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)


 * $$m_{solvente} = V_{solvente} \cdot \rho_{solvente}$$
 * $$m_{solvente} = V_{disolucion} \cdot \frac{V_{solvente}}{V_{disolucion}} \cdot {\rho_{solvente}}$$
 * $$m_{solvente} = 400 mL \cdot \frac{60}{100} \cdot \frac{1 g}{mL}$$
 * $$m_{solvente} = 240 g$$


 * $$\frac{m_{soluto}}{m_{disolucion}} = \frac{20}{100}$$
 * $$\frac{m_{soluto}}{m_{soluto} + m_{solvente}} = \frac{20}{100}$$
 * $$\frac{m_{soluto}}{m_{soluto} + 240 g} = \frac{20}{100}$$
 * $$m_{soluto} = 60 g$$


 * $$V_{soluto} = V_{disolucion} \cdot \frac{V_{soluto}}{V_{disolucion}}$$
 * $$V_{soluto} = 400 mL \cdot \left( \frac{100}{100} - \frac{60}{100} \right)$$
 * $$V_{soluto} = 160 mL$$


 * $$\rho_{soluto} = \frac{m_{soluto}}{V_{soluto}} = \frac{60 g}{160 mL} = 0.375 \frac{g}{mL}$$

3)


 * $$\rho_A = 2 \rho_B$$
 * $$\rho_A V_A + \rho_B V_B = \rho_{disolucion} \cdot V_{disolucion}$$
 * $$2 \rho_B V + \rho_B V = \rho_{disolucion} \cdot 2 V$$
 * $$\rho_B + \frac{\rho_B}{2} = \rho_{disolucion}$$
 * $$2 \frac{g}{mL} + \frac{2 \frac{g}{mL}}{2} = \rho_{disolucion}$$
 * $$3 \frac{g}{mL} = \rho_{disolucion}$$

4)


 * $$m_{soluto} + m_{solvente} = m_{disolucion}$$
 * $$m_B + \rho_{H_2 O} V_{H_2 O} = \rho_{disolucion} \cdot ( V_B + V_{H_2 O} )$$
 * $$m_B + \rho_{H_2 O} V_{H_2 O} = \rho_{disolucion} \cdot \left( \frac{m_B}{\rho_B} + V_{H_2 O} \right)$$
 * $$60 g + \frac{1 g}{mL} 40 mL = \rho_{disolucion} \cdot \left( \frac{60 g}{1.5 \frac{g}{mL}} + 40 mL \right)$$
 * $$\rho_{disolucion} = 1.25 \frac{g}{mL}$$

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)


 * $$m_{regla} \cdot g = m_{desplazada} \cdot g$$
 * $$V_{regla} \cdot \rho_{madera} \cdot g = V_{inmersion} \cdot \rho_{agua} \cdot g$$
 * $$h_{regla} \cdot A \cdot \rho_{madera} \cdot g = h_{inmersion} \cdot A \cdot \rho_{agua} \cdot g$$
 * $$h_{regla} \cdot \rho_{madera} = h_{inmersion} \cdot \rho_{agua}$$
 * $$40 cm \cdot 0.45 \frac{g}{cm^3} = h_{inmersion} \cdot 1 \frac{g}{cm^3}$$
 * $$18 cm = h_{inmersion}$$

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


 * $$h_{regla} \cdot A \cdot \rho_{madera} \cdot g - m_{acero} \cdot g = h_{inmersion} \cdot A \cdot \rho_{agua} \cdot g$$
 * $$h_{regla} \cdot A \cdot \rho_{madera} - m_{acero} = h_{inmersion} \cdot A \cdot \rho_{agua}$$
 * $$h_{regla} \cdot A \cdot \rho_{madera} - h_{inmersion} \cdot A \cdot \rho_{agua} = m_{acero}$$
 * $$40 cm \cdot 3 cm^2 \cdot 0.45 \frac{g}{cm^3} - 15 cm \cdot 3 cm^2 \cdot 1 \frac{g}{cm^3} = m_{acero}$$
 * $$39 g = m_{acero}$$

7)


 * $$capacidad_{agua} = capacidad_{gasolina}$$
 * $$\frac{m_{agua}}{\rho_{agua}} = \frac{m_{gasolina}}{\rho_{gasolina}}$$
 * $$\frac{14.18 kg}{1 \frac{g}{cm^3}} = \frac{28.36 kg}{\rho_{gasolina}}$$
 * $$\rho_{gasolina} = 0.5 \frac{g}{cm^3}$$

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)


 * $$c_{molar} = \frac{n_{H_2 SO_4}}{V_{disolucion}}$$
 * $$c_{molar} = \frac{m_{H_2 SO_4}}{M_{H_2 SO_4} \cdot V_{disolucion}}$$
 * $$c_{molar} = \frac{19.6 g}{98 \frac{g}{mol} 0.5 L}$$
 * $$c_{molar} = 0.4 M$$

10)


