Why do you think sodium bicarbonate is included to neutralize an acidic spill rather than sodium hydroxide?

A triprotic acid is a type of Arrhenius acid that has the ability to donate three protons per molecule during dissociation in aqueous solutions. Thus, the chemical reaction, as outlined in the question, at the third equivalence point, can be expressed as: H3R + 3NaOH ⇒ Na3R + 3H2O, where R denotes the counter ion of the triprotic acid. Consequently, the ratio of reacted acid to base at this point is 1:3.
The moles of NaOH are calculated as 0.106M*0.0352L = 0.003731 mole. Therefore, the amount of H3R is 0.003731mole/3=0.001244mole.
Subsequently, the molar mass of the acid can be determined: 0.307g/0.001244mole=247 g/mol.

Response:B) The octet of the carbon atom is incomplete within the molecule.

Explanation:

K e q equals StartFraction StartBracket upper C EndBracket superscript lower c StartBracket upper D EndBracket superscript lower D over StartBracket upper A EndBracket superscript lower a StartBracket upper B EndBracket superscript lower b EndFraction. To clarify the query, let's express the components in the format of a standard chemical reaction equation: aA + bB ⇄cC + dD. We should bear in mind that the equilibrium constant, commonly represented as Keq, is defined as the ratio of the product concentrations of products raised to the power of their respective molar coefficients to that of reactants raised to their respective molar coefficients. Hence, based on the previous description, we can assert that Keq= [C]^c [D]^d / [A]^a [B]^b. This is expressed in words in the answer provided above.

Density is calculated as Mass divided by Volume.

Volume can also be expressed as Mass divided by Density.

Volume becomes 2 divided by 128.

This results in Volume equal to 1/64.

Therefore, each side measures 4 cm.

a.

b.

c.

d.

Given lithium's atomic mass of 7 and atomic number of 3, we can express it in standard notation with the atomic mass as a superscript and the atomic number as a subscript:

a. The atomic number signifies the total number of protons present in an element. Adding a proton to lithium will increase the atomic number by one. Additionally, modifying the atomic number alters the element, since each element has a distinct atomic number.

It's critical to acknowledge that the mass of an atom is the sum of protons and neutrons; thus, we would also add one to the mass number. This gives us some new species X with:

To identify X, we must locate an element with Z = 4 on the periodic table, which is beryllium:

b. A standard lithium atom possesses an atomic number of 3, indicating it has 3 protons. Since it’s an atom, the positive charge from the protons equals the negative charge from the electrons. Thus, a neutral atom comprises the same number of protons and electrons.

Lithium begins with 3 electrons. If we remove one electron, it results in a lithium cation with a +1 charge, leading to a net charge of +3 from protons and -2 from electrons.

Hence, it can be represented as a lithium cation with a +1 charge:

c. Neutrons, while neutral, contribute to the overall mass of an atom. Therefore, adding a neutron does not affect the overall charge (atomic number) of lithium.

However, this would augment the mass by 1, as each neutron (and proton) counts as 1 atomic mass unit. Since the atomic number remains unchanged, it would still be lithium.

d. By first removing a proton, the atomic number of Li decreases by 1 unit, as the atomic number aligns with the proton count:

The mass will also reduce by 1 unit, accounting for the protons and neutrons:

Following this step, we end up with helium (which has an atomic number of 2) and a mass of 6:

Then, removing 2 neutrons will decrease the mass by 2 units while the charge of He remains unchanged, as this step does not involve protons and neutrons do not alter a nucleus's charge:

Lastly, if an electron is removed, this results in a helium cation with a +1 charge, leaving 2 protons and 1 remaining electron after removing 1 electron from helium which initially had 2 electrons in its atomic state.