Answer:
8.55 moles
Explanation:
The formula for calculating molarity is moles of solute divided by liters of solution. In this case:
moles. Hope this helps!
an element has a metallic gray appearance. it also has the properties of both a metal and a nonmetal. in which section of the pe
riodic table indicated above would the element most likely be found?
1 answer:
If it has properties of metal and non metal it is considered a metalloid there are very few spots in which metalloids are located the spots I circled on a blank periodic tables are where the metalloids are located on an actual periodic table your only options are
boron(B) silicon(si) germanium(Ge) arsenic(As)antimony(Sb)tellurium(Te)astatine(At)those are the only places on the periodic table that has metalloids... I hope this helps
boron(B) silicon(si) germanium(Ge) arsenic(As)antimony(Sb)tellurium(Te)astatine(At)those are the only places on the periodic table that has metalloids... I hope this helps
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Considering the definition of molar mass, the moles of gas used are 10.625 moles.
<h3>Definition of molar mass</h3>The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.
<h3>Amount of moles used</h3>Natural gas has a molar mass of 16.0 g/mole.
You started out the day with a tank containing 200.0 g of natural gas. So, you can apply the following rule of three: If by definition of molar mass 16 grams are contained in 1 mole, 200 grams are contained in how many moles?
<u><em>amount of moles at the beginning= 12.5 moles</em></u>
At the end of the day, your tank contains 30.0 g of natural gas. So, you can apply the following rule of three: If by definition of molar mass 16 grams are contained in 1 mole, 30 grams are contained in how many moles?
<u><em>amount of moles at the end= 1.875 moles</em></u>
The number of moles used will be the difference between the number of moles used initially and the contents at the end of the day.
moles used= amount of moles at the beginning - amount of moles at the end
moles used= 12.5 moles - 1.875 moles
<u><em>moles used= 10.625 moles</em></u>
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Finally, the moles of gas used are 10.625 moles.
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