0

3 years ago

The cell theory was construted by

NeTakaya

Answer:

Matthias Jakob Schleiden

Explanation:

The cell was initially discovered by Robert Hooke, but Schleiden developed the theory.

Hope this helps! :)

4 0

3 years ago

The compound methyl butanoate smells like apples. Its percent composition is 58.8% C, 9.9% H, and 31.4% O. What’s the empirical

blsea [12.9K]

To find the empirical formula you would first need to find the moles of each element:

58.8g/ 12.0g = 4.9 mol C

9.9g/ 1.0g = 9.9 mol H

31.4g/ 16.0g = 1.96 O

Then you divide by the smallest number of moles of each:

4.9/1.96 = 2.5

9.9/1.96 = 6

1.96/1.96 = 1

Since there is 2.5, you find the least number that makes each moles a whole number which is 2.

So the empirical formula is C5H12O2.

6 0

3 years ago

Determine the oxidation number of the element "J" in H3JO2-

igor_vitrenko [27]

Answer:

Oxidation number:

3*1+ oxidation number of J+2*-2= -1

Oxidation number of J = 0

5 0

2 years ago

How many grams of a stock solution that is 92.5 percent H2SO4 by mass would be needed to make 250 grams of a 35.0 percent by mas

IrinaVladis [17]

94.6 g. You must use 94.6 g of 92.5 % H_2SO_4 to make 250 g of 35.0 % H_2SO_4.

We can use a version of the dilution formula

m_1C_1 = m_2C_2

where

m represents the mass and

C represents the percent concentrations

We can rearrange the formula to get

m_2= m_1 × (C_1/C_2)

m_1 = 250 g; C_1 = 35.0 %

m_2 = ?; _____C_2 = 92.5 %

∴ m_2 = 250 g × (35.0 %/92.5 %) = 94.6 g

4 0

3 years ago