They seem to eat a lot ranging from woody plants to vines, though in the rainforest most of their diet is made up of Malaysia and plant material, to be more specific
So I would go with leaves, 50%
Flowers, buds, and insects, 10%
Fruits, 40%
--They also really like figs
Answer:
Molarity = 2.3 M
Explanation:
Molarity can be calculated using the following rule:
Molarity = number of moles of solute / volume of solution
1- getting the number of moles:
We are given that:
mass of solute = 105.96 grams
From the periodic table:
atomic mass of carbon = 12 grams
atomic mass of hydrogen = 1 gram
atomic mass of oxygen = 16 grams
Therefore:
molar mass of C2H6O = 2(12) + 6(1) + 16 = 46 grams
Now, we can get the number of moles as follows:
number of moles = mass / molar mass = 105.96 / 46 = 2.3 moles
2- The volume of solution is given = 1 liter
3- getting the molarity:
molarity = number of moles of solute / volume of solution
molarity = 2.3 / 1
molarity = 2.3 M
Hope this helps :)
Answer:
237.8L of water would need to be added.
Explanation:
The first thing to do is to identify that the equation to be used is M1V1=M2V2. (This equation works because it turns everything into moles which can then be compared).
Then figure out what information you have and what is being found. In this case:
M1 = 54.7 M
V1 = 1092 mL = 1.092 L
M2 = 0.25 M
V2 = unknown
Then solve the equation for whatever you are trying to find.
M1V1=M2V2
V2=M1V1/M2
Now you need to plug everything in.
V2=(54.7M*1.091L)/0.25M
V2=238.93L
That means that the solution needs a volume of 238.7L to gain a molarity of 0.25M but the starting solution already had a volume of 1.092 L meaning that to find the amount of solvent that needs to be added you just subtract the starting volume by the volume that the solution needs to be.
238.93L - 1.091L = 237.8L
Therefore the answer is that 237.8L needs to be added to a 1.092L 54.7M NaCl solution to make the concentration 0.25M.
I hope this helps. Let me know if anything is unclear.
2 H₂ + O₂ = 2 H₂O
Answer B only synthesis.
hope this helps!
