User:5LAY3R95/sandbox

1)

$c_{\frac {m}{m}}={\frac {m_{soluto}}{m_{disolucion}}}$

${\frac {5}{100}}={\frac {500g\cdot {\frac {12}{100}}}{500g+m_{extra}}}$

$m_{extra}=700g$

2)

$m_{solvente}=V_{solvente}\cdot \rho _{solvente}$

$m_{solvente}=V_{disolucion}\cdot {\frac {V_{solvente}}{V_{disolucion}}}\cdot {\rho _{solvente}}$

$m_{solvente}=400mL\cdot {\frac {60}{100}}\cdot {\frac {1g}{mL}}$

$m_{solvente}=240g$

${\frac {m_{soluto}}{m_{disolucion}}}={\frac {20}{100}}$

${\frac {m_{soluto}}{m_{soluto}+m_{solvente}}}={\frac {20}{100}}$

${\frac {m_{soluto}}{m_{soluto}+240g}}={\frac {20}{100}}$

$m_{soluto}=60g$

$V_{soluto}=V_{disolucion}\cdot {\frac {V_{soluto}}{V_{disolucion}}}$

$V_{soluto}=400mL\cdot \left({\frac {100}{100}}-{\frac {60}{100}}\right)$

$V_{soluto}=160mL$

$\rho _{soluto}={\frac {m_{soluto}}{V_{soluto}}}={\frac {60g}{160mL}}=0.375{\frac {g}{mL}}$

3)

$\rho _{A}=2\rho _{B}$

$\rho _{A}V_{A}+\rho _{B}V_{B}=\rho _{disolucion}\cdot V_{disolucion}$

$2\rho _{B}V+\rho _{B}V=\rho _{disolucion}\cdot 2V$

$\rho _{B}+{\frac {\rho _{B}}{2}}=\rho _{disolucion}$

$2{\frac {g}{mL}}+{\frac {2{\frac {g}{mL}}}{2}}=\rho _{disolucion}$

$3{\frac {g}{mL}}=\rho _{disolucion}$

4)

$m_{soluto}+m_{solvente}=m_{disolucion}$

$m_{B}+\rho _{H_{2}O}V_{H_{2}O}=\rho _{disolucion}\cdot (V_{B}+V_{H_{2}O})$

$m_{B}+\rho _{H_{2}O}V_{H_{2}O}=\rho _{disolucion}\cdot \left({\frac {m_{B}}{\rho _{B}}}+V_{H_{2}O}\right)$

$60g+{\frac {1g}{mL}}40mL=\rho _{disolucion}\cdot \left({\frac {60g}{1.5{\frac {g}{mL}}}}+40mL\right)$

$\rho _{disolucion}=1.25{\frac {g}{mL}}$

5)

$V_{soluto}+V_{solvente}=V_{disolucion}$

${\frac {m_{soluto}}{\rho _{soluto}}}+{\frac {m_{solvente}}{\rho _{solvente}}}={\frac {m_{disolucion}}{\rho _{disolucion}}}$

${\frac {m_{soluto}}{\rho _{soluto}}}+{\frac {m_{solvente}}{\rho _{solvente}}}={\frac {m_{soluto}+m_{solvente}}{\rho _{disolucion}}}$

${\frac {m}{\rho _{soluto}}}+{\frac {m}{\rho _{solvente}}}={\frac {2m}{\rho _{disolucion}}}$

${\frac {1}{\rho _{soluto}}}+{\frac {1}{\rho _{solvente}}}={\frac {2}{\rho _{disolucion}}}$

${\frac {1}{\rho _{soluto}}}+{\frac {1}{1{\frac {g}{mL}}}}={\frac {2}{1.5{\frac {g}{mL}}}}$

$\rho _{soluto}=3{\frac {g}{mL}}$

6)

$m_{regla}\cdot g=m_{desplazada}\cdot g$

$V_{regla}\cdot \rho _{madera}\cdot g=V_{inmersion}\cdot \rho _{agua}\cdot g$

$h_{regla}\cdot A\cdot \rho _{madera}\cdot g=h_{inmersion}\cdot A\cdot \rho _{agua}\cdot g$

