Answer:
0.0468 g.
Explanation:
- The decay of radioactive elements obeys first-order kinetics.
- For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).
Where, k is the rate constant of the reaction.
t1/2 is the half-life time of the reaction (t1/2 = 1620 years).
∴ k = ln2/(t1/2) = 0.693/(1620 years) = 4.28 x 10⁻⁴ year⁻¹.
- For first-order reaction: <em>kt = lna/(a-x).</em>
where, k is the rate constant of the reaction (k = 4.28 x 10⁻⁴ year⁻¹).
t is the time of the reaction (t = t1/2 x 8 = 1620 years x 8 = 12960 year).
a is the initial concentration (a = 12.0 g).
(a-x) is the remaining concentration.
∴ kt = lna/(a-x)
(4.28 x 10⁻⁴ year⁻¹)(12960 year) = ln(12)/(a-x).
5.54688 = ln(12)/(a-x).
Taking e for the both sides:
256.34 = (12)/(a-x).
<em>∴ (a-x) = 12/256.34 = 0.0468 g.</em>
Answer:
The correct answer is option B.
Explanation:
As given ,that 30.24 mL of base was neutralize by 86.14 mL of acid which means that moles of base present in 30.24 mL are neutralized by moles of acid present in 86.14 mL.
After dilution of base from 30.24 mL to 50.0 mL .Since, the moles of base are same in the solution as that of the moles in solution before dilution. Moles of acid require to neutralize the base after dilution will same as a that of present moles of acid present in 86.14 mL.
Answer:
An ionic compound is formed when there is a reaction between the elements whose ions are electrostatically attracted.
Explanation:
The ionic compounds form crystalline networks with ionic bonding. Electrostatic attraction is a very strong bond that is very difficult to break. The stability of the ionic compound depends on the lattice energy, the higher it is, the more stable the compound is and the lattice energy is that which is released in the formation. At room temperature they are always in a solid state, because the bonds are very close and as stated before, they are difficult to break (providing a lot of energy)
Answer:
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Explanation:
To start, 1 cubic centimeter = 1 milliliter, so now you have 1.11g/mL.
Now multiply 1.11 by 387 to get the mass of antifreeze in grams, since the mL is canceled out.
387 mL x 1.11g/mL = 429.57 g
