The first factor of production is land, but this includes any natural resource used to produce goods and services. This includes not just land, but anything that comes from the land. Some common land or natural resources are water, oil, copper, natural gas, coal, and forests.
So I’m assuming the answer would be:
C. All natural resources
Nonfoliated is the answer I belive.. Hopefully
Answer:
0.56L
Explanation:
This question requires the Ideal Gas Law:
where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the Ideal Gas constant, and T is the Temperature of the gas.
Since all of the answer choices are given in units of Liters, it will be convenient to use a value for R that contains "Liters" in its units:
Since the conditions are stated to be STP, we must remember that STP is Standard Temperature Pressure, which means
and 
Lastly, we must calculate the number of moles of
there are. Given 0.80g of
, we will need to convert with the molar mass of
. Noting that there are 2 oxygen atoms, we find the atomic mass of O from the periodic table (16g/mol) and multiply by 2: 
Thus, 
Isolating V in the Ideal Gas Law:


...substituting the known values, and simplifying...


So, 0.80g of
would occupy 0.56L at STP.
<span>To raise the liquid temperature to the point of boiling take 1231.776 joules of energy. To convert to a gas takes 5320.645 joules. To raise to 108 degrees Celsius takes 1456.848 joules. Total amount of energy needed (as heat) equals 8009.269 joules or 8.009 kj.</span>
Answer : The mass in grams of calcium sulfate is 0.16 grams.
Explanation :
Molarity : It is defined as the number of moles of solute present in one litre of solution.
Formula used :

Solute is, 
Given:
Molarity of
= 0.0025 mol/L
Molar mass of
= 136 g/mole
Volume of solution = 485 mL
Now put all the given values in the above formula, we get:


Thus, the mass in grams of calcium sulfate is 0.16 grams.