Answer:
2.122×10^25atoms
Explanation:
number of moles=mass/molar mass
7.05moles= mass of pyridine/79
reacting mass of pyridine=556.95
C5H5N= (12×5)+(5)+(14)=79
C5=60
to find the mass of carbon in 556.95g of pyridine we take the stoichometric ratio
60[C5] -----> 79[C5H5N]
x[C5] --------> 556.95g[C5H5N]
cross multiply
x=(60×556.95)/79
x=423g of carbon
moles=mass/molar mass
moles of carbon=423/12
moles=35.25moles of carbon
moles=number of particles/Avogadro's constant
35.25=number of particles/6.02×10^23
number of particles=2.122×10^25atoms of carbon
Answer:
4190.22 L = 4.19 m³.
Explanation:
- For the balanced reaction:
<em>2P₂ + 5O₂ ⇄ 2P₂O₅. </em>
It is clear that 2 mol of P₂ react with <em>5 mol of O₂ </em>to produce <em>2 mol of P₂O₅.</em>
- Firstly, we need to calculate the no. of moles of 6.92 kilograms of P₂O₅ produced through the reaction:
no. of moles of P₂O₅ = mass/molar mass = (6920 g)/(283.88 g/mol) = 24.38 mol.
- Now, we can find the no. of moles of O₂ is needed to produce the proposed amount of P₂O₅:
<u><em>Using cross multiplication:</em></u>
5 mol of O₂ is needed to produce → 2 mol of P₂O₅, from stichiometry.
??? mol of O₂ is needed to produce → 24.38 mol of P₂O₅.
∴ The no. of moles of O₂ needed = (5 mol)(24.38 mol)/(2 mol) = 60.95 mol.
- Finally, we can get the volume of oxygen using the general law of ideal gas:<em> PV = nRT.</em>
where, P is the pressure of the gas in atm (P = 606.1 mm Hg/760 = 0.8 atm).
V is the volume of the gas in L (V = ??? L).
n is the no. of moles of the gas in mol (n = 60.95 mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (396.90°C + 273 = 669.9 K).
∴ V of oxygen needed = nRT/P = (60.95 mol)(0.0821 L.atm/mol.K)(669.9 K)/(0.8 atm) = 4190.22 L/1000 = 4.19 m³.
Answer:
B 1.23 g/cc
Explanation:
For something to float on seawater, the density must be less than 1.03 g/mL. If the object sinks, the density is greater than 1.03 g/mL.
Let’s examine the answer choices. Keep in mind, the ice berg is mostly below the water level.
A. 0.88 g/cc
This is less than 1.03 g/cc, which would result in floating.
B. 1.23 g/cc
This is the best answer choice. The iceberg is mostly beneath the water, but some of it is exposed. The density is greater than 1.03 g/mL, but not so much greater that it would immediately sink.
C. 0.23 g/cc
This is less than 1.03 g/cc, which would produce floating.
D. 4.14 g/cc
This is much greater than 1.03 g/cc and the result would be sinking.
The Volumes can be calculated from Masses by using following Formula,
Density = Mass / Volume
Solving for Volume,
Volume = Mass / Density
Mass of Both Gases = 14.1 g
Density of Argon at S.T.P = 1.784 g/L
Density of Helium at S.T.P = 0.179 g/L
For Argon:
Volume = 14.1 g / 1.784 g/L
Volume = 7.90 L
For Helium:
Volume = 14.1 g / 0.179 g/L
Volume = 78.77 L
Answer: ABC&AMN are congruent by ASA comgruence
angle A=A
sideAN=AB
angle M=C
Explanation:
