Problem 2
You begin with 216 micrograms of Fermium - 253. After three days, the quantity halves, resulting in 108 micrograms left.
Another three days pass. Beginning with 108 micrograms, this amount gets halved again, leaving 54 micrograms.
Finally, after another three-day span, starting from 54 micrograms, you again halve this amount to reach 27 micrograms.
#days Amount in micrograms
0 216
3 108
6 54
9 27
Problem One
Your example is Nitrogen. Begin by completing the table, then formulate some rules to help prepare for possible alternate elements in the test. This approach is quite useful.
Table
Bond Energy Kj/Mol Bond Length pico meters
N - N 167 145
N=N 418 125
N≡N 942 110
Rules
As the number of bonds INCREASES, the energy within the bond also INCREASES
As the number of bonds INCREASES, the distance of the bond DECREASES.
Answer:
The response is provided below.
Explanation:
Numerous aspects can influence the actual results of titration. These factors vary from human error to misjudging measurements, a researcher's interpretation of color changes, and improper techniques during the experimental procedure.
Thus, to mitigate these errors, researchers must engage thoroughly throughout experimentation, and employing gross readings can assist in reducing mistakes when determining the final titre value.
Answer:
The enthalpy of the second intermediate equation is altered by halving its value and changing the sign.
Explanation:
Let's examine both the first and second intermediate reactions alongside the overall equation concerning the examined process;
First reaction;
Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ
Second reaction;
2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ
Thus, the overall reaction becomes;
CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH =?
According to Hess's law, which states that the total heat change in a reaction is equal to the sum of the heat changes for each step, we cannot simply sum the enthalpies for this overall reaction. Instead, we obtain the overall enthalpy by halving the second intermediate reaction's enthalpy and changing its sign before adding, as illustrated below;
Enthalpy of Intermediate reaction 1 + ½(-Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction
The chemical equation can be expressed as:
2H2 + O2 = 2H2O
Given the amounts of the reactants, we need to identify the limiting reactant before calculating the amount of product generated.
4.0 g H2 ( 1 mol / 2.02 g ) = 1.98 mol H2
5.0 g O2 ( 1 mol / 32 g ) = 0.1563 mol O2
The limiting reactant is O2, as it will be fully consumed in the reaction.
0.1563 mol O2 ( 2 mol H2O / 1 mol O2 ) ( 18.02 g / mol ) = 5.6 g H2O will be produced
Answer: 25,200.
Explanation:
1) Starting with: 4.659 × 10⁴ - 2.14 × 10⁴
2) Significant figures need to be considered.
Because the powers are equal (10⁴), the decimal values can be subtracted directly. However, it is essential to first check significant figures and the count of decimal places.
3) The figure 4.659 × 10⁴ has four significant digits (4, 6, 5, and 9), whereas 2.14 × 10⁴ contains three significant digits (2, 1, and 4).
4) When adding or subtracting values with a different number of decimal places, the answer must reflect the same number of decimal places as the number with the least amount of decimal precision.
5) Prior to performing subtraction, it is necessary to round the numbers to the least decimal places. Given that 2.14 has two decimal places and 4.659 has three, round 4.659 to 4.66.
6) Now perform the subtraction 4.66 - 2.14 = 2.52
7) Then multiply by the power of 10: 2.52 × 10⁴ = 25,200. This is the final answer.
