When 85 L of gas at 1.376 atm has its pressure changed to 0.154 atm, what is the new volume?

Chemistry

1 answer:

igor_vitrenko [27]3 years ago

4 0

Answer:

Explanation:

The new volume can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

Since we are finding the new volume

$V_2 = \frac{P_1V_1}{P_2} \\$

From the question we have

$V_2 = \frac{85 \times 1.376}{0.154} = \frac{116.96}{0.154} \\ = 759.480519...$

We have the final answer as

Hope this helps you

You might be interested in

18.2L of gas at 95°C and 760 torr is placed in a 15L container at 80 degrees * C ; what is the new pressure ?

romanna [79]

Answer:

884.56 torr

Explanation:

Formula: $\frac{P_{1}V_{1} }{T_{1}} = \frac{P_{2}V_{2} }{T_{2}}$

P = Pressure

V = Volume

T = Temperature in kelvin (Celsius + 273.15)

$\frac{(760)(18.2) }{368.15} = \frac{P(15) }{353.15}$

$P = \frac{(760)(18.2)(353.15) }{(368.15)(15)}$

P = 884.56169

4 0

3 years ago

What is the composition, in atom percent, of an alloy that contains a) 45.5 lbm of silver, b) 83.7 lbm of gold, and c) 6.3 lbm o

lyudmila [28]

Answer:

$\% atAg=44.6\%\\\% atAu=44.9\%\\\% atCu=10.5\%$

Explanation:

Hello,

In this case, for computing the atom percent, one must obtain the number of atoms of silver, gold and copper as shown below:

$atomsAg=45.5lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{107.87gAg}*\frac{6.022x10^{23}atomsAg}{1molAg}=1.15x10^{26}atomsAg\\atomsAu=83.7lbm*\frac{453.59g}{1lbm}*\frac{1molAu}{196.97gAu}*\frac{6.022x10^{23}atomsAu}{1molAu}=1.16x10^{26}atomsAu\\atomsCu=6.3lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{63.55gCu}*\frac{6.022x10^{23}atomsCu}{1molCu}=2.71x10^{25}atomsCu$

Thus, the atom percent turns out:

$\% atAg=\frac{1.15x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.6\%\\\% atAu=\frac{1.16x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.9\%\\\% atCu=\frac{2.71x10^{25}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =10.5\%$