Calculate the mass of liquid with density of 3.3 g/mL and a volume of 25 oz
1 answer:
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Answer:
21.6 g
Explanation:
The reaction that takes place is:
- CH₄ + 2O₂ → CO₂ + 2H₂O
First we<u> convert the given masses of both reactants into moles</u>, using their <em>respective molar masses</em>:
- 9.6 g CH₄ ÷ 16 g/mol = 0.6 mol CH₄
- 64.9 g O₂ ÷ 32 g/mol = 2.03 mol O₂
0.6 moles of CH₄ would react completely with (2 * 0.6) 1.2 moles of O₂. As there are more O₂ moles than required, O₂ is the reactant in excess and CH₄ is the limiting reactant.
Now we <u>calculate how many moles of water are produced</u>, using the <em>number of moles of the limiting reactant</em>:
- 0.6 mol CH₄ *
= 1.2 mol H₂O
Finally we<u> convert 1.2 moles of water into grams</u>, using its <em>molar mass</em>:
- 1.2 mol * 18 g/mol = 21.6 g
Answer:
- The limiting reagent is N2O4
- 14,09g
Explanation:
- First, we adjust the reaction.
+
⇄
- Second, we assume that the participating moles are equal to the stoichiometric ratios because we do not know the amounts of the reagents.
We can determinate what is the limiting reagent comparing of product amounts which can be formed from each reactant.
Using to form
Using to form
The limiting reagent is N2O4, because can produce only 0, 783 mol of H2O.
This is the minimum measure can be formed of each product.
∴
Explanation:
Let the age to be found in years is y.
Hence,
= (\frac{1}{2})^{\frac{y}{5730yr}}
Solve for y as follows.
0.53913 = (\frac{1}{2})^{\frac{y}{5730yr}}
Now, taking log on both the sides as follows
log 0.53913 =
=
z =
= 5107 years
Thus, we can conclude that the time since death is 5107 years.