<span>A 50-gram sample with a half-life of 12 days will have a remaining mass of 25 grams after its 12-day half-life.
Every cycle of a half-life, the sample will lose half of its mass, so if the half-life, itself, is 12 days and the time period passing is 12 days, one half-life has passed and the material will be halved.</span>
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>
Concentration is the number of moles of solute in a fixed volume of solution
Concentration(c) = number of moles of solute(n) / volume of solution (v)
25.0 mL of water is added to 125 mL of a 0.150 M LiOH solution and solution becomes more diluted.
original solution molarity - 0.150 M
number of moles of LiOH in 1 L - 0.150 mol
number of LiOH moles in 0.125 L - 0.150 mol/ L x 0.125 L = 0.01875 mol
when 25.0 mL is added the number of moles of LiOH will remain constant but volume of the solution increases
new volume - 125 mL + 25 mL = 150 mL
therefore new molarity is
c = 0.01875 mol / 0.150 L = 0.125 M
answer is 0.125 M
If the atom is neutral (meaning, not charged) the number of electron is equal to the number of protons. The mass number of an atom is the sum of the number of proton and the number of neutrons. From the given above, the mass number of gallium is 31 + 39. The answer is letter D. 70.
Answer:
58.6 % by mass of Na₂CO₃
Explanation:
This is the reaction:
Na₂CO₃ + MgCO₃ + 4HCl → MgCl₂ + 2NaCl + 2CO₂ + 2H₂O
Let's find out the moles of CO₂ produced, by the Ideal Gases Law
1.24 atm . 1.67 L = n . 0.082 . 299K
(1.24 atm . 1.67 L / 0.082 . 299K) = n
0.0844 moles = n
Ratio is 2:1, so 2 moles of dioxide were produced by 1 mol of sodium carbonate. Let's make a rule of three:
2 moles of CO₂ were produced by 1 mol of Na₂CO₃
Then, 0.0844 moles of Co₂ would beeen produced by (0.0844 .1)/2 = 0.0422 moles of Na₂CO₃.
Let's convert this moles into mass (mol . molar mass)
0.0422 mol . 106 g/mol = 4.47 g
Finally we can know the mass percent of sodium carbonate in the mixture
(Mass of compound /Total mass) . 100 → (4.47 g / 7.63g) . 100 = 58.6 %
