Answer:
Empirical and molecular formulas are the same, C₅H₁₀O₂.
Explanation:
Hello!
In this case, when determining the empirical and molecular formulas of organic compounds via combustion analysis, we first need to compute the moles of carbon and hydrogen via the yielded mass of carbon dioxide and water:
Next, we need to compute the mass of oxygen by subtracting the mass of carbon and hydrogen to the mass of the sample of the compound:
And consequently the moles:
Now, we need to divide the moles of each atom by the fewest moles, it in this case, those of oxygen to obtain the subscripts in the empirical formula:
Thus, the empirical formula, taken the nearest whole number is:
Now, if we divide the molar mass of the molecular formula (102.1 g/mol) by that of the empirical formula (102.1 g/mol) we infer they are both the same.
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Answer:
The answer to your question is 25 grams
Explanation:
Data
half-life = 5730 years
sample = 200 g
after 3 half-lives
Process
Calculate the amount of sample after one, two and three half-lives.
After each half-life, that of sample is half the previous amount.
Number of half-lives Amount of sample
0 200 g
1 100 g
2 50 g
3 25 g
<h3>
Answer:</h3>
19.3 g/cm³
<h3>
Explanation:</h3>
Density of a substance refers to the mass of the substance per unit volume.
Therefore, Density = Mass ÷ Volume
In this case, we are given;
Mass of the gold bar = 193.0 g
Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm
We are required to get the density of the gold bar
Step 1: Volume of the gold bar
Volume is given by, Length × width × height
Volume = 0.50 cm × 10.0 cm × 2.0 cm
= 10 cm³
Step 2: Density of the gold bar
Density = Mass ÷ volume
Density of the gold bar = 193.0 g ÷ 10 cm³
= 19.3 g/cm³
Thus, the density of the gold bar is 19.3 g/cm³
Answer:
The air will reach a higher final temperature because its specific heat is lower.
Explanation:
Answer:
B. People live longer than they used to
Explanation:
None of the other answers are correct.
