Answer:
A. 8600 J
General Formulas and Concepts:
<u>Thermochemistry</u>
Specific Heat Formula: q = mcΔT
- q is heat (in J)
- m is mass (in g)
- c is specific heat (in J/g °C)
- ΔT is change in temperature (in °C)
Explanation:
<u>Step 1: Define</u>
[Given] <em>m</em> = 1600 g
[Given] ΔT = 214 °C - 202 °C = 12 °C
[Given] <em>c</em> = 0.450 J/g °C
[Solve] <em>q</em>
<u>Step 2: Find Heat</u>
- Substitute in variables [Specific Heat Formula]: q = (1600 g)(0.450 J/g °C)(12 °C)
- Multiply [Cancel out units]: q = (720 J/°C)(12 °C)
- Multiply [Cancel out units]: q = 8640 J
<u>Step 3: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs as our lowest.</em>
8640 J ≈ 8600 J
Topic: AP Chemistry
Unit: Thermodynamics
Answer: The first isotope has a relative abundance of 79% and last isotope has a relative abundance of 11%
Explanation: Given that the average atomic mass(M) of magnesium
= 24.3050amu
Mass of first isotope (M1) = 23.9850amu
Mass of middle isotope (M2)=24.9858amu
Mass of last isotope(M3)= 25.9826amu
Total abundance = 1
Abundance of middle isotope = 0.10
Let abundance of first and last isotope be x and y respectively.
x+0.10+y =1
x = 0.90-y
M = M1 × % abundance of first isotope + M2 × % of middle isotope +M3 ×% of last isotope
24.03050= 23.985× x + 24.9858 ×0.10 + 25.9826×y
Substitute x= 0.90-y
Then
y = 0.11
Since y=0.11, then
x= 0.90-0.11
x=0.79
Therefore the relative abundance of the first isotope = 11% and the relative abundance of the last isotope = 79%
Answer: 4.22 grams of solute is there in 278 ml of 0.038 M 
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
where,
n = moles of solute
= volume of solution in L
Now put all the given values in the formula of molality, we get

mass of
= 
Thus 4.22 grams of solute is there in 278 ml of 0.038 M 
Because the sublevels 1s only has one shape ,and it only contains 2 elections...