Answer: if you think about a tree the older rings would be in the middle, so you’d find objects with older age below/under/inside
Answer:
Intertidal zone
Neritic zone
Open-ocean zone
Note: the correct questions are found below;
In which zone do you find marshes and mangrove forests?
In which zone are plankton plentiful, providing plenty of food for the fish that live there?
In which zone would you find very little plant or animal life compared to other zones?
Explanation:
The intertidal zone, sometimes called the littoral zone, is the area of the marine shoreline that is exposed to air at low tide, and covered with seawater when the tide is high. Intertidal zonation refers to the tendency of plants and animals to form distinct communities between the high and low tide lines. Some microclimates in the littoral zone are moderated by local features and larger plants such as mangroves.
The neritic zone is the region of shallow water (200 meters depth) above the continental shelf where light penetrates to the sea floor.
Due to the abundant supply of sunlight and nutrients such as plankton in this zone, it is the most productive ocean zone supporting the vast majority of marine life.
The open oceans or pelagic ecosystems are the areas away from the coastal boundaries and above the seabed. It encompasses the entire water column and lies beyond the edge of the continental shelf. It extends from the tropics to the polar regions and from the sea surface to the abyssal depths.
Answer:
A) 
B) 0.025 mol (SCN)₂
C) 2 mol (SCN)₂
D) 
E) 3.5504 mol H₂SO₄
Explanation:
2NaSCN +2H₂SO₄ + MnO₂ → (SCN)₂ + 2H₂O + MnSO₄ + Na₂SO₄
A) A conversion factor could be , as it has the units that we want to <u>convert to in the numerator</u>, and the units that we want to <u>convert from in the denominator</u>.
B) 0.05 mol NaSCN * = 0.025 mol (SCN)₂
C) With 4 moles of NaSCN and 3 moles of MnSO₄, the<em> reactant </em>is NaSCN so we use that value to calculate the moles of product formed:
4 mol NaSCN * = 2 mol (SCN)₂
D)
E) 1.7752 mol MnSO₄ * 
= 3.5504 mol H₂SO₄
The reaction between NaOH and Cu(NO₃)₂ is as follows
2NaOH + Cu(NO₃)₂ ---> 2NaNO₃ + Cu(OH)₂
Q1)
stoichiometry of NaOH to Cu(NO₃)₂ is 2:1
this means that 2 mol of NaOH reacts with 1 mol of Cu(NO₃)₂
the mass of Cu(NO₃)₂ reacted - 0.8024 g
molar mass of Cu(NO₃)₂ is 187.56 g/mol
therefore the number of Cu(NO₃)₂ moles that have reacted
- 0.8024 g/ 187.56 g/mol = 0.00427 mol
according to the stoichiometry , number of NaOH moles - 0.00427 mol x 2
then number of NaOH moles that have reacted - 0.00855 mol
In a 3.0 M NaOH solution, 3 moles are in 1000 mL of solution
Then volume required for 0.00855 mol - 1000 x 0.00855 /3 = 2.85 mL
2.85 mL of 3.0 M NaOH is required for this reaction
Q2)
Assuming that there's 100 % yield of Cu(OH)₂ , we can directly calculate the mass of Cu(OH)₂ formed from the number of moles of reactants that were used up.
Stoichiometry of Cu(NO₃)₂ to Cu(OH)₂ is 1:1
this means that 1 mol of Cu(NO₃)₂ gives a yield of 1 mol of Cu(OH)₂
the number of Cu(NO₃)₂ moles that reacted - 0.00427 mol
Therefore an equal amount of moles of Cu(OH)₂ were formed
Then amount of Cu(OH)₂ moles produced - 0.00427 mol
Mass of Cu(OH)₂ formed - 0.00427 mol x 97.56 g/mol = 0.42 g
A mass of 0.42 g of Cu(OH)₂ was formed in this reaction
Speed of light = freq * wavelength
<span>wavelength = speed of light / freq = (3*10^8 m/sec ) / (2.6 * 10^9 /sec) = 1.15 * 10^(-1) m </span>
<span>1.15 * 10^(-1) m *(10^9 nm / m) = 1.15 * 10^8 nm</span>
