Ferns use both sexual and asexual reproduction methods. In sexual reproduction, a haploid spore grows into a haploid gametophyte. ... The sporophyte produces spores, completing the life cycle. Asexual methods of reproduction include apogamy, poliferous frond tips, and rhizome spreading.
Answer:
Hydrogen may not be advantageous as a fuel because...
- Its expensive
- Its difficult to store
- Its highly flammable
- Its dependent on fossil fuels
Explanation:
Its expensive - Not only is hydrogen gas expensive, but it also takes a lot of work to free from other elements. It is both expensive and time-consuming to produce.
Its difficult to store - Moving hydrogen is not an easy task. Moving anything more than small amounts of hydrogen was also very expensive, making it impractical.
Its highly flammable - When exposed to the atmosphere, hydrogen could potentially form explosive mixtures.
Its dependent on fossil fuels - Hydrogen energy itself is renewable. However, the process of separating it from oxygen uses non-renewable sources such as coal and oil.
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<span>XY4Z2-->Square planar (Electron domain geometry: Octahedral) sp3d2
XY4Z-->Seesaw (Electron domain geometry: Trigonal bipyramidal) sp3d
XY5Z-->Square pyramidal (Electron domain geometry: Octahedral) sp3d2
XY2Z3-->Linear (Electron domain geometry: Trigonal bipyramidal) sp3d
XY2Z-->Bent (Electron domain geometry: Trigonal planar) sp2
XY3Z-->Trigonal pyramidal (Electron domain geometry: Tetrahedral) sp3
XY2Z2-->Linear (Electron domain geometry: Tetrahedral) sp3
XY3Z2-->T shaped (Electron domain geometry: Trigonal bipryamidal) sp3d
XY2-->Linear (Electron domain geometry: Linear) sp
XY3 Trigonal planar (Electron geometry: Trigonal planar) sp2
XY4-->Tetrahedral (Electron domain geometry: tetrahedral) sp3
XY5-->Trigonal bipyramidal (Electron domain geometry: Trigonal bipyramidal) sp3d
XY6-->Octahedral (Electron domain geometry: Octahedral) sp3d2</span>
To determine the volume of both concentration of vinegar, we need to set up two equations since we have two unknowns.
For the first equation, we do a mass balance:
mass of 100% vinegar + mass of 13% vinegar = mass of 42% vinegar
Assuming they have the same densities, then we can write this equation in terms of volume.
V(100%) + V(13%) = V(42%)
we let x = V(100%)
y = V(13%)
x + y = 150
For the second equation, we do a component balance:
1.00x + .13y = 150(.42)
x + .13y = 63
The two equations are
x + y = 150
x + .13y = 63
Solving for x and y,
x = 50
y = 100
Therefore, you need to mix 50 mL of the 100% vinegar and 100 mL of the 13% vinegar.