Answer:
70.6 %
Explanation:
First step, we define the reaction:
2P + 3Br₂ → 2PBr₃
We determine the moles of reactant:
35 g . 1mol / 159.8 g = 0.219 moles
We assume, the P is in excess, so the bromine is the limiting reagent.
3 moles of Br₂ can produce 2 moles of phophorous tribromide
Then, 0.219 moles may produce (0.219 . 2) /3 = 0.146 moles of PBr₃
We convert moles to mass:
0.146 mol . 270.67 g /mol = 39.5 g
That's the 100 % yield reaction, also called theoretical yield. The way to determine the % yield is:
(Yield produced / Thoeretical yield) . 100
(27.9 / 39.5) . 100 = 70.6 %
Answer: The mass is 980.6g of Gold.
Explanation:
We begin by looking for the number of moles equivalent to 3.0 x 10^24 gold atoms.
Using the Avogadro's number,
6.02 x 10^23 atoms of gold make up 1 mole of gold.
3.0 x 10^24 atoms would make up: 1 / 6.02 x 10^23 x 3.0 x 10^24 = 4.98moles.
Now that we know the number of moles, we can then look for the mass using the formular:
Moles = mass/ molar mass
4.98 = mass / 196.9 (atomic mass of gold)
Making "mass" the subject of formula : mass = 4.98 x 196.9= 980.6g
Answer:
120mph
Explanation:
Google
divide the speed value by 1.467
or
176 times 60 second in a minute times 60 minutes in an hour
than divide by 5280 the amount of feet in a mile
Answer:
vHe / vNe = 2.24
Explanation:
To obtain the velocity of an ideal gas you must use the formula:
v = √3RT / √M
Where R is gas constant (8.314 kgm²/s²molK); T is temperature and M is molar mass of the gas (4x10⁻³kg/mol for helium and 20,18x10⁻³ kg/mol for neon). Thus:
vHe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol
vNe = √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol
The ratio is:
vHe / vNe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol / √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol
vHe / vNe = √20.18x10⁻³kg/mol / √4x10⁻³kg/mol
<em>vHe / vNe = 2.24</em>
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I hope it helps!
Answer:
P = 164 Atm
Explanation:
PV = nRT => P = nRT/V
n = 10.0 moles
R = 0.08206 L·Atm/mol·K
T = 27.0°C = 300 K
V = 1.50 Liters
P = (10.0 mol)(0.08206 L·Atm/mol·K )(300 K)/(1.50 Liters) = 164.12 Atm ≅ 164 Atm (3 sig. figs.)
