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Montano1993 [528]
3 years ago
11

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:

<h2>759.48 L</h2>

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

<h3>759.48 L</h3>

Hope this helps you

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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\%

Best regards.

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