Answer:
Such molecule must have molecular formula of C15N3H15
Explanation:
Mass of carbon in such molecule
The atomic mass of carbon is 12.01 g/mol, so in 182.28 g of carbon there is 15.18 mols of carbon.
Mass of Nitrogen in such molecule
The atomic mass of nitrogen is 14.01 g/mol, so in 42.53g of nitrogen there is 3.04 mols of nitrogen.
Mass of Hydrogen in such molecule
The atomic mass of Hydrogen is 1.00 g/mol, so in 15.19 g of Hydrogen there is 15.19 mols of Hydrogen.
Such molecule must have molecular formula of C15N3H15
Answer:
Explanation:
We want to convert from moles to grams, so we must use the molar mass.
<h3>1. Molar Mass</h3>
The molar mass is the mass of 1 mole of a substance. It is the same as the atomic masses on the Periodic Table, but the units are grams per mole (g/mol) instead of atomic mass units (amu).
We are given the compound PI₃ or phosphorus triiodide. Look up the molar masses of the individual elements.
- Phosphorus (P): 30.973762 g/mol
- Iodine (I): 126.9045 g/mol
Note that there is a subscript of 3 after the I in the formula. This means there are 3 moles of iodine in 1 mole of the compound PI₃. We should multiply iodine's molar mass by 3, then add phosphorus's molar mass.
- I₃: 126.9045 * 3=380.7135 g/mol
- PI₃: 30.973762 + 380.7135 = 411.687262 g/mol
<h3>2. Convert Moles to Grams</h3>
Use the molar mass as a ratio.
We want to convert 3.14 moles to grams, so we multiply by that value.
The units of moles of PI₃ cancel.
<h3>3. Round</h3>
The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated, that is the tens place.
The 2 in the ones place tells us to leave the 9.
3.14 moles of phosphorous triiodide is approximately equal to <u>1290 grams of phosphorus triodide.</u>
part 1 : the final volume : 1.404 L
part 2 : the initial concentration : 4.06 M
<h3>Further explanation
</h3>
Dilution is the process of adding a solvent to get a more dilute solution.
The moles(n) before and after dilution are the same.
Can be formulated :
M₁V₁=M₂V₂
M₁ = Molarity of the solution before dilution
V₁ = volume of the solution before dilution
M₂ = Molarity of the solution after dilution
V₂ = Molarity volume of the solution after dilution
part 1 :
M₁=44.8%
V₁=0.73 L
M₂=23.3%
part 2 :
V₁=739 ml=0.739 L
V₂=1.5 L
M₂=2
Answer:
Given: 42 g of N2
Solve for O2 mass that contains the same number of molecules to 42 g of N2.
Solve for the number of moles in 42 g of N2
1 mole of N2 = (14 * 2) g = 28 g so the number of moles in 42 g of N2 is equal to 42 g / 28 g per mole = 1.5 moles
Solve for mass of 1 mole of oxygen
1 mole of O2 = 16 g * 2 = 32 g per mole
Solve for the mass of 1.5 moles of oxygen
mass of 1.5 moles of O2 = 32 g per mole * 1.5 moles
mass of 1.5 moles of O2 = 48 g
So 48 g of O2 contains the same number of molecules as 42 g of N2
The top one is different from the bottom because of is curvature shape while the bottom one is a square shape i think the bottom will heat up faster because of the nice even area inside where heat waves can evenly flow
