Answer:
Explanation:
2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O
n/mol: 4.3
13 mol of O₂ react with 2 mol of 2C₄H₁₀
In a combustion of a hydrocarbon compound, 2 reactions are happening per element:
C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O
Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.
1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C
Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H
The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(<span>0.1510 g) * 100
</span>Percent composition = 99.88%
2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:
Amount of C = 0.01138 mol
Amount of H = 0.014244 mol
Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25
The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5
Thus, the empirical formula of the hydrocarbon is C₄H₅.
3. The molar mass of the empirical formula is
Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.
Molecular Formula = C₈H₁₀
<span>Gallium-72 that is what i got when i did the math for it</span>
Answer:
2.82 L
T₁ = 303 K
T₂ = 263 K
The final volume is smaller.
Explanation:
Step 1: Given data
- Initial temperature (T₁): 30 °C
- Initial volume (V₁): 3.25 L
- Final temperature (T₂): -10 °C
Step 2: Convert the temperatures to Kelvin
We will use the following expression.
K = °C + 273.15
T₁: K = 30°C + 273.15 = 303 K
T₂: K = -10°C + 273.15 = 263 K
Step 3: Calculate the final volume of the balloon
Assuming constant pressure and ideal behavior, we can calculate the final volume using Charles' law. Since the temperature is smaller, the volume must be smaller as well.
V₁/T₁ = V₂/T₂
V₂ = V₁ × T₂/T₁
V₂ = 3.25 L × 263 K/303 K = 2.82 L
The given statement is true .
<h3>What is Rutherford’s gold foil experiment?</h3>
- A piece of gold foil was hit with alpha particles, which have a favorable charge. Most alpha particles went right around. This showed that the gold particles were mostly space.
- The Rutherford gold leaf investigation supposed that most (99%) of all the mass of an atom is in the middle of the atom, that the nucleus is very small (105 times small than the length of the atom) and that is positively captured.
- For the distribution experiment, Rutherford enjoyed a metal sheet that could be as thin as practicable. Gold is the most malleable of all known metals. It can easily be converted into very thin sheets. Hence, Rutherford established a gold foil for his alpha-ray scattering experimentation.
To learn more about Rutherford’s gold foil experiment, refer to:
brainly.com/question/4113533
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