Answer:
Explanation:
mass ratio of oxygen and nitrogen in air at Miami
= 21 : 79
ratio of their moles
= ( mol weight of oxygen is 32 and of nitrogen is 28 )
= .65625 : 2.8214
= 1 : 4.3
This ratio will also be maintained in the air of Denver though total pressure decreases there.
Partial pressure of oxygen in air at both the places
mole fraction of oxygen
=
= .18868
partial pressure of oxygen at Denver
= .18868 x .83 x 760
= 119 mm.
Answer:- 38.2 g.
Solution:- The equation used for solving this type of calorimetry problems is:
where, q is the heat energy, m is mass, c is specific heat and delta T is the change in temperature.
Water temperature is increasing from 14.5 degree C to 50.0 degree C.
= 35.5 degree C
q is given as 5680 J and specific heat value is .
The equation could be rearranged for m as:
Let's plug in the values in it:
m = 38.2 g
So, the mass of water in the kettle is 38.2 g.
Answer:
Explanation:
Hello,
In this case, for the calculation of the temperature in degree Celsius we subtract 273.15 to the given temperature in kelvins:
Next, by applying the following equation we compute it in degree Fahrenheit:
Clearly, since the initial unit has two significant figures the computed units also show two significant figures.
Regards.
Answer:
a) 11.55 moles of carbon dioxide gas are formed from 3.85 mole of propane.
15.4 moles of water are formed from 3.85 mole of propane.
b)0.5176 moles of water are formed from 0.647 mole of oxygen gas.
0.1294 moles of propane are consumed.
Explanation:
a) Moles of propane = 3.85 moles
According to reaction, 1 mole of propane gives 3 moles of carbon dioxide gas.
Then 3.85 moles of propane will give:
of carbon dioxide gas.
11.55 moles of carbon dioxide gas are formed from 3.85 mole of propane.
According to reaction, 1 mole of propane gives 4 moles of water gas.
Then 3.85 moles of propane will give:
of water .
15.4 moles of water are formed from 3.85 mole of propane.
b) Moles of oxygen gas = 0.647 moles
According to reaction, 5 mole of oxygen gas gives 4 moles of water.
Then 0.647 moles of oxygen will give:
of water.
0.5176 moles of water are formed from 0.647 mole of oxygen gas.
According to reaction, 5 mole of oxygen gas reacts with 1 mole of propane.
Then 0.647 moles of oxygen will give:
of propane.
0.1294 moles of propane are consumed.