<span>The maximum number of electrons in a single d-subshell is:
10</span>
Answer: -
71
Explanation: -
From the diagram, we see that volume goes from 70 to 75 in 5 markings.
Each marking is for 1.
It is the most accuracy possible as there is no smaller marking.
Significant figures expresses the required amount of accuracy.
Thus the volume indicated from the diagram is 71 .
The statement 'whether people should take medicine or if they should seek alternative treatments' describes an ethical dilemma that drug designers face.
<h3>What is drug development?</h3>
Drug development refers to all the processes from target drug identification to drug validation and commercialization.
Drug development involves different stages of development including preclinical and clinical trials.
Ethical dilemmas in drug development include the release of a drug that is ineffective when compared to parallel treatments.
Learn more about drug development here:
brainly.com/question/8187660
<span>0.0687 m
The balanced equation is
BaCl2 + Na2SO4 ==> BaSO4 + 2 NaCl
Looking at the equation, it indicates that there's a 1 to 1 ratio of BaCl2 and Na2SO4 in the reaction. So the number of moles of each will be equal. Now calculate the number of moles of Na2SO4 we had. Start by looking up atomic weights.
Atomic weight sodium = 22.989769
Atomic weight sulfur = 32.065
Atomic weight oxygen = 15.999
Molar mass Na2SO4 = 2 * 22.989769 + 32.065 + 4 * 15.999 = 142.040538 g/mol
Moles Na2SO4 = 0.554 g / 142.040538 g/mol = 0.003900295 mol
Molarity is defined as moles per liter, so let's do the division.
0.003900295 mol / 0.0568 l = 0.068667165 mol/l = 0.068667165 m
Rounding to 3 significant figures gives 0.0687 m</span>
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%
