When the water starts boiling out with bubbles...
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>
[H+][OH-] = 1x10-14 = Kw (Memorize this relationship)
(2.70x10-2)[OH-] = 1x10-14
[OH-] = 1x10-14/2.70x10-2
[OH-] = 3.70x10-13 M (assuming you ignore the autoionization of H2O)
Answer:
hope this helps you......
We can use the ideal gas law equation to find the pressure
PV = nRT
where;
P - pressure
V - volume - 3.9 x 10⁻³ m³
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 45 °C + 273 = 318 K
n - number of moles
there are 6.022 x 10²³ molecules in 1 mol of H₂
Therefore when theres 2.6 x 10²³ H₂ molecules - 1 / (6.022 x 10²³) x 2.6 x 10²³
Number of H₂ moles - 0.43 mol
substituting these values in the equation,
P x 3.9 x 10⁻³ m³ = 0.43 mol x 8.314 Jmol⁻¹K⁻¹ x 318 K
P = 291.5 kPa