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DedPeter [7]
3 years ago
13

Alanna spots a bird in her back yard. The bird is sitting on a tree. Explain how the outer coverings of the bird and tree are di

fferent and have different functions.
Chemistry
2 answers:
AfilCa [17]3 years ago
8 0

Answer:

A bird's feathers are supposed to keep the bird warm and help them fly. A tree's bark is supposed to protect the inside of the tree.

Explanation:

If a tree has no bark, the inside of it is exposed. If a bird has no feathers, the bird is cold and cant fly.

geniusboy [140]3 years ago
6 0

Answer:

...............................U

Explanation:

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Lies just beyond the continental slope is the open ocean zone and it has three subzones namely the epipelagic(sunlit zone), mesopelagic(disphotic zone) and bathypelagic zones(aphotic zone). The oceanic zone covers 65% of the ocean's water and where different types of terrains can be found. From deep trenches, deep sea volcanoes and basins. A variety of sea creatures can also be found on each subzone.
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The quantity 0.0000064 g expressed in Scientific notation is
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How does a foam-filled winter coat help keep a person warm?
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Answer:

It slows down the transfer of thermal energy from outside to inside the coat.

Explanation:

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The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I
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Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:  

Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\

Equation:  

3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\

Calculating the amount of MnCO_3

= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\=  13.1 \ kg

For point b:

Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\  of\  Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\

=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg

Equation:  

3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\

Calculating the amount of MnO_2:  

= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\=  6516 \ g \\\\=6.52 \ kg\\\\

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