 * $$c_{normal} = \phi \cdot c_{molar}$$
 * $$c_{normal} = \phi \cdot 1.5 \frac{g_{disolucion}}{cm^3} \cdot \frac{60 g_{NaOH} }{100 g_{disolucion}} \cdot \frac{1 mol_{NaOH}}{40 g_{NaOH}}$$
 * $$c_{normal} = 1 \frac{N}{M} 0.0225 \frac{mol}{cm^3}$$
 * $$c_{normal} = 1 \frac{N}{M} 0.0225 \frac{mol}{cm^3} \frac{1000 cm^3}{1 L}$$
 * $$c_{normal} = 22.5 N$$

11)


 * $$\frac{1.84 g_{disolucion}}{cm^3} \cdot \frac{98 g_{H_2 SO_4} }{100 g_{disolucion}} \cdot \frac{1 mol_{H_2 SO_4} }{ 98 g_{H_2 SO_4} } \cdot V_{disolucion} = 1L \cdot 1M$$
 * $$V_{disolucion} = 54.34 mL$$

12)


 * $$\frac{1.066 g_{disolucion}}{cm^3} \cdot \frac{10 g_{H_2 SO_4} }{100 g_{disolucion}} \cdot \frac{1 mol_{H_2 SO_4} }{ 98 g_{H_2 SO_4} } \cdot\frac{2N}{M} \cdot \frac{1000 cm^3}{1 L}= c_{normal}$$
 * $$2.176 N = c_{normal}$$

13)


 * $$\frac{1.54 mol_{etanol}}{1 kg_{solvente}} \cdot \frac{46 g_{etanol}}{1 mol_{etanol}} \cdot 2.5 L_{solvente} \cdot \frac{1 kg_{solvente}}{1 L_{solvente}} = m_{etanol}$$
 * $$177.1 g = m_{etanol}$$

14)


 * $$2L \cdot 4 M = V_{total} \cdot 1.5 N \cdot \frac{1 M}{N}$$
 * $$V_{total} = 5.3 L = 2L + V_{agregar}$$
 * $$V_{agregar} = 3.3 L$$

15)


 * $$500 mL \cdot 6.8 M = V_{cc} \cdot 8 M$$
 * $$V_{cc} = 425 mL$$


 * $$V_{total} = V_{cc} + V_{agua}$$
 * $$V_{agua} = 0.075 L$$

16)


 * $$\frac{70 mg_{Na}}{mL} \cdot 50 mL \cdot \frac{1 g_{Na}}{1000 mg_{Na}} \cdot \frac{85 g_{NaNO_3}}{23 g_{Na}} = m_{NaNO_3}$$
 * $$12.93 g = m_{NaNO_3}$$

17)


 * $$P = P^0 \cdot \chi$$
 * $$\frac{n_{etilenglicol}+n_{etanol}}{n_{etanol}} = \frac{P^0}{P}$$
 * $$1 + \frac{n_{etilenglicol}}{n_{etanol}} = \frac{P^0}{P}$$
 * $$1 + \frac{m_{etilenglicol}}{m_{etanol}} \cdot \frac{M_{etanol}}{M_{etilenglicol}} = \frac{P^0}{P}$$
 * $$1 + \frac{m_{etilenglicol}}{1000 g} \cdot \frac{46 \frac{g}{mol}}{62 \frac{g}{mol}} = \frac{100 tor}{100 torr - 13.2 torr}$$
 * $$m_{etilenglicol} = 205 g$$

18)


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


 * $$\chi_{agua} = \frac{n_{agua}}{n_{agua}+n_{etanol}} = \frac{1}{1+\frac{n_{etanol}}{n_{agua}}}$$
 * $$\chi_{agua} = \frac{1}{1+\frac{m_{etanol}}{m_{agua}} \frac{M_{agua}}{M_{etanol}} }$$
 * $$\chi_{agua} = \frac{1}{1+ \frac{18 \frac{g}{mol}}{46 \frac{g}{mol}} }$$
 * $$\chi_{agua} = 0.71875$$


 * $$P_{disolucion} = 175 torr \cdot 0.71875 + 400 torr \cdot (1-0.71875)$$
 * $$P_{disolucion} = 175 torr \cdot 0.71875 + 400 torr \cdot (1-0.71875)$$
 * $$P_{disolucion} = 238 torr$$

19)


 * $$\pi V = nRT$$
 * $$\pi = \frac{mRT}{MV}$$
 * $$\pi = \frac{2.02 g \cdot 0.082 \frac{atm \cdot L}{mol \cdot K} \cdot 293.15 K}{60 \frac{g}{mol} 0.145 L}$$
 * $$\pi = 5.58 atm$$

20)


 * $$\pi V = nRT$$
 * $$\pi = \frac{mRT}{MV}$$
 * $$\pi = \frac{c_{\frac{m}{v}}RT}{M}$$
 * $$\pi = \frac{3.4 \frac{g}{L} \cdot 0.082 \frac{atm \cdot L}{mol \cdot K} \cdot 293.15 K }{58 \frac{g}{mol}}$$
 * $$\pi = 1.41 atm$$