Answer:
9.91 mL
Explanation:
Using the combined gas law equation as follows;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (torr)
P2 = final pressure (torr)
V1 = initial volume (mL)
V2 = final volume (mL)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question;
V1 = 15.0mL
V2 = ?
P1 = 760 torr
P2 = 1252 torr
T1 = 10°C = 10 + 273 = 283K
T2 = 35°C = 35 + 273 = 308K
Using P1V1/T1 = P2V2/T2
760 × 15/283 = 1252 × V2/308
11400/283 = 1252V2/308
Cross multiply
11400 × 308 = 283 × 1252V2
3511200 = 354316V2
V2 = 3511200 ÷ 354316
V2 = 9.91 mL
Started out as some other type of rock, but have been substantially changed from their original igneous, sedimentary, or earlier metamorphic form. Metamorphic rocks form when rocks are subjected to high heat, high pressure, hot mineral-rich fluids or, more commonly, some combination of these factors.
Answer:
21.6 g
Explanation:
The reaction that takes place is:
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
One reaction that destroys o3 molecules in the stratosphere is when an ozone molecule combines with an oxygen atom to form two oxygen molecules, or through certain chemical reactions involving molecules containing hydrogen, nitrogen, chlorine, or bromine atoms. The atmosphere maintains a natural balance between ozone formation and destruction.
