The correct answer to the question above is heat. Most of the energy from a lower trophic level is converted into heat. When an organism from a higher trophic level consumed an organism from a lower trophic level, it is mostly heat that is being converted to.
At STP, or standard temperature and pressure, 1 mol of any gas will take up 22.4 liters of space. Assuming STP, 4.5 moles of H2 will take up 100.8L.
Answer:
We'll have 1 mol Al2O3 and 3 moles H2
Explanation:
Step 1: data given
Numer of moles of aluminium = 2 moles
Number of moles of H2O = 6 moles
Step 2: The balanced equation
2Al + 3H2O → Al2O3 + 3H2
Step 3: Calculate the limiting reactant
For 2 moles Al we need 3 moles H2O to produce 1 mol Al2O3 and 3 moles H2
Aluminium is the limiting reactant. It will completely be consumed (2 moles).
H2O is in excess. There will react 3/2 * 2 = 3 moles
There will remain 6 - 3 = 3 moles
Step 4: Calculate moles products
For 2 moles Al we need 3 moles H2O to produce 1 mol Al2O3 and 3 moles H2
For 2 moles Al we'll have 2/1 = 1 mol Al2O3
For 2 moles Al We'll have 3/2 * 2 = 3 moles H2
We'll have 1 mol Al2O3 and 3 moles H2
Answer:
36.66%
Explanation:
Step 1: Given data
- Mass of the sample: 2.875 g
Step 2: Calculate the mass of salt
The mass of the sample is equal to the sum of the masses of the components.
m(sample) = m(iron) + m(sand) + m(salt)
m(salt) = m(sample) - m(iron) - m(sand)
m(salt) = 2.875 g - 0.660 g - 1.161 g
m(salt) = 1.054 g
Step 3: Calculate the percent of salt in the sample
We will use the following expression.
%(salt) = m(salt) / m(sample) × 100%
%(salt) = 1.054 g / 2.875 g × 100% = 36.66%
'The Sedimentary rock formed from years of sediments piling on top of it and being compressed.'
