The fraction of Earth's radius (6371 km) relative to the thickness of the oceanic (7.5 km) and continental crust (35 km) is 0.12 and 0.55, respectively.
What we know:
- The average radius of Earth (E) = 6371 km
- The average thickness of oceanic crust (O) = 7.5 km
- The average thickness of continental crust (C) = 35 km
We need to convert all the above units from kilometers to miles:
Now, we can calculate the fraction of Earth's radius relative to each type of crust, with the given equation:
- <u>For the oceanic crust (O)</u>:
- <u>For the continental crust (C)</u>:
Therefore, the fraction of Earth's radius relative to the oceanic and continental crust is 0.12 and 0.55, respectively.
You can see another example of calculation of fractions of Earth's radius here: brainly.com/question/4675868?referrer=searchResults
I hope it helps you!
Answer:
Flood barriers
Explanation:
Many towns in flood-prone areas are surrounded by flood barriers that are built to hold back rising water during seasonal highs.
- Flood barriers are water resistant walls that are used to prevent breaking water during a high period of inundation from their channels.
- They are usually vertical artificial structures which can be temporary or permanent.
- They help to hold or contain the flood water during the high season.
- These walls prevents flood plains and other nearby areas from getting flooded during the rainy season.
Hello!
Determine the mass of 5.20 moles of C6H12 (gram-formula mass = 84.2 grams/mole).
We have the following data:
m (mass) = ?
n (number of moles) = 5.20 moles
MM (Molar mass of C6H12) ≈ 84.2 g/mol
Now, let's find the mass, knowing that:
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I Hope this helps, greetings ... Dexteright02! =)
The complete balanced chemical
equation is:
4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g)
In statement form: 4mol NH3 reacts with 5 mol O2 to produce 6
mol H2O
First let us find for the limiting reactant:
>molar mass NH3 = 17 g/mol
moles NH3 = 54/17 = 3.18 mol NH3
This will react with 3.18*5/4 = 3.97 mol O2
>molar mass O2 = 32g/mol
moles O2 = 54/32 = 1.69 mol O2
We have insufficient O2 therefore this is the limiting
reactant
From the balanced equation:
For every 5.0 mol O2, we get 6.0 mol H2O, therefore
moles H2O formed = 1.69
mol O2 * 6/5 = 2.025 mol
Molar mass H2O = 18g/mol
<span>mass H2O formed = 2.025*18 = 36.45 grams H2O produced</span>
Answer:
Supersaturated
Explanation:
The tea has absorbed and dissolved as much sugar as it could. If there is sugar left at the bottom, it means the solution is supersaturated because it can't absorb any more.
