A 100.0 g sample of an unknown metal at 90.0 degree C is thrown into a calorimet

Question

A 100.0 g sample of an unknown metal at 90.0 degree C is thrown into a calorimeter containing 50.0 g of water at 25.0 degree C. After the system comes to equilibrium, the temperature is 31 6 degree C. If no heat is lost to the environment or the calorimeter, calculate the heat capacity of the metal The heal capacity of water is 4.184 J/gK. A gas occupies a volume of 0.600 L at 36 degree C and 1.00 atm pressure. What will its volume be at 0 degree C and 150.0 mmHg? Calculate the mass density of a sample of argon gas at -15.0 degree C and. 10.0 atm in a 1.00 L flask.

Explanation / Answer

8) the gas occupies volume of 0.600 L at 36 0c and 1.0 atm pressure ,

now we will calculate the Volume at 00c and 150 mm of hg pressure,

we know the ideal gas low = PV=nRT,

here we will first caculate the n (moles) using the formula as,

n= PV/RT,

BY PUTTING THE vALUES TO THE FORMULA,

n= 1.0*0.600L/0.0820*309 k, ( 36+273=309 K),

n=0.600/25.338,

n=0.02367 moles ,

NOW WE CAN CALCULATE THE Volume at 0 0 cand at 150.0 mm oF hg,

Here we need to convert mm of hg to ATM,

150 MM OF HG = 0.1973 ATM,

V= nRT/P, BY putting the values to equation,

V=0.02367*0.0820*273 K/0.1973,

V= 0.5298/0.1973,

v=2.6856 L,

9) by using ideal gas low ,

PV=nRT,

WE will calculate the moles first , (n),

n= PV/RT,

PUTTING THE VALUES ,

n= 10.0*1L/0.0820*258 k, (-15 0 C=258 K),

n= 10.0 /21.156,

n= 0.4726 moles of Argon,

from That we can calculate the mas sof argon as,

mol wt of argon = 39.948 ,

mass = moles*mol wt,

mass = 0.4726*39.948,

= 18.47 gms of argon ,

then we can calculate the density as,

density = mass / Volume,

density= 18.47/1.0 L,

= 18.47 gm/L

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