1) Ca-37, with a half-life of 181.1(10) ms.
Answer:
C8H17N
Explanation:
Mass of the unknown compound = 5.024 mg
Mass of CO2 = 13.90 mg
Mass of H2O = 6.048 mg
Next, we shall determine the mass of carbon, hydrogen and nitrogen present in the compound. This is illustrated below:
For carbon, C:
Molar mass of CO2 = 12 + (2x16) = 44g/mol
Mass of C = 12/44 x 13.90 = 3.791 mg
For hydrogen, H:
Molar mass of H2O = (2x1) + 16 = 18g/mol
Mass of H = 2/18 x 6.048 = 0.672 mg
For nitrogen, N:
Mass N = mass of unknown – (mass of C + mass of H)
Mass of N = 5.024 – (3.791 + 0.672)
Mass of N = 0.561 mg
Now, we can obtain the empirical formula for the compound as follow:
C = 3.791 mg
H = 0.672 mg
N = 0.561 mg
Divide each by their molar mass
C = 3.791 / 12 = 0.316
H = 0.672 / 1 = 0.672
N = 0.561 / 14 = 0.040
Divide by the smallest
C = 0.316 / 0.04 = 8
H = 0.672 / 0.04 = 17
N = 0.040 / 0.04 = 1
Therefore, the empirical formula for the compound is C8H17N
Mole represents the a huge Avogadro number (way larger than atom), so we can eliminate option A and D.
And we know that Mercury is a solid , which mean it has greater mass than chlorine
So the answer is : B. 1.0 mol mercury (Hg) Atom
When water changes state in the water cycle, the total number of water particles remains the same. The changes of state include melting, sublimation, evaporation, freezing, condensation, and deposition. All changes of state involve the transfer of energy.
Answer:
2 HC₂H₃O₂(aq) + Sr(OH)₂(aq) ⇒ Sr(C₂H₃O₂)₂(aq) + 2 H₂O
Explanation:
Let's consider the reaction between acetic acid and strontium hydroxide. This is a neutralization reaction, in which an acid reacts with a base to form salt and water. The unbalanced equation is:
HC₂H₃O₂(aq) + Sr(OH)₂(aq) ⇒ Sr(C₂H₃O₂)₂(aq) + H₂O
We have 1 acetate ion to the left and 2 to the right, so we will multiply HC₂H₃O₂(aq) by 2.
2 HC₂H₃O₂(aq) + Sr(OH)₂(aq) ⇒ Sr(C₂H₃O₂)₂(aq) + H₂O
Finally, we multiply water by 2 to get the balanced equation.
2 HC₂H₃O₂(aq) + Sr(OH)₂(aq) ⇒ Sr(C₂H₃O₂)₂(aq) + 2 H₂O
