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>
Answer:
19
Explanation:
The total distance traveled by the toy cay would be 19 cm.
The total distance traveled should not be mistaken for total displacement. While displacement measures the distance and direction from the starting position of the toy car relative to its final position, the total distance traveled is calculated by adding all the movements of the toy car together. Hence;
Total distance traveled = 9 + 4 + 6 = 19 cm
it´s actually Lithium and fluorine / Magnesium and Chlorine / Beryllium and Nitrogen
N₀ is the number of C-14 atoms per kg of carbon in the original sample at time = Os when its carbon was of the same kind as that present in the atmosphere today. After time ts, due to radioactive decay, the number of C-14 atoms per kg of carbon is the same sample which has decreased to N. λ is the radioactive decay constant.
Therefore N = N₀e-λt which is the radioactive decay equation,
N₀/N = eλt In (N₀.N= λt. This is the equation 1
The mass of carbon which is present in the sample os mc kg. So the sample has a radioactivity of A/mc decay is/kg. r is the mass of C-14 in original sample at t= 0 per total mass of carbon in a sample which is equal to [(total number of C-14 atoms in the sample at t m=m 0) × ma]/ total mass of carbon in the sample.
Now that the total number of C-14 atoms in the sample at t= 0/ total mass of carbon in sample = N₀ then r = N₀×ma
So N₀ = r/ma. this equation 2.
The activity of the radioactive substance is directly proportional to the number of atoms present at the time.
Activity = A number of decays/ sec = dN/dt = λ(number of atoms of C-14 present at time t) =
λ₁(N×mc). By rearranging we get N = A/(λmc) this is equation 3.
By plugging in equation 2 and 3 and solve t to get
t = 1/λ In (rλmc/m₀A).
Answer is: there are ten atoms in one molecule of antifreeze.
One molecule of ethylene glycol (C₂H₄(OH)₂) has two carbon atoms, six hydrogen atoms (4 + 2 · 1) and two oxygen atoms (2 · 1). So there are:
2 + 6 + 2 = 10 atoms.
Ethylene glycol (C₂H₄(OH)₂) is an odorless, sweet-tasting, colorless viscous liquid.
