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3 years ago

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6.3 kJ of energy is required to increase the temperature of some water from 60 °C to 75 °C. The specific heat capacity of water

is 4200 J per kg per °C. What is the mass of the water?

1 answer:

Andreas93 [3]3 years ago

4 0

Answer:

Mass of the water = 0.1kg is 100g

Explanation:

Mathematically;

Quantity of heat = mass of water * specific heat capacity * temperature change

From the question, we identify the following;

Quantity of heat = 6.3 KJ

1 KJ = 1000 J, so 6.3 KJ = 6.3 * 1000 = 6300 J

Mass of water = ?

Specific heat capacity = 4200

Temperature change is the difference between the final and initial temperature. = 75-60 = 15

6300 = m * 15 * 4200

63 = m * 15 * 42

63 = 630 m

m = 63/630

m = 0.1 kg

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Teniendo en cuenta los siguientes fenómenos: ebullición del agua- movimiento de un cuerpo- disolución de sal en agua- combustión

miv72 [106K]

Answer:

Las siguientes son reacciones químicas;

combustión de leña

oxidación del hierro

descomposición del agua en hidrógeno y oxígeno

Explanation:

Una reacción química da como resultado la formación de una (s) sustancia (s) nueva (s), mientras que un cambio físico no conduce a la formación de una sustancia nueva.

Las siguientes son reacciones químicas;

combustión de leña: la combustión de madera implica la oxidación del carbono según la reacción; C (s) + O2 (g) -------> CO2 (g)

oxidación del hierro: La oxidación del hierro conduce a la formación de óxidos de hierro. Como; 2Fe (s) + O2 (g) ----> 2FeO (s)

descomposición del agua en hidrógeno y oxígeno: esta es una reacción química en la que el agua se descompone de la siguiente manera; 2H2O (l) -----> 2H2 (g) + O2 (g)

Todos estos procesos enumerados anteriormente conducen a la formación de nuevas sustancias, por lo tanto, son reacciones químicas.

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2 years ago

Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t

myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16 g/mol}=0.5 mol$

Moles of ethane gas = $n_2=\frac{18.0 g}{30 g/mol}=0.6 mol$

Moles of propane gas = $n_3$

$n=n_1+n_2+n_3$

$2.22=0.5 mol +0.6 mol+ n_3$

$n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol$

Mole fraction of methane = $\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}$

$\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252$

Similarly, mole fraction of ethane and propane :

$\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703$

$\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

Partial pressure of methane gas:

$p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm$

Partial pressure of ethane gas:

$p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm$

Partial pressure of propane gas:

$p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm$

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = $n_1=\frac{37.8 g g}{28g/mol}=1.35 mol$

Moles of oxygen gas = $n_2=\frac{62.2 g}{32 g/mol}=1.94 mol$

Mole fraction of nitrogen= $\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103$

Similarly, mole fraction of oxygen

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

$p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg$

Partial pressure of oxygen gas:

$p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg$

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