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fomenos
3 years ago
9

93.0 mL of O2 gas is collected over water at 0.930 atm and 10.0C what would be the volume of this dry gas at STP

Chemistry
1 answer:
Dahasolnce [82]3 years ago
3 0
The  volume  of  the  dry   gas  at     stp  is  calculated as follows

calculate  the  number  on  moles  by use  of   PV =nRT  where  n  is  the number  of  moles

n  is therefore  = Pv/RT
P  = 0.930  atm
R(gas  contant=  0.0821  L.atm/k.mol
V= 93ml  to  liters =  93/1000= 0.093L
T=  10  +  273.15 = 283.15k

n=  (0.930  x0.093)  /(0.0821  x283.15) =  3.  72  x10^-3  moles

At  STp    1  mole  =  22.4L
what about  3.72  x10^-3 moles

by  cross  multiplication

volume =  (3.72  x10^-3)mole  x  22.4L/  1  moles  = 0.083 L   or  83.3 Ml
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1 year ago
.325 mol KOH to grams
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2 years ago
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2 years ago
At a festival, spherical balloons with a radius of 280. Cm are to be inflated with hot air and released. The air at the festival
evablogger [386]

Answer:

8608.18 balloons

Explanation:

Hello! Let's solve this!

Data needed:

Enthalpy of propane formation: 103.85kJ / mol

Specific heat capacity of air: 1.009J · g ° C

Density of air at 100 ° C: 0.946kg / m3

Density of propane at 100 ° C: 1.440kg / m3

First we will calculate the propane heat (C3H8)

3000g * (1mol / 44g) * (103.85kJ / mol) * (1000J / 1kJ) = 7.08068 * 10 ^ 6 J

Then we can calculate the mass of the air with the heat formula

Q = mc delta T

m = Q / c delta T = (7.08068 * 10 ^ 6 J) / (1.009J / kg ° C * (100-25) ° C) =

m = 93566.96kg

We now calculate the volume of a balloon.

V = 4/3 * pi * r ^ 3 = 4/3 * 3.14 * 1.4m ^ 3 = 11.49m ^ 3

Now we calculate the mass of the balloon

mg = 0.946kg / m3 * 11.49m ^ 3 = 10.87kg

The amount of balloons is

93566.96kg / 10.87kg = 8608.18 balloons

5 0
3 years ago
Read 2 more answers
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