The epidermis is a dermal tissue<span> that is usually a single layer of cells covering the younger parts of a plant. It secretes a waxy layer called the cuticle that inhibits water loss.</span>
Hello!
The answer is C. Hibernate during the cold winter months.
Why?
Alpine marmots are known for having a long hibernation duration which starts in October (winther) and ends in April (summer) (about 7 months). During this long period, they are able to reduce their bear beats from 200 per minute to just 30 or 38 beats, and their breaths from 60 breaths/minute to 1-3 breaths/minute, guaranteeing an extreme energy saving process.
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The Lewis structure the student drew is incorrect. The total number of electrons in bonds and lone pairs is equal to the total number of valence electrons available, then a valid Lewis structure is produced.
The correct structure is the following:
The structure that was drawn by the student was incorrect because the total number of electrons around Oxygen were incorrect, there were 6 instead of 8, which was against the octet rule. The correct structure should have 8 electrons around each atom.
Sexually tranmsitted? No. Genetic? You can be more suseptible to certain
cancers for genetic reasons, but you can't have cancer in your genetic
makeup. Cancer is indeed a mutation of the cells, where they forget to
naturally die off, and spread.
Answer:
11.43g of Aluminum Hydroxide
Explanation:
Since we know that the sulfuric acid is the limiting reactant in this chemical reaction, we know that we are going to be left with excess aluminum hydroxide. So to find the amount of leftover aluminum hydroxide we are going to need to convert the given amount of sulfuric acid to the amount of aluminum hydroxide needed to react with the sulfuric acid.
![\frac{35g H_{2}SO_{4}}{1}*\frac{1 mole H_{2}SO_{4}}{98.079 g H_{2}SO_{4}} *\frac{2 moles Al(OH)_{3} }{3 moles H_{2}SO_{4}} * \frac{78.003 g Al(OH)_{3} }{1 mole Al(OH)_{3} } = 18.557 g Al(OH)_{3}](https://tex.z-dn.net/?f=%5Cfrac%7B35g%20H_%7B2%7DSO_%7B4%7D%7D%7B1%7D%2A%5Cfrac%7B1%20mole%20H_%7B2%7DSO_%7B4%7D%7D%7B98.079%20g%20H_%7B2%7DSO_%7B4%7D%7D%20%2A%5Cfrac%7B2%20moles%20Al%28OH%29_%7B3%7D%20%7D%7B3%20moles%20H_%7B2%7DSO_%7B4%7D%7D%20%2A%20%5Cfrac%7B78.003%20g%20Al%28OH%29_%7B3%7D%20%7D%7B1%20mole%20Al%28OH%29_%7B3%7D%20%7D%20%3D%2018.557%20g%20Al%28OH%29_%7B3%7D)
Once you do that, you need to subtract that number from the amount of aluminum hydroxide given to get the amount of left over aluminum hydroxide.
![30 g Al(OH)_{3} - 18.557 g Al(OH)_{3} = 11.43 g Al(OH)_{3}](https://tex.z-dn.net/?f=30%20g%20Al%28OH%29_%7B3%7D%20-%2018.557%20g%20Al%28OH%29_%7B3%7D%20%3D%2011.43%20g%20Al%28OH%29_%7B3%7D)
Hope this helps!