The correct answer is letter C
<h3>
Answer:</h3>
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
- Analyzing Reactions RxN
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 2C + O₂ → 2CO₂
[Given] 0.25 moles O₂
[Solve] moles CO₂
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol O₂ → 2 mol CO₂
<u>Step 3: Stoichiometry</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units]:
Answer:
% = 76.75%
Explanation:
To solve this problem, we just need to use the expressions of half life and it's relation with the concentration or mass of a compound. That expression is the following:
A = A₀ e^(-kt) (1)
Where:
A and A₀: concentrations or mass of the compounds, (final and initial)
k: constant decay of the compound
t: given time
Now to get the value of k, we should use the following expression:
k = ln2 / t₁/₂ (2)
You should note that this expression is valid when the reaction is of order 1 or first order. In this kind of exercises, we can assume it's a first order because we are not using the isotope for a reaction.
Now, let's calculate k:
k = ln2 / 956.3
k = 7.25x10⁻⁴ d⁻¹
With this value, we just replace it in (1) to get the final mass of the isotope. The given time is 1 year or 365 days so:
A = 250 e^(-7.25x10⁻⁴ * 365)
A = 250 e^(-0.7675)
A = 191.87 g
However, the question is the percentage left after 1 year so:
% = (191.87 / 250) * 100
<h2>
% = 76.75%</h2><h2>
And this is the % of isotope after 1 year</h2>
The molecular formula : C₁₈H₁₈N₈
<h3>Further explanation</h3>
Given
62.41% C, 5.24% H, and 32.36% N
Required
The molecular formula
Solution
mol ratio
C : 62.41/12.0096 = 5.1967
H : 5.24/1.00784 = 5.1992
N : 32.36/14.0067 = 2.310
Divide by 2.310(smallest)
C : 5.1967/2.31=2.25
H : 5.1992/2.31 = 2.25
N : 2.31/2.31 = 1
Multiplied by 4
C : H : N = 9 : 9 : 4
The empirical formula : C₉H₉N₄
(C₉H₉N₄)n=346.40 g/mol
(12.0096 x 9 + 1.00784 x 9 + 14.0067 x 4)n=346.4
(108.0864+9.07056+56.0268)n=346.4
(173.184)n=346.4
n=2
<em>The molecular formula : C₁₈H₁₈N₈</em>
<span>6.50x10^3 calories.
Now we have 4 pieces of data and want a single result. The data is:
Mass: 100.0 g
Starting temperature: 25.0°C
Ending temperature: 31.5°C
Specific heat: 1.00 cal/(g*°C)
And we want a result with the unit "cal". Now you need to figure out what set of math operations will give you the desired result. Turns out this is quite simple. First, you need to remember that you can only add or subtract things that have the same units. You may multiply or divide data items with different units and the units can combine or cancel each other. So let's solve this:
Let's start with specific heat with the unit "cal/(g*°C)". The cal is what we want, but we'ld like to get rid of the "/(g*°C)" part. So let's multiply by the mass:
1.00 cal/(g*°C) * 100.0 g = 100.0 cal/°C
We now have a simpler unit of "cal/°C", so we're getting closer. Just need to cancel out the "/°C" part, which we can do with a multiplication. But we have 2 pieces of data using "°C". We can't multiply both of them, that would give us "cal*°C" which we don't want. But we need to use both pieces. And since we're interested in the temperature change, let's subtract them. So
31.5°C - 25.0°C = 6.5°C
So we have a 6.5°C change in temperature. Now let's multiply:
6.5°C * 100.0 cal/°C = 6500.0 cal
Since we only have 3 significant digits in our least precise piece of data, we need to round the result to 3 significant figures. 6500 only has 2 significant digits, and 6500. has 4. But we can use scientific notation to express the result as 6.50x10^3 which has the desired 3 digits of significance. So the result is 6.50x10^3 calories.
Just remember to pay attention to the units in the data you have. They will pretty much tell you exactly what to add, subtract, multiply, or divide.</span>