Question

hola necesito yuda con estos ejercicios de quimica de la G a la K porfa

Answer 1

$G) \\ V=87L \\ T=27+273=300K \\ P=740mmHg* \frac{1atm}{760mmHg} = \frac{37}{38} atm \\ \\ \\ PV=nRT \\ \frac{37}{38} *87=n*0,082*300 \\ \\ n= \frac{ \frac{37}{38}*87 }{0,082*300} \\ \\ n=3,4435moles$

Ahora que conocemos el número de moles que teníamos en el primer momento, podemos calcular el volumen que ocupará en las nuevas condiciones:

$T=5+273=278K \\ P= \frac{37}{38} atm \\ \\ \\ PV=nRT \\ V= \frac{nRT}{P} \\ \\ V= \frac{3,4435*0,082*278}{ \frac{37}{38} } \\ \\ V=80,62L \\ \\ V=80,62L* \frac{1000mL}{1L} =80.620mL$


$I) \\ V_{1} =79,5mL \\ T_{1} = 273+45=318K \\ \\ P_{cte}=\ \textgreater \ \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}} =\ \textgreater \ \frac{79,5}{318} = \frac{V_{2}}{273} \\ \\V_{2} = \frac{79,5*273}{318} =68,25mL$

$J) \\ V_{1}=200mL \\ T_{1}=30+273=303K \\ \\ P_{cte}=\ \textgreater \ \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}} =\ \textgreater \ \frac{200}{303} = \frac{260}{T_{2}} \\ \\T_{2} = \frac{260*303}{200} =393,9K \\ \\ 393K=\ \textgreater \ F=1,8*(393,3-273)+32=248,54F$

$K) \\ V_{1}=45mL \\ T_{1}=27+273=300K \\ \\ P_{cte}=\ \textgreater \ \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}} =\ \textgreater \ \frac{45}{300} = \frac{6000}{T_{2}} \\ \\T_{2} = \frac{6000*300}{45} =40.000K \\ \\ 40.000K-273=39.727C$