This problem is providing us with the volume of nitric acid that is titrated with 0.18 L of 0.1-M sodium hydroxide and asks for the concentration of the acid. At the end, the result turns out to be 0.045M, according to the following.
<h3>Acid-base titrations:</h3>
In chemistry, acid-base titrations allow us to quantify the volume or concentration of an acid or base via the following equation:
Where the subscript A stands for the acid and B for the base; which means one can calculate any of the variables there by knowing the other three. This equation is based on the balanced neutralization chemical equation, which takes place between the acid and the base.
Thus, we can write the reaction between NaOH and HNO3 as:
In such a way, we can solve for the concentration of the acid as shown below:
Learn more about titration: brainly.com/question/25485091
Answer:
2.76 × 10⁻¹¹
Explanation:
I don’t have access to the ALEKS Data resource, so I used a different source. The number may be different from yours.
1. Calculate the free energy of formation of CCl₄
C(s)+ 2Cl₂(g)→ CCl₄(g)
ΔG°/ mol·L⁻¹: 0 0 -65.3
ΔᵣG° = ΔG°f(products) - ΔG°f(reactants) = -65.3 kJ·mol⁻¹
2. Calculate K
T = (25.0 + 273.15) K = 298.15 K
Answer: Atoms with 11 protons, 10 neutrons and 11 electrons belong to the same element with 11 protons, 12 neutrons and 11 electrons.
Explanation:
Elements that contain same number of valence electrons belong to the same group. This is because they will have same reactivity (or properties) due to which they lie in the same group.
For example, element with 11 protons, 10 neutrons and 11 electrons is same as the element with 11 protons, 12 neutrons and 11 electrons.
Hence, both these atoms belong to the same element.
Thus, we can conclude that atoms with 11 protons, 10 neutrons and 11 electrons belong to the same element with 11 protons, 12 neutrons and 11 electrons.
Ammonification refers to chemical reactions in which amino groups (NH2) associated with organic forms of nitrogen are converted into ammonia (NH3) or ammonium (NH4+).