$h_{regla}\cdot \rho _{madera}=h_{inmersion}\cdot \rho _{agua}$

$40cm\cdot 0.45{\frac {g}{cm^{3}}}=h_{inmersion}\cdot 1{\frac {g}{cm^{3}}}$

$18cm=h_{inmersion}$

Se necesita reducir el peso efectivo de la regla de madera usando la bola de acero

$h_{regla}\cdot A\cdot \rho _{madera}\cdot g-m_{acero}\cdot g=h_{inmersion}\cdot A\cdot \rho _{agua}\cdot g$

$h_{regla}\cdot A\cdot \rho _{madera}-m_{acero}=h_{inmersion}\cdot A\cdot \rho _{agua}$

$h_{regla}\cdot A\cdot \rho _{madera}-h_{inmersion}\cdot A\cdot \rho _{agua}=m_{acero}$

$40cm\cdot 3cm^{2}\cdot 0.45{\frac {g}{cm^{3}}}-15cm\cdot 3cm^{2}\cdot 1{\frac {g}{cm^{3}}}=m_{acero}$

$39g=m_{acero}$

7)

$capacidad_{agua}=capacidad_{gasolina}$

${\frac {m_{agua}}{\rho _{agua}}}={\frac {m_{gasolina}}{\rho _{gasolina}}}$

${\frac {14.18kg}{1{\frac {g}{cm^{3}}}}}={\frac {28.36kg}{\rho _{gasolina}}}$

$\rho _{gasolina}=0.5{\frac {g}{cm^{3}}}$

8)

$m_{agua}+m_{acido}=m_{disolucion}$

$V_{agua}\rho _{agua}+V_{acido}\rho _{acido}=V_{disolucion}\cdot \rho _{disolucion}$

$V\rho _{agua}+V\rho _{acido}=2V\cdot \rho _{disolucion}$

$\rho _{agua}+\rho _{acido}=2\cdot \rho _{disolucion}$

$\rho _{acido}=0.4{\frac {g}{cm^{3}}}$

${\frac {m_{acido}}{m_{disolucion}}}={\frac {V_{acido}\cdot \rho _{acido}}{V_{disolucion}\cdot \rho _{disolucion}}}$

${\frac {m_{acido}}{m_{disolucion}}}={\frac {V\cdot \rho _{acido}}{2V\cdot \rho _{disolucion}}}$

${\frac {m_{acido}}{m_{disolucion}}}={\frac {\rho _{acido}}{2\cdot \rho _{disolucion}}}$

${\frac {m_{acido}}{m_{disolucion}}}=0.2857=28.57\%$

9)

$c_{molar}={\frac {n_{H_{2}SO_{4}}}{V_{disolucion}}}$

$c_{molar}={\frac {m_{H_{2}SO_{4}}}{M_{H_{2}SO_{4}}\cdot V_{disolucion}}}$

$c_{molar}={\frac {19.6g}{98{\frac {g}{mol}}0.5L}}$

$c_{molar}=0.4M$

10)

$c_{normal}=\phi \cdot c_{molar}$

$c_{normal}=\phi \cdot 1.5{\frac {g_{disolucion}}{cm^{3}}}\cdot {\frac {60g_{NaOH}}{100g_{disolucion}}}\cdot {\frac {1mol_{NaOH}}{40g_{NaOH}}}$

$c_{normal}=1{\frac {N}{M}}0.0225{\frac {mol}{cm^{3}}}$

$c_{normal}=1{\frac {N}{M}}0.0225{\frac {mol}{cm^{3}}}{\frac {1000cm^{3}}{1L}}$

$c_{normal}=22.5N$

11)

${\frac {1.84g_{disolucion}}{cm^{3}}}\cdot {\frac {98g_{H_{2}SO_{4}}}{100g_{disolucion}}}\cdot {\frac {1mol_{H_{2}SO_{4}}}{98g_{H_{2}SO_{4}}}}\cdot V_{disolucion}=1L\cdot 1M$

$V_{disolucion}=54.34mL$

12)

