Answer:
Sulfur hexafluoride, also known as sulfur(VI) fluoride, is a chemical compound. Its chemical formula is SF6. It contains sulfur in its +6 oxidation state.
False
Explanation:
Sound waves are longitudinal mechanical waves.
Mechanical waves are waves that requires a medium to propagate them.
- A transverse wave is a wave that is propagated perpendicularly to its source.
- An example is electromagnetic radiation.
- A longitudinal wave is a wave that is directed parallel to their source.
- Sound wave is a longitudinal wave.
- It has series of rarefaction and compression along its travel path.
- Rarefaction are areas of sparse particles.
- Compression are areas with dense particles.
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Answer:
0.4694 moles of CrCl₃
Explanation:
The balanced equation is:
Cr₂O₃(s) + 3CCl₄(l) → 2CrCl₃(s) + 3COCl₂(aq)
The stoichiometry of the equation is how much moles of the substances must react to form the products, and it's represented by the coefficients of the balanced equation. So, 1 mol of Cr₂O₃ must react with 3 moles of CCl₄ to form 2 moles of CrCl₃ and 3 moles of COCl₂.
The stoichiometry calculus must be on a moles basis. The compounds of interest are Cr₂O₃ and CrCl₃. The molar masses of the elements are:
MCr = 52 g/mol
MCl = 35.5 g/mol
MO = 16 g/mol
So, the molar mass of the Cr₂O₃ is = 2x52 + 3x35.5 = 210.5 g/mol.
The number of moles is the mass divided by the molar mass, so:
n = 49.4/210.5 = 0.2347 mol of Cr₂O₃.
For the stoichiometry:
1 mol of Cr₂O₃ ------------------- 2 moles of CrCl₃
0.2347 mol of Cr₂O₃----------- x
By a simple direct three rule:
x = 0.4694 moles of CrCl₃
<h3><u>Answer</u>;</h3>
1.0875 x 10-2 atm
<h3><u>Explanation;</u></h3>
2O3(g) → 3O2(g)
rate = -(1/2)∆[O3]/∆t = +(1/3)∆[O2)/∆t
The average rate of disappearance of ozone ... is found to
be 7.25 × 10–3 atm over a certain interval of time.
This means (ignoring time)
∆[O3]/∆t = -7.25 × 10^–3 atm
(it is disappearing, thus the negative sign)
rate = -(1/2)∆[O3]/∆t
rate = -(1/2)*(-7.25 × 10^–3 atm)
= 3.625 × 10^–3 atm
Now use the other part of the expression:
rate = +(1/3)∆[O2)∆t
3.625 × 10–3 atm = +(1/3)∆[O2)/t
∆[O2)/∆t = (3)*(3.625× 10^–3 atm)
= 1.0875 x 10-2 atm over the same time interval