Partial pressure=mole fraction×Pt
x=0.044÷44(maolarmass of CO2)×Pt
x=0.044÷(44)2×Pt
x=5×10^-4×Pt
x=5×10^-4×Pt
where Pt:1atm=760mmHg
xatm=750mmHg
750×1÷760=0.99
now;5×10^-4×099=4.95×10^-4.
Pt=4.95×10^-4
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Answer:
<em>so mass in gram=560grams</em>
Explanation:
number of moles=10moles
molar mass=56grams/moles
mass in gram of Fe=?
as we know that

<em>evaluating the formula</em>
<em>number of moles×molar mass=mass in gram</em>
<em>mass in gram=10moles×56grams/moles</em>
<em>mass in gram=560grams</em>
<em>i hope this will help you :)</em>
128 ml is the voume of the balloon if the temperature of the gas increases to 320.0k.
Explanation:
given that:
T1 (initial temperature) = 300K
V1 ( initial volume) = 120ml
T2 (final temperature) = 320 K
V2 (final volume) = ?
Pressure remained constant throughout the process.
From the equation
= 
Since pressure is constant the equation will be:
= 
V2 = 
Putting the values in the above formula:
V2 = 
= 128 ml
128 ml is the volume of the gas if temperature increases from 3OO K to 320k