${\frac {1.066g_{disolucion}}{cm^{3}}}\cdot {\frac {10g_{H_{2}SO_{4}}}{100g_{disolucion}}}\cdot {\frac {1mol_{H_{2}SO_{4}}}{98g_{H_{2}SO_{4}}}}\cdot {\frac {2N}{M}}\cdot {\frac {1000cm^{3}}{1L}}=c_{normal}$

$2.176N=c_{normal}$

13)

${\frac {1.54mol_{etanol}}{1kg_{solvente}}}\cdot {\frac {46g_{etanol}}{1mol_{etanol}}}\cdot 2.5L_{solvente}\cdot {\frac {1kg_{solvente}}{1L_{solvente}}}=m_{etanol}$

$177.1g=m_{etanol}$

14)

$2L\cdot 4M=V_{total}\cdot 1.5N\cdot {\frac {1M}{N}}$

$V_{total}=5.3L=2L+V_{agregar}$

$V_{agregar}=3.3L$

15)

$500mL\cdot 6.8M=V_{cc}\cdot 8M$

$V_{cc}=425mL$

$V_{total}=V_{cc}+V_{agua}$

$V_{agua}=0.075L$

16)

${\frac {70mg_{Na}}{mL}}\cdot 50mL\cdot {\frac {1g_{Na}}{1000mg_{Na}}}\cdot {\frac {85g_{NaNO_{3}}}{23g_{Na}}}=m_{NaNO_{3}}$

$12.93g=m_{NaNO_{3}}$

17)

$P=P^{0}\cdot \chi$

${\frac {n_{etilenglicol}+n_{etanol}}{n_{etanol}}}={\frac {P^{0}}{P}}$

$1+{\frac {n_{etilenglicol}}{n_{etanol}}}={\frac {P^{0}}{P}}$

$1+{\frac {m_{etilenglicol}}{m_{etanol}}}\cdot {\frac {M_{etanol}}{M_{etilenglicol}}}={\frac {P^{0}}{P}}$

$1+{\frac {m_{etilenglicol}}{1000g}}\cdot {\frac {46{\frac {g}{mol}}}{62{\frac {g}{mol}}}}={\frac {100tor}{100torr-13.2torr}}$

$m_{etilenglicol}=205g$

18)

$P_{disolucion}=P_{agua}^{0}\cdot \chi _{agua}+P_{etanol}^{0}\cdot \chi _{etanol}$

$P_{disolucion}=P_{agua}^{0}\cdot \chi _{agua}+P_{etanol}^{0}\cdot (1-\chi _{agua})$

$\chi _{agua}={\frac {n_{agua}}{n_{agua}+n_{etanol}}}={\frac {1}{1+{\frac {n_{etanol}}{n_{agua}}}}}$

$\chi _{agua}={\frac {1}{1+{\frac {m_{etanol}}{m_{agua}}}{\frac {M_{agua}}{M_{etanol}}}}}$

$\chi _{agua}={\frac {1}{1+{\frac {18{\frac {g}{mol}}}{46{\frac {g}{mol}}}}}}$

$\chi _{agua}=0.71875$

$P_{disolucion}=175torr\cdot 0.71875+400torr\cdot (1-0.71875)$

$P_{disolucion}=175torr\cdot 0.71875+400torr\cdot (1-0.71875)$

$P_{disolucion}=238torr$

19)

$\pi V=nRT$

$\pi ={\frac {mRT}{MV}}$

$\pi ={\frac {2.02g\cdot 0.082{\frac {atm\cdot L}{mol\cdot K}}\cdot 293.15K}{60{\frac {g}{mol}}0.145L}}$

$\pi =5.58atm$

20)

$\pi V=nRT$

$\pi ={\frac {mRT}{MV}}$

$\pi ={\frac {c_{\frac {m}{v}}RT}{M}}$

$\pi ={\frac {3.4{\frac {g}{L}}\cdot 0.082{\frac {atm\cdot L}{mol\cdot K}}\cdot 293.15K}{58{\frac {g}{mol}}}}$

$\pi =1.41atm$