Answer:
324.18 g/mol
Explanation:
Let the molecular mass of the antimalarial drug, Quinine is x g/mol
According to question,
Nitrogen present in the drug is 8.63% of x
So, mass of nitrogen =
Also, according to the question,
2 atoms are present in 1 molecule of the drug.
Mass of nitrogen = 14.01 amu = 14.01 g/mol (grams for 1 mole)
So, mass of nitrogen = 14.01×2 = 28.02
These 2 must be equal so,
solving for x, we get:
<u>x = 324.18 g/mol</u>
The answer is 2 electrons.
The electron configuration of calcium is 2:8:8:2
Calcium has two electrons in its outermost shell. These are its valence electrons and are the ones used in bonding with other elements. Valence electrons of an atom are those electrons that are in its outer energy shell or that are available for bonding.
Calcium is a metal. When metals react with non-metals, electrons are transferred from the metal atoms to the non-metal atoms forming ions. The resulting compound is known as an ionic compound.
For example, when calcium metal reacts with chlorine gas, calcium gives up its two valence electrons and Chlorine accepts them resulting in a new substance called calcium chloride in which the two elements have ended up forming ionic bonds.
Answer:
A one-step mechanism involving a transition state that has a carbon partially bonded to both chlorine and oxygen
Explanation:
The compound CH3Cl is methyl chloride. This is a nucleophilic substitution reaction that proceeds by an SN2 mechanism. The SN2 mechanism is a concerted reaction mechanism. This means that the departure of the leaving group is assisted by the incoming nucleophile. The both species are partially bonded to opposite sides of the carbon atom in the transition state.
Recall that an SN2 reaction is driven by the attraction between the negative charge of the nucleophile (OH^-) and the positive charge of the electrophile (the partial positive charge on the carbon atom bearing the chlorine leaving group).
Answer:
Qsp > Ksp, BaCO3 will precipitate
Explanation:
The equation of the reaction is;
Na2CO3 + BaBr2 -------> 2NaBr + BaCO3
Since BaCO3 may form a precipitate we can determine the Qsp of the system.
Number of moles of Na2CO3 = 0.96g/106 g/mol = 9.1 * 10^-3 moles
concentration of NaCO3 = number of moles/volume of solution = 9.1 * 10^-3 moles/10 L = 9.1 * 10^-4 M
Number of moles of BaBr2 = 0.20g/297 g/mol = 6.7 * 10^-4 moles
concentration of BaBr2 = 6.7 * 10^-4 moles/10 L = 6.7 * 10^-5 M
Hence;
[Ba^2+] = 6.7 * 10^-5 M
[CO3^2-] = 9.1 * 10^-4 M
Qsp = [6.7 * 10^-5] [9.1 * 10^-4]
Qsp = 6.1 * 10^-8
But, Ksp for BaCO3 is 5.1*10^-9.
Since Qsp > Ksp, BaCO3 will precipitate
<h2>Answer:</h2>
pOH = 4.6
<h2>Explanations:</h2>
The sum of pH and pOH of a solution is 14as shown:
Given the following parameter
pH = 9.4
Substitute the given parameters into the formula to have:
Hence the pOH of the solution of pH of 9.4 is 4.6