The combustion of 30.0 g of glucose at room temperature and pressure produces 24.0 L of carbon dioxide.
<h3>What is combustion?</h3>
It is a reaction in which a substance burns with oxygen to form carbon dioxide and water.
Let's consider the combustion of glucose.
C₆H₁₂O₆ + 6 O₂ ⇒ 6 CO₂ + 6 H₂O
First, let's convert 30.0 g of glucose to moles using its molar mass.
30.0 g × 1 mol/180.16 g = 0.167 mol
The molar ratio of C₆H₁₂O₆ to CO₂ is 1:6. The moles of carbon dioxide produced are:
0.167 mol Glucose × (6 mol CO₂/1 mol Glucose) = 1.00 mol CO₂
1 mol of an ideal gas at room temperature and pressure occupies 24.0 L.
The combustion of 30.0 g of glucose at room temperature and pressure produces 24.0 L of carbon dioxide.
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Answer:
Final pressure = 362.7 Pa
Explanation:
Given that,
Initial volume, V₁ = 930 ml
Initial pressure P₁ = 156 Pa
Final volume, V₂ = 400 mL
We need to find the final pressure. We know that the relation between volume and pressure is inverse i.e.
So, the final pressure is equal to 362.7 Pa.
<span>C7H8
First, determine the number of relative moles of each element we have and the molar masses of the products.
atomic mass of carbon = 12.0107
atomic mass of hydrogen = 1.00794
atomic mass of oxygen = 15.999
Molar mass of CO2 = 12.0107 + 2 * 15.999 = 44.0087
Molar mass of H2O = 2 * 1.00794 + 15.999 = 18.01488
We have 5.27 mg of CO2, so
5.27 / 44.0087 = 0.119749 milli moles of CO2
And we have 1.23 mg of H2O, so
1.23 / 18.01488 = 0.068277 milli moles of H2O
Since there's 1 carbon atom per CO2 molecule, we have
0.119749 milli moles of carbon.
Since there's 2 hydrogen atoms per H2O molecules, we have
2 * 0.068277 = 0.136554 milli moles of hydrogen atoms.
Now we need to find a simple integer ratio that's close to
0.119749 / 0.136554 = 0.876937
Looking at all fractions n/m where n ranges from 1 to 10 and m ranges from 1 to 10, I find a closest match at 7/8 = 0.875 with an error of only 0.001937, the next closest match has an error over 6 times larger. So let's go with the 7/8 ratio.
The numerator in the ratio was for carbon atoms, and the denominator was for hydrogen. So the empirical formula for toluene is C7H8.</span>
Answer:
- <u>b. three half-lives</u>
Explanation:
The number of half-lives elapsed, n, is calculated dividing the time by the half-life time:
- n = time / half-life time
<u>a. A sample of Ce-141 with a half-life of 32.5 days after 32.5 days</u>
- n = 32.5 days / 32.5 days = 1 half-life
<u />
<u>b. A sample of F-18 with a half-life of 110 min after 330 min</u>
- n = 330 min / 110 min = 3 half-lives
<u />
<u>c. A sample of Au-198 with a half-life of 2.7 days after 5.4 days</u>
- n = 5.4 days / 2.7 days = 2 half-lives
What’s the difference between 2 percent milk and whole milk
