Answer:
68133080.02 g
Explanation:
I believe that the question is to find the mass of air in the room and not the molar mass of air since the molar mass of air was already given in the question as 28.97 g/mol.
Now, if 1 mole of a gas occupies 22.4 L
x moles of air occupies 52,681,428.8 Liters
x = 1 * 52,681,428.8 /22.4
x = 2351849.5 moles of air
Now, number of moles = mass/ molar mass
but molar mass = 28.97 g/mol
2351849.5 = mass/28.97
mass = 2351849.5 * 28.97
mass = 68133080.02 g
Non metals form ionic bonds by gaining and losing electrons to complete their valence she'll and increase stability.
Metals form covalent bonds by sharing electrons.
Answer:
Concept: Chemical Analysis
- Start by taking inventory of the elements that you have
- Make a list, one for the right side and another for the left side
- Then add coefficients to the elements to the right or left side to balance out the equation
Answer:
...BY CHOOSING A PROPER INDICATOR, SCIENTISTS CAN MINIMIZE THE DIFFERENCE IN THESE TWO NUMBERS,...
Explanation:
A chemical indicator is an agent or substance which gives a visible sign especially a color change when introduced to a solution of base indicating the threshold of the concentration of acid used in the titration procedure. These indicators include methyl orange, methyl red, phenolpthalein. They give a color change when the concentration of an acid as reached a critical limit in order for the reaction to be stopped. These indicators are involved in acid-base titrations, oxidation- reduction reactions and so on. More accurate results of the volume of the required acid or base is obtained by the introduction of these indicators.
Answer:
0.11%
Explanation:
Without mincing words, let us dive straight into the solution to the question/problem. The first step to solve this question is to write out the chemical reaction, that is the reaction showing the dissociation of acetic acid.
CH3COOH <=======================================> CH3COO⁻ + H⁺
Initially, the amount present in the acetic acid which is = 12M, the concentration for CH3COO⁻ and H⁺ is 0 respectively.
At equilibrium, the amount present in the acetic acid which is = 12 - x, the concentration for CH3COO⁻ = x and H⁺ = x respectively. Note that the ka for acetic acid = 1.8 × 10⁻⁵.
1.8 × 10⁻⁵ = x²/ 14 - x. Therefore, x = 0.0158 M.
The next thing to do is to calculate for the percentage of dissociation, this can be done as given below:
percentage of dissociation = x/14 × 100. Recall that the value that we got for x = 0.0158 M. Hence, the percentage of dissociation = 0.0158 M/ 14m × 100 = 0.11%
